Algebraic Expressions
2. Let's say and write these expressions in their square forms.
a) x2 + 2xy + y2 = ......................... b) t2 – 2tr + r2 = .........................
c) x2 + 2.x.3 + 32 = ......................... d) y2 – 2.x.3 + 32 = .........................
e) a2 + 2.a.5 + 52 = ......................... f) b2 – 2.b.7 + 72 = .........................
g) p2 – 4p + 4 = ......................... h) r2 + 4r + 4 = .........................
Creative Section - A
3. Let’s study these diagrams and write the areas in algebraic expression forms
as shown in the example.
Dx 1C
x x2 1.x x+1 Area of ABCD = x2 + 1.x + 1.x + 12
(x + 1)2 = x2 + 2x + 1
1 1.x 12
A x+1 B
D x 2C S x 3R M x yE
x2 2.x x2 x.y
a) x+2 b) c) x+y
x x x
2 2.x y x.y
A x+2 B
PQ P x+y R
Area of ABCD
Area of PQRS Area of PREM
4. Let's find the squares of the following expressions (i) by geometrical process
(ii) without using formula (iii) using formula.
a) (x + 1) b) (x + 2) c) (x + 3)
5. Let's find the squares of the following expressions.
a) (x + 3) b) (x – 4) c) (2y + 3) d) (x – 5y) e) (6m – 5n)
f) (4x + 3y ) g) (a2 + x2) 1 1
h) (x + 1 ) i) y y j) a 2a
x
6. Let's expand.
a) (a + 3) 2 b) (2a – 5)2 c) (2x + 3y)2
d) 2t + 1 2 e) 3x + 1 2 f) c2 − 1 2
2t 3x c
7. Let's simplify. b) (p – q)2 + 2pq c) (x + 3)2 – 3 (2x + 3)
a) (a + b)2 – 2ab
d) (3m – 4)2 + 8(3m – 2) e) (a – t)2 + (a + t)2 f) (x + y)2 – (x – y)2
g) (2x – 3y)2 – (2x + 3y)2 h) (4b + 5c)2 – (4b – 5c)2 i) (3a – 5b)2 – (5a + 3b)2
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8. Let's express the following expressions in the square forms.
a) x2 + 6x + 9 b) x2 – 8x + 16 c) 4a2 + 4ab + b2
d) p2 – 6pq + 9q2 e) 4x2 + 12xy + 9y2 f) 25x2 – 40xy + 16y2
g) 49a2 – 42ab + 9b2 h) x2 + 2 + 1 i) 4p2 – 2 + 1
x2 4p2
9. Find the squares of the following numbers by using the formula of (a + b)2 or
(a–b)2.
a) 49 b) 51 c) 99 d) 101 e) 999 f) 1001
Creative Section - B
10. a) If a + b = 3 and ab = 2, find the value of a2 + b2.
b) If p + q = 7 and pq = 12, find the value of p2 + q2.
c) If x – y = 1 and xy = 6, find the value of x2 + y2.
d) If m – n = 5 and mn = 14, find the value of m2 – n2.
11. a) If p + 1 = 4, find the value of p2 + 1 and p– 1 2.
p p2 p
b) If m + 1 = 5, find the value of m2 + 1 and m– 1 2.
m m2 m
c) If x − 1 = 3, find the value of x2 + 1 and x + 1 2.
x x2 x
d) If y − 1 = 6, find the value of y2 + 1 and y + 1 2.
y y2 y
It's your time – Project Work!
12. a) Let's write the formulae you learnt in this lesson in a chart paper and
b) compare it to your friends' work.
c)
Let's write any three binomial expressions in the form of (a + b) and find
the squares of your expressions.
Let's write any three binomial expressions in the form of (a – b) and find
the square of your expressions.
(iii) The product of (a + b) and (a – b)
Lets multiply (a – b) by (a + b).
(a + b) × (a – b) = a (a – b) + b (a – b) = a2 – ab + ab – b2 = a2 – b2
Thus, (a + b) (a – b) = a2 – b2
Da CD C Da bC
a–b (a + b) (a – b)
a–b
ab a2 – b2
b EF A a+b B
AB AB Area of ABCD
Area of ABCD Area of ABCDEF a2 – b2 = (a + b) (a – b)
= a2 = a2 – b2
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Worked-out examples
Example 1: Find the products of the following expressions.
a) (4x + 5) (4x – 5) b) (a2 + b2) (a2 – b2)
Solution:
a) (4x + 5) (4x – 5) Consider, 4x = a and 5 = b
= (4x)2 – 52 Now, (a + b) (a – b) = a2 – b2
= 16x2 – 25 ∴(4x + 5) (4x – 5) = (4x)2 – 52.
b) (a2 + b2) (a2 – b2) Consider, a2 = a and b2 = b
= (a2)2 – (b2)2 Now, (a + b) (a – b) = a2 – b2
= a4 – b4 ∴(a2 + b2) (a2 – b2) = (a2)2 – (b2)2
Example 2: Find the product of (2x + 3y) (2x – 3y) (4x2 + 9y2).
Solution:
(2x + 3y) (2x – 3y) (4x2 + 9y2) = [(2x)2 – (3y2)] (4x2 + 9y2)
= (4x2 – 9y2) (4x2 + 9y2) = (4x2)2 – (9y2)2 = 16x4 – 81y4
Example 3: Find the product of
a) 51 × 49 b) 102 × 98 by using the formula (a + b) (a – b) = a2 – b2.
Solution:
a) 51 × 49 = (50 + 1) (50 – 1) b) 102 × 98 = (100 + 2) (100 – 2)
= (50)2 – (1)2 = (100)2 – (2)2
= 2500 – 1 = 2499 = 10000 – 4 = 9996
2.6 × 2.6 − 1.4 × 1.4
Example 4: Simplify 2.6 − 1.4 Vedanta ICT Corner
Solution: Please! Scan this QR code or
browse the link given below:
2.6 × 2.6 – 1.4 × 1.4 = (2.6)2 – (1.4)2 https://www.geogebra.org/m/bsyhwthh
2.6 − 1.4 2.6 − 1.4
= (2.6 + 1.4) (2.6 – 1.4)
2.6 – 1.4
= 2.6 + 1.4 = 4
EXERCISE 10.5
General Section - Classwork
1. Let's express as the product of two binomials.
a) x2 – y2 = .................................. b) p2 – q2 = ..................................
c) a2 – 4 = .................................. d) 9 – b2 = ..................................
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2. Let's say and write the products as quickly as possible.
a) (a + x) (a − x) = .......................... b) (x + 1) (x – 1) = ..........................
c) (a – 3) (a + 3) = .......................... d) (2 + p) (2 − p) = ..........................
Creative Section - A
3. Let's find the area of the following rectangles.
a) b) c)
x–2 x+3 a–5
x+2 a+5
x–3
4. Let's find the products of the following expression by using formula.
a) (x + 2) (x –2) b) (a +3) (a –3)
c) (2p + 3) (2p – 3) d) (2a + 5b) (2a – 5b)
e) x+ 1 x– 1 f) y+ 2 y– 2
2 2 3 3
g) (x2 + y2)(x2 –y2) h) (x2 + 9)(x2 – 9)
5. Let's simplify.
a) (a + 1) (a – 1) (a2 + 1) b) (x + 2) (x – 2) (x2 + 4)
c) (p + 3) (p –3) (p2 + 9) d) (x +y) (x –y) (x2 + y2)
e) (a + 2y) (a – 2y) (a2 + 4y2) f) (2x + 3y) (2x – 3y) (4x2 + 9y2)
6. Let's find the products by using the formula (a + b) (a – b) = a2 – b2
a) 11 × 9 b) 41 × 39 c) 52 × 48 d) 102 × 98
7. Let's simplify.
a) 1.8 × 1.8 – 1.2 × 1.2 b) 1.6 × 1.6 – 0.9 × 0.9
1.8 + 1.2 1.6 + 0.9
c) 2.4 × 2.4 – 1.6 × 1.6 d) 4.6 × 4.6 – 2.5 × 2.5
2.4 – 1.6 4.6 – 2.5
e) 0.7 × 0.7 – 0.3 × 0.3 f) 0.009 × 0.009 – 0.006 × 0.006
0.7 + 0.3 0.009 – 0.006
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It's your time - Project work!
8. a) Let’s take a few rectangular sheet of papers (photocopy paper). Then fold
them to get square sheet of papers as shown in the diagrams.
b) Again, let’s fold each square sheet of paper as shown in the diagram and complete
the sums.
(i) x x x x1
x x2 x2 x x2–12 x–1 x2–12 = ....x.2..–...1..2.... = ..........................
x
(ii) x 12 1 x x+1
1 xy
x
x x2 x2 x x2–y2 x–1 x2–y2 = ................ = ..........................
x
y2 y x+y
y
9. a) Let's find the products of any three pairs of binomials of the form (a + b)
(a – b). Then show that (a + b) (a – b) is always a2 – b2.
b) Let's find the product of any two sets of three binomials of the form (a + b)
(a + b) (a + b).
c) Let's find the products of any two sets of three binomials of the form (a – b)
(a – b) (a – b).
10.9 Division of algebraic expressions
(i) Division of monomials by monomials
While dividing a monomial by another monomial, we divide the coefficient of
dividend by the coefficient of divisor. Then, subtract the power of the base of
divisor from the power of the same base of dividend. For example:
Example 1: Divide a) 28x4y3 by 7x2y2 b) 42a5b4 by 6a4b4
Solution: ÷ 7x2y2 =14278xx24yy23 = 4x4 – 2 y3 – 2 = 4x2y
a) 28x4y3
7
b) 42a5b4 ÷ 6a4b4 = 42a5b4 = 7a5 – 4 b4 – 4 = 7ab° = 7a
16a4b4
(ii) Division of polynomials by monomials
While dividing a polynomial by a monomial, we divide each term of the
polynomial by the monomial separately. For example,
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Example 2: Divide a) 8x4 – 6x3 by 2x2 b) 35a3b2 + 15a2b3 by 5ab
Solution:
a) (8x4 – 6x3) ÷ 2x2 Alternative process Checking:
2x2 ) 8x4 – 6x3 ( 4x2 – 3x 2x2 (4x2 − 3x)
= 8x4 – 6x3 = 2x2 × 4x2 − 2x2 × 3x
2x2 2x2 +8x4 = 8x4 − 6x3
−
= 4x4 – 2 – 3x3 – 2
− 6x3
= 4x2 – 3x −+ 6x3
∴ Quotient = (4x2 – 3x) 0
b) (35a3b2 + 15a2b3) ÷ 5ab 5ab ) 35a3b2 + 15a2b3 (7a2b + 3ab2
35a3b2 15a2b3
= 5ab + 5ab ±35a3b2 Checking:
5ab (7a2b + 3ab2)
= 7a3 – 1 b2 – 1 + 3a2 – 1 b3 – 1 15a2b3 = 5ab × 7a2b + 5ab × 3ab2
± 15a2b3
= 7a2b + 3ab2
0
= 35a3b2 + 15a2b3
∴ Quotient 7a2b + 3ab2
(iii) Division of polynomials by a binomial
While dividing a polynomial by a binomial at first we should arrange the
terms of divisor and dividend in descending (or ascending) order of power
of common bases. Then we should start the process of division dividing the
term of dividend with the highest power. For example,
Example 3: Divide (x2 + 2x – 15) by (x – 3).
Solution:
x – 3 ) x2 + 2x – 15 ( x + 5 Divide x2 by x → x2 ÷ x = x (quotient)
Multiply the divisor (x – 3) by the quotient x.
+x2 – 3x x (x – 3) = x2 – 3x
–+ Subtract the product from the dividend.
x2 + 2x – 15
5x – 15 ±
±5x 15 ±x2 3x
0 5x – 15
Again, the remainder is the new dividend and repeat
the process till the remainder is not divisible by divisor.
Example 4: The length of a rectangular field is (x + 10) m and its area is
(2x2 + 17x – 30) sq.m.
(i) Find the breadth of the field.
(ii) If x = 10 m, find the actual length, breadth and area of the field.
Solution:
Here,
Length of a rectangular field (l) = (x + 10) m
Area of the field (A) = (2x2 + 17x – 30) sq.m
Breadth (b) = ?
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Now, breadth = Area ÷ length
So, x + 10) 2x2 + 17x – 30 (2x – 3 (i) 2x2 ÷ x = 2x
(ii) 2x(x + 10) = 2x2 + 20x
– 2x2 +– 20x (iii) 2x2 + 17x – 2x2 – 20x = – 3x
– 3x – 30 (iv) – 3x ÷ x = – 3
(v) – 3(x + 10) = – 3x – 30
+– 3x +– 30 (vi) – 3x – 30 + 3x + 30 = 0
0
∴ Breadth (b) = (2x – 3)m
Again,
when x = 10 m, then actual length (l) = (x + 10)m = (10 + 10)m = 20m
breadth (b) = (2x – 3)m = (2 × 10 – 3) m = 17m
and area (A) = (2x2 + 17x – 30)m2
= (2 × 102 + 17 × 10 – 30) m2
= (200 + 170 – 30) m2 = 340m2 l × b = 20 m × 17 m = 340 m2
Example 5: Divide (x4 – y4) by (x − y) x4 ÷ x = x3
Solution: Then, x3(x − y) = x4 − x3y
x − y ) x4 – y4 ( x3 + x2y + xy2 + y3 And, x4 – y4 – (x4 – x3y) = +x3y – y4
± x4 x3y
Again, x3y ÷ x = x2y,
+ x3y – y4
± x3y x2y2 Then, x2y(x – y) = x3y – x2y2
+ x2y2 – y4 And, x3y – y4 – (x3y – x2y2) = x2y2 – y4
± x2y2 xy3
Again, x2y2 ÷ x = xy2
+ xy3 – y4
± xy3 y4 Then, xy2 (x – y) = x2y2 – xy3
0 And, x2y2 – y4 – (x2y2 – xy3) = xy3 – y4
Again, xy3 ÷ x = y3
Then, y3(x – y) = xy3 – y4
And, xy3 – y4 – (xy3 – y4) = 0
EXERCISE 10.6
General Section A – Classwork
Let's say and write the quotients as quickly as possible.
1. a) x2 ÷ x = ........ b) y2 ÷ y2 = ........ c) a4 ÷ a = ........
d) p4 ÷ p2 = ........ e) b3 ÷ b2 = ........ f) m4 ÷ m3 = ........
2. a) x2y2 ÷ xy = ........ b) a3b3 ÷ ab = ........ c) p3q3 ÷ p2q2 = ........
d) p2q3 ÷ pq2 = ........ e) m4n4 ÷ mn2 = ........ f) x4y3 ÷ x2y3 = ........
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3. a) 4x2 ÷ 2x = ........ b) 10y3 ÷ 5y = ........ c) 25a5 ÷ 7a3 = ........
d) 40a3b2 ÷ 4ab = ........ e) 20c2d3 ÷ 5c2d3 = ........ f) 55x4y3 ÷ 11x2y2 = ........
Creative Section - A b) 24x2 ÷ 6x2 c)18a2 ÷ 9a2
4. Let's find the quotients.
e) 66x4y5 ÷ 11x3y2 f) – 20p4q5 ÷ 5p3q4
a) 10x3 ÷ 5x
d) 32a3b2 ÷ 8ab
g) 45m7n6 ÷ ( – 9m2n4) h) –28a3t8 ÷ (– 7at5) i) − 56a2b3c ÷ (− 8ab3c)
5. Let's divide and find the quotients.
a) (6x2 + 9x ) ÷ 3x b) (8a3 – 12a2) ÷ 4a
c) (25y6 – 30y5) ÷ 5y4 d) (2p + 4p2 – 6p3) ÷ 2p
e) (12x3 – 9x2 + 15x) ÷ 3x f) (14m4 – 21m3 – 28m2) ÷ 7m2
g) (10a4b3 – 15a3b4 + 20a2b2) ÷ 5ab
h) (36x5y4 + 45x4y5 – 63x3y3) ÷ (–9x3y3)
i) (44p3q4 – 55p4q3 + 66p5q2) ÷ ( – 11p3q2)
j) (6xyz – 9x2y2z2 – 12x3y3z3) ÷ (– 3xyz) b) (a2 – 4) ÷ (a – 2)
6. Let's find the quotients. d) (9x2 – 1) ÷ (3x – 1)
a) (a2 – 4) ÷ (a + 2) f) (x2 + 6x + 8) ÷ (x + 4)
c) (9x2 – 1) ÷ (3x + 1) h) (y2 – 11y + 28) ÷ (y – 7)
e) (x2 + 5x + 6) ÷ ( x + 2) j) (p2 – 5p – 24) ÷ (p + 3)
g) (y2 + y – 12) ÷ (y – 3) l) (5m2 + 13m – 6) ÷ (m + 3)
i) (p2 – 3p – 10) ÷ (p + 2) n) (6x2 + x – 12) ÷ (2x + 3)
k) (4m2 + 5m – 6) ÷ (m + 2) p) (16p2 – 40pq + 25q2 ) ÷ (4p – 5q)
m) (6x2 + 5x – 6) ÷ (3x – 2)
o) (4p2 + 12pq + 9q2 ) ÷ ( 2p + 3q)
7. a) The product of two algebraic expressions is 4x2 – 9. If one of the expressions is
(2x + 3), find the other expression.
b) The product of two algebraic expressions is (x2 – 4x – 32). If one of the
expressions is (x + 4), find the other expression.
c) The area of a rectangle is (x2 – 16) sq. cm. If its breadth is (x – 4) cm, find its
length.
d) The area of a rectangular field is (a2 – 25) sq. m. If its length is (a + 5) m, find
its breadth.
8. a) If a = 4x3y2, b = 3x2y3 and c = 6xy, find the value of ab .
c
21a4a4b64b,6s, hshowowththatatzxp+rq
b) If p = 7a3b3, q = 6a4b4 and r = = 3ab.
c) If x = 8a4b5, y = 4a5b4 and z =
y = 4b + 2a.
z
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Creative Section - B
9. Let's divide and find the quotients.
a) (2x3 – 5x2 – 24x – 18) ÷ (2x + 3) b) (3x2 – 10x2 + 7x + 10) ÷ (3x + 2)
c) (a4 – b4) ÷ (a + b) d) (a4 – b4) ÷ (a – b)
e) (a5 + b5) ÷ (a + b) f) (a5 – b5) ÷ (a – b)
g) (x6 – y6) ÷ (x2 – y2) h) (x6 – y6) ÷ (x3 – y3)
10. a) The length of a rectangular carpet is (x – 5)m and its area is (x2 – 12x + 35) sq. m.
(i) Find its breadth
(ii) If x = 10m, find its actual length, breadth and area of the carpet.
b) The area of a rectangular garden is (x2 – 4x – 32) sq. m and its breadth is (x – 8)m.
(i) Find its length.
(ii) If x = 15m, find the actual length, breadth, and area of the garden.
11. a) Sunayana distributed (3x2 + 20x + 25) sweets equally among (3x + 5) friends
on her birthday.
(i) How many sweets did each of the friends receive?
(ii) If x = 5, find the total number of sweets distributed among friends, actual
number of friends and share of sweets.
b) On Children's Day Hari Narayan distributed (2x2 + 13x – 24) copies equally
among (2x – 3) number of children of a child care centre.
(i) Find the number of copies received by each child.
(ii) If x = 20, find the actual number of copies, students, and share of each
child.
It's your time - Project work!
12. a) Let's write any three pairs of monomial expressions with the same bases with
higher power of dividend. Divide the dividend by the divisor in each pair
and show that: Dividend = Quotient × Divisor + Remainder
b) Let's write any three pairs of binomial dividend and divisor in the forms of
(a2 – b2) and (a + b) or (a2 – b2) and (a – b).
Then, divide the dividend by the divisor and find quotient in each pair.
c) Let's find the products of any three pairs of binomial expressions. Then,
divide each product by one of the expressions in each pair and get another
expression.
E.g.: find → (x + 2) (x + 3) = x2 + 5x + 6, then find
(x2 + 5x + 6) ÷ (x + 2) = x + 3
10.10 Algebraic products and factors
We know that, in 2 × 5 = 10 → 10 is the product, 2, and 5 are the factors.
in 3 × 4 = 12 → 12 is the product, 3, and 4 are the factors.
Similarly, in x × x = x2 → x2 is the product, x, and x are the factors.
in 2a(a + b) = 2a2 + 2ab → 2a2 + 2ab is the product, 2a and (a + b)
are factors.
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Thus, when two or more algebraic expressions are multiplied, the result is called
the product and each expression is called the factor of the product. The process of
finding out factors of an algebraic expression is known as factorisation. It is also
known as resolution of the expression into its factors.
Of course, the process of factorisation is the reverse process of multiplication. For
example:
Multiplication of 2x(x – 3) = 2x2 – 6x and factorisation of 2x2 – 6x = 2x (x – 3)
Multiplication of (a + b) (a – b) = a2 – b2 and factorisation a2 – b2 = (a + b) (a – b)
Now, let’s learn the process of factorisation of different types of expressions.
(i) To find factors of expressions which have a common factor in each of its
term
In this type of expression, the factor which is present in each term is taken as
common and each term of the expression is divided by the common factor. The
product form of the common factor and the quotient represents the factorisation
of the expression.
Worked-out examples
Example 1: Write each expression as the product of its factors.
a) ax + bx b) 2px2 – 6p2x c) 4ax2 + 6a2x – 8ax
Solution:
a) ax + bx x is present in each term. So, it is the common factor.
= x (a + b) x is taken as common, then ax ÷ x = a and bx ÷ x = b
b) 2px2 – 6p2x 2 px2 – 6p2x Direct process
= 2px (x – 3p) 2px2 ÷ 2px = x and 6p2x ÷ 2px = 3p
= 2px × x – 2 × 3p × px
= 2px (x – 3p)
c) 4ax2 + 6a2x – 8ax Remember!
= 2ax (2x + 3a – 4) Among the factors which are present in each term, the
factor with the least power is taken as a common factor.
Example 2: Resolve into factors 3a (x + y) – 4b (x + y)
Solution: (x + y) is the common factor.
3a (x + y) – 4b (x + y)
= (x + y) (3a – 4b) 3a(x + y) ÷ (x + y) = 3a and 4b(x + y) ÷ (x + y) = 4b
(ii) To find factors of expressions having a common factor in the groups of
terms
In this case, terms of the given expression are arranged in groups in such a way
that each group has a common factor. For example:
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Algebraic Expressions
Example 3: Write each expression as the product of its factors.
a) a2 + ab + ca + bc b) x2 – 3a + 3x – ax
Solution:
a) a2 + ab + ca + bc Terms are arranged in groups.
= a (a + b) + c (a + b) a is common in a2 + ab and c is common in ca + bc
= (a + b) (a + c) (a + b) is common.
b) x2 – 3a + 3x – ax Arranging the terms in suitable groups
= x2 + 3x – ax – 3a
= x (x + 3) – a (x + 3) x is common in x2 + 3x and a is common in ax – 3a
= (x + 3) (x – a)
(iii) To find factors of expressions having the difference of two squared terms
The algebraic expression a2 – b2 is the difference of two squared terms. Here,
a2 and b2 are the squared terms and a and b are their square roots respectively.
We have learnt that a2 – b2 is the product of (a + b) and (a – b)
∴ a2 – b2 = (a + b) (a – b) Square root of a2
a2 – b2 = (a + b) (a – b)
Thus, (a + b) and (a – b) are the factors of a2 – b2.
Square root of b2
To factorise such expressions, we should re-write
the given terms in the form of a2 – b2.
Then, a2 – b2 = (a + b) (a – b) represents the factorisation of the expression.
Example 4: Find the factors of a) 4x2 – 9y2 b) 81x4 – 16y4.
Solution: Square root of 4x2 = 2x and square root of 9y2 = 3y
a) 4x2 – 9y2
= (2x)2 – (3y)2
= (2x + 3y) (2x – 3y) Using a2 – b2 = (a + b) (a – b)
b) 81x4 – 16y4 Square root of 81x4 and 16y4 are 9x2 and 4y2
= (9x2)2 – (4y2)2
= (9x2 + 4y2) (9x2 – 4y2) Using a2 – b2 = (a + b) (a – b)
= (9x2 + 4y2) [(3x)2 – (2y)2] 9x2 – 4y2 is still in the form a2 – b2.
So, it is further factorised.
= (9x2 + 4y2) (3x + 2y) (3x – 2y)
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Algebraic Expressions
Example 5: Simplify by factorisation process.
a) 24 × 40 – 20 × 24 b) 652 – 552 c) 49 × 51
Solution:
a) 24 × 40 – 20 × 24 b) 652 – 552 c) 49 × 51
= 24 (40 – 20) = (65 + 55) (65 – 55) = (50 – 1) (50 + 1)
= 24 × 20 = 120 × 10 = 502 – 12
= 480 = 1200 = 2500 – 1 = 2499
(iv) To find the factors of expressions of the form x2 + px + q
While factorising a trinomial expression of the form x2 + px + q, we should
search any two numbers a and b such that a + b = p and ab = q. Clearly, a
and b must be the factors of q. Then px is expanded in the form ax + bx and
factorisation is performed by grouping.
Example 6: Express as the product of factors. a) x2 + 5x + 6 b) a2 – 3a – 28
Solution:
a) x2 + 5x + 6 = x2 + (2 + 3)x + 6 b) a2 – 3a – 28 = a2 – (7 – 4)a – 28
= x2 + 2x + 3x + 6 = a2 – 7a + 4a – 28
= x(x + 2) + 3(x + 2) = a(a – 7) + 4(a – 7)
= (x + 2) (x + 3) = (a – 7) (a + 4)
EXERCISE 10.7
General Section - Classwork
1. Let's say and write the factors and products of these expressions as quickly as
possible.
a) a × a2 factors are .............................................. product is .........................
b) 2x2 × x factors are ............................................ product is .........................
c) 3p (p + 1) factors are ........................................... product is .........................
d) y2 (y – 2 ) factors are ............................................ product is .........................
2. Let's find the factors, say and write the factors as quickly as possible.
a) ax + ay = ............................. b) ax – ay = .............................
(c) ak + bk = ............................. (d) ak – bk = .............................
(e) a2 + a = ............................. (f) x2 – x = .............................
(g) x2 + xy = ............................. (h) p3 – p2q = .............................
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Laws of Indices Algebraic Expressions
3. a) a2 – b2 = ............................. b) b2 – c2 = .............................
c) x2 – 22 = ............................. d) y2 – 32 = .............................
e) p2 – 16 = ............................. f) r2 – 25 = .............................
4. Let's study the following tricky ways of factorisation carefully.
3+6 3×6 –3–6 –3×(–6)
x2 + 9x + 18 = (x + 3) (x + 6) x2 – 9x + 18 = (x – 3) (x – 6)
3–6 3×(–6)
6–3 6×(-3)
x2 + 3x – 18 = (x + 6) (x – 3) x2 – 3x – 18 = (x + 3 ) (x – 6)
Could you investigate the idea of tricky ways of factorisation. Apply your
investigation, say and write the factors as quickly as possible.
a) x2 + 5x + 6 = ............................. b) x2 – 5x + 6 = .............................
c) x2 + x – 12 = ............................. d) x2 – x – 6 = .............................
Creative Section
5. Let's write the following expressions as the product of their factors.
a) 2ax + 2bx b) 2x2y – 2xy2 c) 2abx – 4aby
d) 3mx2 – 6mx e) ax2 + a2x – ax f) 4x5y4 – 2x4y5 – 8x3y3
g) a(x – y) + b(x – y) h) x (a + b) – y(a + b) i) 2x (x – y) + y(x – y)
6. Let's find the factors of the following products.
a) ax + bx + ay + by b) ma + na + mb + nb c) xy + xz + y2 + yz
d) 3ax – bx – 3ay + by e) 4ax – 3bx – 8ay + 6by f) x2 – zx + xy – yz
g) a2b + ca – ab2c – bc2 h) a2bc + c2a – ab2 – bc i) x2 + 4a + 4x + ax
7. Let's express each binomial as the product of its factors.
a) a2 – 4 b) x2 – 9 c) m2 – 16 d) p2 – 25
h) 64b2 – 9c2
e) 4a2 – 9b2 f) 25x2 – 36y2 g) 49p2 – 81q2 l) 80a3 – 5ab2
i) 100m2 – 49n2 j) 8x2y2 – 18 k) 75a2b2 – 27x2y2
8. Let's find the factors of each of the following binomials.
a) a4 – 16 b) x4 – 81 c) y4 – 625 d) 16p4 – q4
h) 16p4 – 81q4
e) 81m4 – n4 f) x4 – 16y4 g) a4 – b4
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Algebraic Expressions
9. Let's express each trinomial as the product of its factors.
a) x2 + 3x + 2 b) a2 + 7a + 12 c) p2 + 6p + 8 d) x2 + 7x + 10
h) b2 – 11b + 30
e) y2 – 8y + 15 f) x2 – 9x + 20 g) m2 – 9m + 14 l) x2 + 5x – 24
i) x2 – 2x - 15 j) x2 – 3x - 28 k) a2 + 4a – 12
10. Let's simplify by factorisation process.
a)13 × 50 – 40 × 13 b) 28 × 60 – 30 × 28 c)36 × 24 + 26 × 36
d)152 – 52 e)272 – 172 f)752 – 552
g)21 × 19 h)79 × 81 i)101 × 99
It's your time - Project work!
11. a) Let's write any five pairs of difference of square numbers in the form
a2 – b2, where a2 > b2. Then, factorise and simplify the each pair.
[E.g. 52 – 32 = (5 + 3) (5 – 3) = 8 × 2 = 16]
b) Let's write any five binomial expressions of the form x2 – a2, where a2 is a
square number. Then factorise your expressions.
12. a) Let's find the products of any four pairs of binomials of the forms
(i) (x + a) (x + b) (ii) (x + a) (x – b) (iii) (x – a) (x + b) (iv) (x – a) (x – b),
where a and b are any natural numbers and a > b.
b) Again, factorise each product to get the factors.
[E.g. (x + 1) (x + 2) = x2 + 3x + 2 and
x2 + 3x + 2 = x2 + (1 + 2)x + 2 = x2 + x + 2x + 2 = x(x + 1) + 2(x + 1)
= (x + 1) (x + 2)
10.11 Simplification of rational expressions
We know that 1, 2, 1 ,– 3 , etc. are the rational numbers. Similarly,
2 5
x x2 x + 2
3 , y2 , x – 7 , etc. are called rational expressions. Here, we shall discuss about
addition, subtraction, multiplication, and division of rational expressions.
Multiplication and division of rational expressions
In multiplication, we simplify the numerical coefficients as in case of
multiplication of fraction. In the case of variables we apply the product and
quotient rules of indices. In division, we should multiply the dividend by the
reciprocal of divisor.
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Algebraic Expressions
Worked-out examples
Example 1: Multiply 4a3b2 × 10x4y3 .
5x3y2 12a4b3
Solution:
4a3b2 10x4y3 = 14a3b2 × 120 x4y3 = 2x4 – 3 × y3 – 2 = 2xy
5x3y2 × 12a4b3 5x3y2 × 12a4b3 3a4 – 3 b3 – 2 3ab
13
Example 2: Divide: 5x2y ÷ 15a3y3
8ax3 16a2x2
Solution: 126a2x2
15a3y3
5x2y ÷ 15a3y3 = 15x2y × = 2a2x4y = 2x4 – 3 = 2x
8ax3 16a2x2 81ax3 3 3a4x3y3 3a4 – 2 y3 – 1 3a2y2
Example 3: Simplify: 6y2 × 2x3 ÷ 5xy
4x2 3y3 6ab
Solution:
6y2 × 2x3 ÷ 5xy = 261y2 × 12x3 × 6ab = 6abx3y2 = 6ab = 6ab
4x2 3y3 6ab 4x2 3y3 5xy 5x3y4 5y4 – 2 5y2
21 1
EXERCISE 10.8
General Section - Classwork
1. Let's say and write the answers as quickly as possible.
a) x2 × a = ................................... b) x3 × a = ...................................
a2 x a3 x
c) x2 × y = ................................... d) x3 × y2 = ...................................
y2 x y3 x2
2. a) a2 ÷ a = ................................... b) a3 ÷ a2 = ...................................
x2 x x3 x2
c) x2 ÷ x = ................................... d) x4 ÷ y2 = ...................................
y3 y y4 x2
Creative Section
3. Simplify.
a) 2x3 × 9a2 b) 4y4 × 10z4 c) x4y3 × 15ab2
3a3 8x2 5z3 12y3 5a3b3 9x2y
d) 4a4 ÷ 8a3 e) 6y5 ÷ 18y3 f) x3y2 ÷ x2y
3b3 9b2 5x4 10x6 a3b2 a2b
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4. Simplify.
a) 2x2 × 10y4 × 9x4 b) 4a3 × 9b6 × 6a c) x2y3 × a2b3 ÷ ab
3y3 6x3 4y4 3b2 8a5 15b3 a3b2 x3y2 xy
d) 3a4b3 ÷ 6a2b2 × 8x2y e) x2y2 × abxy ÷ xyz f) a3 × c3x ÷ a2c2
5x3y4 15x2 9ab2 a3b3 x3y3 a2b2 x3y2 ab2 b2xyz
OBJECTIVE QUESTION
Let’s tick (√) the correct alternative.
1. What are the terms of the expression 2x + 3y?
(A) 2, x, 3 and y (B) 2x, 3 and y (C) 2x and 3y (D) 2, x and 3y
2. The expression consisting of 2 unlike terms is called
(A) monomial (B) binomial (C) trinomial (D) multinomial
3. Which of the following expression is not a monomial?
(A) 5xyz (B) 7pq + 8pq (C) (3mn)2 – 5m2n2 (D) ax2 +a2x
4. Which of following expressions is a binomial?
(A) 3a2b – ab2 + a2b (B) pq + qr + pr (C) xy + z – 1 (iv) x2y –3x – x2
5. The degree of the polynomial 3x2y5z is
(A) 7 (B) 8 (C) 10 (D) 11
6. If p = 2 and q = 4, what is the value of – 2 (p – q)?
(A) – 8 (B) 4 (C) –4 (D) 8
7. What should be added to 3ax to get 7ax + 1?
(A) 4ax (B) 4ax + 1 (C) -4ax + 1 (D) -4ax – 1
8. What should be subtracted from a3 + b3 to get a3+ab?
(A) b3 + ab (B) b3 – ab (C) ab – b3 (D) b3
9. To what expression must x2 – 3x + 1 be added to make the sum zero?
(A) 0 (B) x2 – 3x + 1 (C) -x2 + 3x + 1 (D) -x2 + 3x – 1
10. From what expression must x2 + 5x + 8 be subtracted to make the difference unity?
(A) x2 + 5x + 7 (B) x2 + 5x + 9 (C) -x2 – 5x – 7 (D) -x2 – 5x + 9
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Unit 11 Equation, inequality and Graph
11. 1 Open statement and equation - Looking back
Classwork - Exercise
1. Let's say and write whether the following statements are 'true', 'false' or open
statements.
a) The sum of 2 and 5 is 7. It is ................................. statement.
b) The difference of 9 and 4 is 6. It is ................................. statement.
c) The sum of x and 4 is 8. It is ................................. statement.
d) The difference of y and 3 is 9. It is ................................. statement.
e) The product of 5 and p is 10. It is ................................. statement.
2. Let's say and write the equations and find the values of the variables.
a) The sum of x and 2 is 5, equation is ........................ and x = .........
b) The difference of y and 1 is 6, equation is ........................ and y = .........
c) The product of 3 and p is 6, equation is ........................ and p = .........
d) The quotient of x divided by 2 is 4, equation is ........................ and x = .........
3. Let's say and write the values of variables that make the following open
statements (or equations) true.
a) x + 2 = 7, x = ...................... b) x – 2 = 7, x = ......................
c) 4y = 8, y = ...................... d) m = 3, m = ......................
2
Let's suppose, the sum of x and 5 is 9. i.e. x + 5 = 9. It is a mathematical statement.
Here, unless x is replaced by any number, we cannot say whether x + 5 = 9
is a true or a false statement. For example:
When x is replaced by 1, then 1 + 5 = 6, which is false.
When x is replaced by 2, then 2 + 5 = 7, which is false.
When x is replaced by 3, then 3 + 5 = 8 which is false.
When x is replaced by 4, then 4 + 5 = 9 which is true.
Thus, the mathematical statements which cannot be predicted as true or false
statements until the variable is replaced by any number are known as Open
Statements.
x > 3, x + 2 < 7, x + 3 = 7, 2x = 10, etc. are also the examples of open statements.
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11.2 Linear equations in one variable
Let’s consider any two open statements: x + 3 > 7 and x + 3 = 7.
The open statement x + 3 > 7 can be true for many values of x. However, the
other open statement x + 3 = 7 can be true only for a fixed value of x. Such open
statement containing ‘equal to’ (=) sign and can be true only for a fixed value of
variable is called an equation.
Furthermore, in x + 3 = 7, the equation has only one variable which is x. So, it is
the equation in one variable. Also, the variable x has power (or exponent) 1. So, it is
called a linear equation. Thus, x + 3 = 7 is a linear equation in one variable.
11.3 Solution to equations
Let’s consider an equation, x + 3 = 7.
This equation can be true only for a fixed value of x which is 4. So, 4 is called the
solution (or root) of the equation. The process of getting a solution to an equation
is called solving equation.
Worked-out Examples
Example 1: Solve a) x + 2 = 8 b) 7a = 21 c) t =5
Solution: 2
a) x + 2 = 8 Direct process
x + 2 = 8
or, x + 2 – 2 = 8 – 2 Subtracting 2 from both sides
or, x = 8 – 2
or, x = 6
or, x = 6
b) 7a = 21 Direct process
or, 7a = 21 Dividing both sides by 7 7a = 21
7 7 21
or, a = 3 or, a = 37
or, a =
c) t =5 Direct process
2 t
2 = 5
or, t × 2 = 5 × 2 Multiplying both sides by 2
2 or, t = 5 × 2
or, t = 10
or, t = 10
Example 2: Solve the equations and check the solutions.
9x 2 3
a) 7 (x – 10) = 10 – 3x b) 11 + 7 = 43 c) y + 1 = 2y − 1
Solution: Checking:
a) 7 (x – 10) = 10 – 3x Putting x = 8 in the given equation:
or, 7x – 70 = 10 – 3x 7(x – 10) = 10 – 3x
or. 7 (8 – 10) = 10 – 3 × 8
or, 7x + 3x = 10 + 70
or, 7 ( – 2) = 10 – 24
or, 10x = 80 or, – 14 = – 14
or, LHS = RHS
x = 80 =8 ∴ x = 8 is the required solution.
10
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Equation, Inequality and Graph
b) 9x + 7 = 43 Checking
11
9x Substituting x = 44 in the given equation:
11
or, = 43 − 7 9x + 7 = 43
11
9x × 444
or, 11 = 36 or, 9 11 + 7 = 43
or, 9x = 11 × 36 or, 9 × 4 + 7 = 43
or, x = 11 ×9136 4
or, x = 11 × 4 or, 36 + 7 = 43
43 = 43 ∴ LHS = RHS
Hence, x = 44 is the required solution.
or, x = 44 Checking
Substituting y = 5 in the given equation:
2 3
c) 2 1 = 3 1 y + 1 = 2y – 1
+ −
y 2y or, 5 2 1 = 2 3 – 1
+ ×5
or, 2(2y − 1) = 3(y + 1) 62 = 3
or, 10 –
or, 4y − 2 = 3y + 3 1
13 = 3
or, 4y − 3y = 3 + 2 or, 9
or, y = 5 31 = 1 ∴ LHS = RHS
3
Hence, y = 5 is the required solution.
Example 3: Solve a) 40% of x = Rs 80 b) x + 25 % of x = Rs 555.
Solution: b) x + 25 % of x =Rs 555
a) 40 % of x =Rs 80 or, x + 25 × x = Rs 555
100
40 x
or, 100 × x = Rs 80 or, x+ 4 = Rs 555
or, 2x = Rs 80 or, 4x + x = Rs 555
5 4
or, 2x = 5 × Rs 80 or, 5x = Rs 555
4
or, x = 5 × Rs 80 = Rs 200
2 or, 5x = 4 × Rs 555
∴ x = Rs 200. or, x = 4 × Rs 555
5
∴ x= 4 × Rs 111 = Rs 444.
EXERCISE 11.1
General Section - Classwork
1. Let's say and write 'true', 'false', or 'open' statement in the blank spaces.
a) The sum of 4 and 5 is 10. It is ............................ statement.
b) The difference of x and 5 is 3. It is ............................ statement.
c) The product of 6 and 7 is 42. It is ............................ statement.
d) The quotient of x divided by 3 is 2. It is ............................ statement.
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Equation, Inequality and Graph
2. Let's make equations. Then say and write solutions of the equations.
a) The sum of x and 6 is 9. Equation is ...................... solution is ........
b) The product of y and 3 is 15. Equation is ...................... solution is ........
c) The difference if a and 2 is 10. Equation is ...................... solution is ........
d) The quotient of x divided by 4 is 3. Equation is ...................... solution is ........
Let's say and write the values of letters as quickly as possible.
3. a) x + 2 = 7, x = ............... b) x + 5 = 9, x = ...............
c) x – 4 = 10, x = ............... d) x – 2 = 5, x = ...............
e) 2x = 8, x = ............... f) 3x = 15, x = ...............
g) x = 5, x = ............... h) x = 4, x = ...............
2 3
4. a) 2x – 1 = 5, x = ............... b) 3x + 2 = 14, x = ...............
c) 5x – 3 = 7, x = ............... d) 3x = 6, x = ...............
5
e) x + 1 = 3, x = ............... f) x–2 = 5, x = ...............
2 3
Creative Section - A
5. a) D efine open mathematical statements with examples.
b) W rite the difference in the meaning of x + 2 > 5 and x + 2 = 5.
c) W hat is an equation? Write with an example.
d) Define solution of an equation with an example.
6. Let's solve the equations and check the solutions.
a) 3x + 1 = 7 b) 6x – 13 = 5
c) 8x + 13 = 6x + 25 d) 5x – 18 = 2x + 3
e) 2(3x + 4) = 5(2 + x) f) 9(7x – 2) = 5 (10x – 1)
g) 3x + 2(x + 2) = 22 – 3 (2x – 5) h) 6(y – 3) – 5(y – 8) = 48 – 3(y – 2)
i) 5.8x – 7.3 = 3.8 x + 2.7 j) 1.4x – 9.8 = 10.2 – 3.6x
7. Solve each of the following equations and check the answer in each case.
a) x = 2 b) 5x = 15 c) 7x = – 14
3 6 8
d) 3x = – 6 e) 2x = 8 f) 3x = 9
−5 9 3 4 2
g) x−1 = 3 h) 4x − 5 = 5 i) x−1 = x+2
2 3 2 3
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Equation, Inequality and Graph
j) 3x − 1 = 2x + 3 k) 4 = 5 l) 3 = 7
5 7 x−1 x+1 x+1 3x − 1
m) x+3 = 4 n) 2x – 1 = 1
3x − 5 5 x+4 2
Creative Section - A
8. Let's make equations and solve the equations.
a) 5% of x = Rs 10 b) 10% of x = Rs 50 c) 20% of x = 60 kg
f) 90% of x = 360 students
d) 30% of x = 90 km e) 40% of x = Rs 200
9. Let's make equations and solve the equations.
a) x + 10% of x = 55 kg b) x + 20% of x = Rs 96 c) x + 25% of x = Rs 125
d) x – 15% of x = 34 km e) x – 30% of x = Rs 70 f) x – 75% of x = Rs 75
g) x = 160 + 20% of x h) x = Rs 300 + 50% of x i) x = 600 − 20% of x
10. Let's solve.
a) 3x – 1 = 1 b) 5x + 1 = 2 c) x – x =2
8 4 2 12 4 3 2 3
d) x + x = 7 e) x+1 + x+6 = 5 f) 2x + 1 + x+3 =2
3 4 2 3 3 4
g) 2 + 3 = 5 h) 3 + 2 = 7 i) 5 – 1 = 1
x x 2x x 4n 2n 8
It's your time - Project work!
11. a) Let's write any integer to the right hand side of each of the following equation
such that the solution of each equation will be an integer. Also, check the
solution.
(i) x + 7 = ........ (ii) x – 5 = ........ (iii) 3x =+2 2....=............((ivvi)ii4)x = ........
(v) 2x + 1 = ........ (vi) 3x – 8 = ........ (vii) x
x–3 = .......
5
b) Let's write any integer in the blank spaces such that the solution must be
an integer. Then solve the equations and find the solutions. Also, check the
solution.
(i) x + ........ = 9 (ii) x – ........ = –2
(iii) 2x + ........... = 5 (iv) 3x – ........... = 2
c) Let's make linear equations of your own and solve each equation to get the
given solutions.
(i) x = 1 (ii) x = 3 (iii) x = –2 (iv) x = 4 (v) x = –1
11.4 Applications of equations
We use equations to find the unknown value of any quantity. For this, we should
consider the unknown value of the given verbal problems as the variables like
x, y, z, a, b, c, etc. Then, the verbal problems should be translated into mathematical
sentences in the form of equations. By solving the equations, we obtain the required
values.
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Worked-out examples
Example 1: The sum of two numbers is 50. If one of the numbers is 35, find the
other number.
Solution: Answer checking:
Let the other number be x. If the required number is 15,
Now, x + 35 = 50 then, 35 + 15 = 50
or, x = 50 – 35 = 15 which is given in the question.
Hence, the required number is 15.
Example 2: There are 555 students in a school. If the number of boys is 55
more than that of girls, find the number of boys and girls.
Solution:
Let the number of girls be x.
Then, the number of boys = (x + 55)
Now, number of girls + number of boys = total number of students
or, x + (x + 55) = 555 Answer checking:
Number of boys = 305 and number of girls = 250
or, 2x + 55 = 555 Then, 305 + 250 = 555
which is given in the question.
or, 2x = 555 − 55
or, x = 500 = 250
2
∴ Number of girls = x = 250
Also number of boys = (x + 55) = (250 + 55) = 305
Example 3: If the sum of three consecutive even numbers is 36, find the numbers.
Solution:
Let the smallest even number be x.
Then, the second consecutive even number = x + 2
And, the third consecutive even number = x + 4
Now, x + (x + 2) + (x + 4) = 36 Answer checking:
or, 3x + 6 = 36 If the required consecutive even
or, 3x = 36 – 6 numbers are 10, 12 and 14, then
or, 10 + 12 + 14 = 36
x = 30 = 10 which is given in the question.
3
∴ The first even number = x = 10
The second even number = x + 2 = 10 + 2 = 12
The third even number = x + 4 = 10 + 4 = 14
Hence, the required consecutive even numbers are 10, 12 and 14.
Example 4: Mother divides Rs 1,30,000 between her son and daughter in the
ratio 6 : 7. Find the shares of the son and the daughter.
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Solution: Answer checking:
Let the share of the son = Rs 6x Rs 60,000 + Rs 70,000 = Rs 1,30,000
And, 60,000 : 70,000 = 6 : 7
The share of the daughter = Rs 7x which are given in the question.
Now, 6x + 7x = Rs 1,30,000
or, 13x = Rs 1,30,000
or, x = Rs 1,30,000 = Rs 10,000
13
∴ The share of the son = Rs 6x = Rs (6 × 10,000) = Rs 60,000
The share of the daughter = Rs 7x = Rs (7 × 10,000) = Rs 70,000
Hence, the shares of the son and the daughter are Rs 60,000 and Rs 70,000 respectively.
Example 5 : The length of a rectangular field is 10m longer than its breadth
and its perimeter is 100m. Find the length and the breadth of the
field.
Solution:
Let, the breadth (b) of the rectangular field be x m.
Then, the length (l) of the field = (x + 10) m
Now, the perimeter of the field = 100 m Answer checking:
or, 2 (l + b) = 100 m l = 30 and b = 20 m
or, 2 (x + 10 + x) = 100 m Thus, l is 10 m longer than b
or, 2 (2x + 10) = 100 m Also, perimeter = 2(l + b)
or, 4x + 20 = 100 m = 2(30 m + 20m)
= 2 × 50 m
or, 4x = (100 – 20) m = 100 m
or, x = 80 m = 20 m which is given in the question.
4
∴ The breadth of the field (b) = x m = 20 m
The length of the field (l) = (x + 10) m = (20 + 10) m = 30 m
Example 6: One-fourth part of a pole is inside the mud, two-fifth part is inside
the water and the remaining length of 7 m is above the surface of
the water. Find the length of the pole.
Solution:
Let the length of the pole be x m.
The length of the pole inside the mud = 1 of x m = x m.
4 4
2 2x
The length of the pole inside the water = 5 of x m = 5 m.
The length of the pole above the surface of the water = 7 m
x 2x
Now, x = 4 + 5 + 7
or, x = 5x + 8x + 140
20
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or, x = 13x + 140
20
or, 20x = 13x + 140 Answer checking:
or, 20x – 13x = 140 x = 20 = 5 m and 2x = 2 × 20 = 8 m
4 4 5 5
or, 7x = 140 Then, 5 m + 8 m = 13 m and 20 m – 13 m = 7 m
140 which is given in the question.
7
or, x = = 20
Hence, the required length of the pole is 20 m.
EXERCISE 11.2
General Section - Classwork
1. Let's say and write the equation corresponding to the following mathematical
statements where x is the unknown number. Also, find the value of x.
Mathematical Statement Corresponding Equation Value of x
a) The sum of two numbers is 10
and one of them is 3.
b) The difference of two numbers
is 6 and the smaller one is 9.
c) The difference of two numbers
is 7 and the greater one is 12.
d) The product of two numbers is
18 and one of them is 9.
e) Six times a number is 42.
f) Two times a number increased
by 7 is 17.
g) 4 less than three times a number
is 11.
h) A number exceeds the other by 5
and their sum is 11.
i) The sum of two consecutive odd
numbers is 12.
j) The quotient of dividing 56 by x
is 8.
Creative Section - A
Let's solve these problems by making linear equations.
2. a) The sum of two numbers is 25. If one of the numbers is 10, find the other
number.
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b) The difference of two numbers is 16. If the greater number is 40, find the
smaller one.
c) The difference of two numbers is 28. If the smaller number is 10, find the
greater one.
3. a) Two complementary angles differ by 10°. Find the angels.
b) There are 32 students in a class and 18 of them are girls. Find the number of
boys.
c) There are 3 more boys than girls in a class and the total number of students
is 33, find the number of girls and boys.
d) There are 2 less boys than girls in a class and the total number of students is
34, find the number of boys and girls.
4. a) If the sum of two consecutive numbers is 23, find them.
b) If the sum of two consecutive even numbers is 34, find them.
c) If the sum of two consecutive odd numbers is 12, find them.
d) If the sum of three consecutive odd numbers is 33, find them.
5. a) Find two numbers whose sum is 35 and the greater number exceeds the
smaller one by 9.
b) The sum of two numbers is 20. If the smaller number is 4 less than the bigger
one, find the numbers.
c) Find two numbers whose sum is 15 and difference is 3.
d) A sum of Rs 100 is divided into two parts. If the greater part exceeds the
smaller one by Rs 20, find the parts of the sum.
e) A sum of Rs 180 is divided into two parts. If the smaller part is Rs 50 less
than the greater part, find the parts of the sum.
f) Sunayana and Anamol donated some money in a Coronavirus relief fund.
The amount donated by Anamol is Rs 250 more than that of Sunayana. If
they donated Rs 1000 in total, find the amount donated by each of them.
6. a) If 7 books cost Rs. 300 more than 5 same books, find the cost of a book.
b) Six times a number is 21 less than nine times the same number, find the
number.
c) Shashwat's weekly earning is earns Rs. 4,500 more than the earning of
4 days. What is his daily earning?
d) After 42 years, Akanksha will be four times as old as she is now. Find her
present age.
e) If a number is as many greater than 79 as it is less than 97, find the number.
7. a) If 2 part of the distance between two places is 34 km, find the distance
3
between the places.
b) 3 part of the total number of students of a school are girls. If there are 240
4
boys, find the number of girls.
c) If 10% discount on the price of an item amounts to Rs 100, find the price of
the item.
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d) If 13% VAT on the selling price of an article amounts to Rs 130, find the
selling price of the article.
e) If 20% of profit on the cost price of an item amounts to Rs 400, find the cost
price of the item.
8. a) Divide Rs 650 in the ratio of 2 : 3.
b) A father divides Rs 54,000 between his son and daughter in the ratio of 4 : 5.
Find the shares of each of them.
c) Pratik, Devashish and Bishwant invest a sum of Rs 1,50,000 on a business in
the ratio of 2 : 3 : 5. Find the share of each of their investment.
d) If the angles of a triangle are in the ratio 1 : 2 : 3, find the size of each angle.
e) If two complementary angles are in the ratio of 2 : 3, find the angles.
Creative Section - B
9. a) 1 part of a pole is inside the mud, 2 part is inside the water, and the remaining
5 3
length of 8 m is above the surface of the water. Find the length of the pole.
s15pepnat r43t
b) Bipin had some money. He part of his money to buy goods for his
birthday, he gave his sister of the money, and he gave the rest of
Rs 120 to his mother. How much money did he have in the beginning?
10. a) The length of a rectangular ground exceeds its breadth by 5m. If its
perimeter is 110m, find its length and breadth.
b) The breadth of a rectangular handkerchief is 10cm shorter than its length. If
the perimeter of the handkerchief is 100cm, find its length and breadth.
c) The length of a rectangular lawn is two times of its breadth. If its perimeter is
120 m, find its length and breadth.
d) The length and breadth of a rectangular park are in the ratio 5 : 2. If its
perimeter is 168 m, find its length and breadth.
It's your time - Project work!
11. a) How many students are there in your class and how many girls are there?
Make an equation and find the number of boys.
b) How many students are there in your class and how many boys are there?
Make an equation and find the number of girls.
c) How many teachers are there in your school and how many are male
teachers? Make an equation and find the number of lady teachers.
Let's have a fun!
12. Let's ask your friends to think of any number and multiply the
number by 4. Then, tell them to add 6 to the product and divide the sum by 2.
At last, tell them to subtract two times the original number from the quotient
and ask their answer. Are they surprised? Solve it by making an equation.
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11.5 Linear equation with two variables
Let, there are x number of boys and y number of girls in a class. The total number
of students in the class is 27.
(i) Can you express the total number of students in an equation?
(ii) If there are 18 boys, how many girls are there in the class?
(iii) If there are 9 girls, how many boys are there in the class?
Here, we can express the total number of students in the class as:
x + y = 27
This equation has two variables: x and y. Each variable has the exponent 1. Such
equations are called linear equations with two variables.
In x + y = 27, if x = 18, then y = 27 – 18 = 9.
Also, if y = 9, then x = 27 – 9 = 18.
Furthermore, to solve linear equations with two variables, a pair of equations are
always required. For example:
x + y = 27 ......... equation (i) x + y = 27 ......... equation (i)
x = 18 ......... equation (ii) or,
y = 9 ......... equation (ii)
Now, let's study the following examples and learn more about solving linear equation
with two variables.
Worked-out examples
Example 1: The sum of a number and three times another number is 27.
a) Write an equation to represent the sum.
b) If the first number is 9, find the second number.
Solution:
Let, the first number be x and the second number be y.
Then, 3 times the second number = 3y
a) The required equation is: x + 3y = 27 ...... (i)
b) Also, x = 9 .....(ii)
Putting the value of x in equation (i) we get,
9 + 3y = 27
or, 3y = 27 – 9
or, y = 18 = 6
3
Hence, the second number is 6.
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Example 2: x and y are any two natural numbers, where x = 3y. The difference
of these numbers is 8.
a) Write an equation to represent the difference.
b) Find the values of x and y.
Solution:
a) Here, the required equation is x – y = 8 ................ (i)
b) Also, x = 3y ................. (ii)
Putting x = 3y in equation (i), we get, 3y – y = 8
or, 2y = 8
or, y = 8 =4
2
Now, putting the value of y in x = 3y, we get
x = 3y = 3 × 4 = 12
Hence, x = 12 and y = 4.
Example 3: The sum of the age of a father and two times the age of his son is
58 years.
a) Write an equation to represent the sum.
b) If the son is 11 years old, find the age of the father.
Solution:
Let, the age of the father be x years and the age of the son be y years.
Here, two times the age of the son = 2y
a) The required equation is: x + 2y = 58 ............ (i)
b) Also, y = 11 ....................... (ii)
Putting the value of y in equation (i), we get,
x + 2 × 11 = 58
or, x + 22 = 58
or, x = 58 – 22 = 36
Hence, the age of the father is 36 years.
Example 4: The difference of two numbers is 4. If six times the smaller number is
equal to five times the bigger one, find the numbers.
Solution:
Let the smaller number be x and the bigger one be y.
From the first condition, When the bigger number is y, the smaller
y – x = 4 number is x and the difference is 4, then
or, y = x + 4 .................... (i) y–x=4
From the second condition,
6x = 5y .................... (ii) 6 times the smaller number is 6x
5 times the bigger number is 5y
Substituting the value of y from
equation (i) in equation (ii), we get,
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6x = 5 (x + 4) I can mentally check the
or, 6x = 5x + 20 answers !
or, 6x – 5x = 20 24 – 20 = 4
or, x = 20 6 × 20 = 120 and 5 × 24 = 120
Substituting the value of x in equation (i), we get,
y = 20 + 4 = 24
So, the required numbers are 20 and 24.
Example 5: The total cost of 2 exercise books and 4 pens is Rs 150. If the cost of
a exercise book is same as the cost of 3 pens, find the cost of 1 exercise
book and 1 pen.
Solution:
Let the cost of 1 exercise book be Rs x and that of 1 pen is Rs y .
Then, the cost of 2 exercise book is Rs 2x and 4 pens is Rs 4y.
From the first condition, The cost of 1 exercise books = Rs x
2x + 4y = 150 .................. (i)
The cost of 3 exercise books = Rs 3x
From the second condition,
x = 3y .................... (ii) The cost of 1 pen = Rs y
The cost of 4 pens = Rs 4y
Substituting the value of x from equation (ii) in equation (i), we get,
2 × 3y + 4y = 150
or, 6y + 4y = 150
or, 10y = 150
or, y = 15
Putting y = 15 in equation (ii), we get
x = 3y = 3 × 15 = 45
Hence, the cost of 1 exercise book is Rs 45 and that of 1 pen is Rs 15.
Example 6: The length of a rectangular ground is two times its breadth. If the
Solution: perimeter of the ground is 96 m, find the length and breadth of the
ground.
Let, the length of the ground be x m and the breadth be y m.
From the first condition, From the second condition,
Perimeter = 96 m
length = 2 × breadth 2 (l + b) = 96
x = 2y .................... (i)
96
(x + y) = 2
x + y = 48 .................... (ii)
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Substituting the value of x from equation (i) in equation (ii), we get,
2y + y = 48
or, 3y = 48
or, y = 48 =16
3
Substituting the value of y in equation (i), we get,
x = 2 × 16 = 32
Hence, the length of the ground is 32 m and the breadth is 16 m.
EXERCISE 11.3
General Section - Classwork
1. Let’s say and write the values of the variables as quickly as possible.
a) In x + y = 5, if x = 3, then y = .......... b) In x – y = 4, if x = 1, then y = ..........
c) In 2x + y = 9, if y = 1, then x = ....... d) In 3x – 2y = 4, if y = 1, then x = .....
2. Let’s write the value of x or y of your own choice, then, find the value of unknown
variable.
a) x + y = 10, x = .............. and y = ..............
b) x – y = 3, x = .............. and y = ..............
c) x = 2y, x = .............. and y = ..............
d) 3x = y, x = .............. and y = ..............
Creative Section - A
3. Let's solve the following linear equations with two variables.
a) x + y = 3 and x = 2 b) x + y = 5 and y = 1
c) x – y = 7 and x = 10 d) x + y = 6 and x = 2y
e) x + y = 8 and y = 3x f) x + y = 9 and x = 1 y
2
g) y = x – 2 and x + y = 4 h) y = x – 2 and x + y = 6
i) y = x + 2 and x + y = 8 j) y = x + 6 and x + y = 14
k) x – y = 5 and y = 11 – x l) 2x – y = 3 and y = 9 – x
4. a) x and y are any two natural numbers and their sum is 9.
(i) Write an equation to represent the sum of numbers.
(ii) If x = 4, find the value of y.
b) p and q are any two natural numbers and their sum is 21.
(i) Write an equation to represent the sum of numbers.
(ii) If q = 2p, find the values of p and q.
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c) The difference of two numbers x and y is 3.
(i) Write an equation to represent the difference of the numbers.
2x
(ii) If y = 3 , find the values of x and y.
d) The total number of students in a class is 36 and the number of girls are two
times the number of boys.
(i) Write an equation to represent the total number of students.
(ii) Write an equation to represent the number of girls and boys.
(iii) Find the number of girls and boys in the class.
e) The preimeter of a rectangular school ground is 220 m.
(i) Write an equation to represent the perimeter of the ground.
(ii) If the length of the ground is 60 m, find the breadth of the ground.
f) The perimeter of a rectangular garden is 120 m.
(i) Write an equation to represent the perimeter of the garden.
(ii) If the length of the garden is three times its breadth, find the length and
breadth of the garden.
5. a) The sum of the age of a father and two times the age of his daughter is 50 years.
(i) Write an equation to represent the sum of the ages.
(ii) If the daughter is 8 years old, find the age of the father.
b) The sum of the age of a mother and three times the age of her son is 51 years.
(i) Write an equation to represent the sum of the ages.
(ii) If the mother is 30 years old, find the age of the son.
6. a) The sum of two numbers is 18 and their difference is 4. Find the numbers.
b) The sum of two numbers is 39. If the greater number is 5 more than the smaller
one, find the numbers.
c) The sum of two numbers is 54. If the smaller number is 6 less than the greater
one, find the numbers.
d) The difference of two numbers is 6. If three times the smaller number is equal
to two times the greater one, find the numbers.
Creative Section - B
7. a) The total number of students in a class is 36. If the number of girls is 6 more
than the number of boys, find the number of boys and girls.
b) The total cost of a box and a pen is Rs 100. If the cost of a box is same as the cost
of 3 pens, find the cost of a box and a pen.
8 . a) The sum of the age of a father and his son is 40 years. If the father is 28 years
older than his son, find their age.
b) The sum of the age of two sisters is 20 years. If one of them is 4 years younger
than other, find their age.
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c) Father is three times as old as his son. If the difference of their age is 24 years,
find their age.
9. a) The length of a rectangular garden is two times its breadth. If the perimeter of
the garden is 72 m, find the length and breadth of the garden.
b) The perimeter of a rectangular field is 120 m. If the field is 10 m longer than its
breadth, find the length and breadth of the field.
It’s your time - Project work
10. a) Let's make the pair of linear equations with variables x and y. Solve them to get
the following values of variables. For example: (i) x + y = 5... (i) x = 2 ... (ii)
(i) x = 2 (ii) x = 5 (iii) x = –3 (iv) x = –1
y=3 y = –1 y=6 y = –2
b) Let’s find the unit cost of each of the following pairs of items in your local
market. Consider the unit cost of one item as Rs x and another item as Rs y.
Then, make a pair of linear equations with two variables in each case. Solve the
equations and find the values of x and y.
(i) 1 pencil and 1 eraser (ii) 1 pen and 1 exercise book
(iii) 1 kg of potatoes and 1 kg of tomatoes (iv) 1 kg of sugar and 1 kg of rice
11.6 Trichotomy – Review
Let’s consider two whole numbers 3 and 7.
There is only one way to compare these two numbers.
Either, 3 < 7 (3 is less than 7) or 7 > 3 (7 is greater than 3).
But, 3 = 7 or 3 > 7 or, 7 < 3 are not true comparisons.
Thus, if a and b are any two whole numbers, only one comparison from the following
can be true comparison between them.
Either, a = b or, a < b or, a > b.
Such a property of whole numbers is known as Trichotomy property. The sign
‘=’ (equal to), ‘<’ (less than) and ‘>’ (greater than) are the trichotomy signs.
Negation of trichotomy
Let’s consider any two numbers 4 and 9. <
<
Here, 4 < 9 or 9 > 4 are true comparisons.
<
But, 4 </ 9 (4 is not less than 9) or 9 / 4 (9 is not greater than 4) are false comparisons.
Here, 4 </ 9 is the negation of 4 < 9 and 9 / 4 is the negation of 9 > 4.
Thus, ‘</ ‘ (is not less than) is the negation of ‘<’ (is less than), / (is not greater than)
is the negation of ‘>’ (is greater than) and ‘≠' (is not equal to) is the negation of '='
(is equal to).
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Trichotomy rules
1. Let's suppose that a, b, and c are any three whole numbers, where a > b, then
(i) a + c > b + c [Addition axiom] eg, 9 > 4, then 9 + 3 > 4 + 3
(ii) a – c > b – c [Subtraction axiom] eg, 8 > 5, then 8 – 2 > 5 – 2
(iii) a × c > b × c [Multiplication axiom] eg, 4 > 3, then 4 × 2 > 3 × 2
(iv) ac > b [Division axiom] eg, 9 > 6, then 9 > 6
c 3 3
Thus, when an equal positive number is added to or subtracted from or
multiplied or divided by both sides of trichotomy sign the sign remains the
same.
2. Let's suppose that, a and b are any two whole numbers and c is a negative
integer.
(i) If a > b, then a × (– c) < b × (– c) eg, 4 > 2, then 4 × (– 2) < 2 × (– 2)
(ii) If a > b, then a < b eg, 10 > 8, then 10 < 8
–c –c –2 –2
Thus, when both sides of trichotomy sign are multiplied or divided by an equal
negative number, the sign ‘<’ is changed to ‘>’ and the sign ‘>’ is changed to ‘<’.
EXERCISE 11.4
General Section - Classwork
1. Let's say and write ‘true’ or ‘false’ in the blank spaces.
a) –2 < – 7 ............... b) 0 > – 5 ............... c) 6 < – 6 ...............
d) 2 – 4 = 8 – 6 ............... e) 10 < -55 ............... f) –6 > –9 ...............
–5 3 3
2. Let's insert the appropriate trichotomy sign ( <, > or = ) in the boxes.
a) 9 –9 b) –9 9 c) 7 – 11 4–8
d) 4 × (–2)
–4 × (–2) f) 10 –20 g) 15 –18
–4 8 –3 3
Creative Section
3. a) Define trichotomy property of whole numbers with examples.
b) Define negation of trichotomy with examples.
c) State the addition, subtraction, multiplication, and division axioms of the
trichotomy rules with examples.
d) If x > y, then (i) x × (–a) < y (–a) and (ii) x < y . Justify these facts with
one example of each. –a –a
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4. Let's rewrite the following trichotomy statements using negation signs.
a) x > 5 b) x < 8 c) – a = – 7 d) p – 4 < 10
e) 12 = y + 4 f) – 9 > a – 4 g) x + 3 > – 6 h) y – 5 < 2
5. Write the following statements using trichotomy signs. Also, write the
negation of each statement. Rewrite each negation by using signs.
a) x is less than 3 b) y is greater than – 5 c) – a is equal to – 2
d) x + 4 is less than 6 e) 10 is greater than p – 9 f) p + r is less than q + r
11.7 Inequalities
'The sum of x and 5 is 9' is an open mathematical statement.
Here, x + 5 = 9 is the open mathematical statement containing 'equal to' (=) sign.
Such type of open mathematical statement is called an equation.
On the other hand, if the open mathematical statement contains trichotomy sign
such as < (less than), > (greater than), ≤ (less than and equal to), ≥ (greater than
and equal to), it is an inequality. Inequality is also called inequation. For example:
x > 5 (x is greater than 5), y < 7 (y is less than 7), x ≤ 9 (x is less than and equal to 9),
p ≥ – 2 (p is greater than and equal to – 2), etc. are a few examples of inequalities.
11.8 Replacement set and solution set
Let’s consider a set of natural number less than 5. Then, N = {1, 2, 3, 4}.
Consider an inequality x < 3.
Now, let’s replace x by the natural numbers of the set N.
When x = 1, 1 < 3 It is true. When x = 2, 2 < 3 It is true.
When x = 3, 3 < 3 It is false. When x = 4, 4 < 3 It is false.
Thus, the inequality x < 3 is true only for a certain values of x taken from the set of
natural numbers N, less than 5.
Here, the set of values of x that makes the inequality true is {1, 2} and it is known
as Solution set. Similarly, the set of natural numbers, N = {1, 2, 3, 4} from which
numbers are used to replace x in the inequality is known as the replacement set.
11.9 Graphical representation of solution sets
We use number lines to show the solution sets of the given inequalities graphically.
Let's study the following illustrations.
x>3 x≤3
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
The circle at 3 shows that 3 is not The solid circle shows that 3 is
included in the solution set. included in the solution set.
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-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 x≤ –2
–5< x <5 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 –7≤ x ≤ 7
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Worked-out examples
Example 1: If R = {0, 1, 2, 3, 4, 5} be the replacement set of the inequality
2x + 3 < 9, find its solution set and illustrate graphically.
Solution:
Here, the replacement set, R = {0, 1, 2, 3, 4, 5}
T orh, e2gxiv+2exn3+i–n33eq u<<al99it –y 3is , Iwwnhhxeenn< xx3== 01,, 01 << 33 iiss ttrruuee
or, 2x < 6
when x = 2, 2 < 3 is true
or, 2x < 6 when x = 3, 3 < 3 is false
2 2
or, x < 3 ∴ The solution set = {0, 1, 2}
0≤x≤2
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Example 2: Find the solution sets of the following inequalities and represent
them graphically. (i) 3x – 1 > x + 9 (ii) 4 – 5x ≤ 2x + 18
Solution:
(i) 3x – 1 > x + 9
or, 3x – 1 – x > 9
or, 2x – 1 > 9 1 is added to both sides.
or, 2x – 1 + 1 > 9 + 1
or, 2x > 10
or, 2x > 10 Both sides are divided by 2.
or, 2 2
x>5
x > 5
∴ Solution set = {6, 7, 8, …} -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
(ii) 4 – 5x ≤ 2x + 18
or, 4 – 5x – 2x ≤ 18
or,
4 – 7x ≤ 18
or, 4 – 4 – 7x ≤ 18 – 4 4 is subtracted from both sides.
or,
– 7x ≤ 14
or, ––7x7x ≥ 14 Both sides are divided by – 7. So, the sign ≤ is changed into ≥.
or, ≥ ––27 x ≥ -2
∴ Solution set = {– 2, – 1, 0, 1, 2,…}
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
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Equation, Inequality and Graph
Example 3: Solve 4 < x + 2 ≤ 9 and represent the solution set graphically.
Solution: 2 is subtracted from all sides.
4<x+2≤9
or, 4 – 2 < x + 2 – 2 ≤ 9 – 2 2<x≤7
or, 2 < x ≤ 7
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Example 4: The difference of two times a number and 5 is less than and equal to –9.
(i) Express it using trichotomy sign.
(ii) Solve the inequality and draw a graph to represent it.
Solution:
(i) Let the number be x.
According to the given statement, 2x – 5 ≤ – 9
(ii) or, 2x – 5 + 5 ≤ – 9 + 5 5 is added to both sides.
or, 2x ≤ – 4
or, 22x ≤ –22 Both sides are divided by 3.
or, x ≤ – 2 x ≤ –2
∴ Solution set = {– 2, – 3, – 4, …}
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Example 5: When one-third of a number is subtracted from 2, the difference is
greater than and equal to 4.
(i) Express it using trichotomy sign.
(ii) Solve the inequality and represent it graphically.
Solution:
(i) Let the number be x. x
3
According to the given statement, 2 – ≥4
(ii) 2– x ≥ 4
3
or, 2–2– x ≥ 4 – 2 2 is subtracted from both sides.
3
or, – x ≥ 2
3
– 3 × (– x ( Both sides are multiplied by –3.
or, 3 ≤ 2 × (– 3) So, the sign ≥ is changed into ≤.
or, x ≤ –6
∴ Solution set = {–6, –7, –8, ...}
x ≤– 6
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
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Equation, Inequality and Graph
EXERCISE 11.5
General Section - Classwork
1. Let's say and write the inequalities represented by each of the following graphs.
b)
a)
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
c) d)
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
e) f)
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
g) h)
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
2. Let's represent these inequalities in the number lines.
a) x > – 4 b) x ≤ 2
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
c) – 4 < x < 4 d) – 6 ≤ x ≤ 7
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Creative Section
3. a) What is an inequality? Write with an example.
b) In what way x + 2 < 7 is different from x + 2 = 7?
c) Define replacement set and solution set of an inequality with examples.
4. Let's draw graphs and represent the following inequalities.
a) x > 3 b) x < 4 c) x < – 2 d) x ≥ – 3
e) 2 < x < 7 f) – 6 ≤ x ≤ 6 g) – 5 < x ≤3 h) – 2 ≤ x < 6
5. If the replacement set is W = {0, 1, 2, 3, 4, 5}, let's find the solution set of the
following inequalities and represent them graphically.
a) x < 4 b) x ≥ 1 c) x ≤ 5 d) x + 1 < 5
e) x – 2 ≥ 1 f) x < 5 g) 2x > 4 h) x – 3 ≤ 1
3 3 5 5 4 4
6. Let's find the solution sets of the following inequalities and represent them
graphically.
a) x + 1 < 5 b) x + 2 > 5 c) x – 1 ≤ 2 d) x – 4 ≥ – 2
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Equation, Inequality and Graph
e) 3 < x + 9 f) 10 > x + 7 g) 7 ≥ x + 10 h) 3 ≤ x – 3
i) 3 – x ≤ – 2 j) 2x > 6 k) 3x – 4 < 2 l) 2x – 5 >, 10 − x
m) x < – 1 n) 2x > – 4 o) x – 1 ≤ – 3 p) 2x –3 ≥ 1
2 3 2 – 6 2
7. Let's solve the following inequalities and show the solutions graphically.
a) 5 < x + 1 < 9 b) 3 < x + 2 < 10 c) – 1 < x + 3 < 7
d) – 2 < x – 1 < 5 e) – 4 ≤ 2x ≤ 6 f) – 9 ≤ 3x ≤ 15
8. a) The sum of two times a number and 2 is less than 12.
(i) Express the statement by using trichotomy sign.
(ii) Solve the inequality and draw a graph to show the solution.
b) The sum of three times a number and 7 is greater than and equal to – 8.
(i) Express the statement by using trichotomy sign.
(ii) Solve the inequality and represent the solution graphically.
c) The difference of four times a number and 5 is less than and equal to 3.
(i) Express the statement by using trichotomy sign.
(ii) Solve the inequality and show the solution graphically.
d) When 5 times a number is subtracted from 7 the result is greater than and
equal to –3.
(i) Express the statement in a mathematical sentence.
(ii) Solve it and show the solution in a number line.
e) The difference of 5 and one-third of a number is less than and equal to 4.
(i) Express the statement in a mathematical sentence.
(ii) Solve it and show the solution in a number line.
9. a) The length a line segment is always greater than its x+1 x−3
BC
each part. On the basis of this fact, write down two A
inequalities from the figure along side. Solve them
and represent the solution graphically.
b) The sum of the length of two sides of any triangle is always A
greater than the length of third side. On the basis of this fact, 3x 2x
write down three inequalities from the triangle given below.
Solve them and represent the solution graphically. B 3x + 4 C
11.10 Graph of linear equation
Consider a linear equation x + y = 3. (0, 3)
It can be written as y = 3 – x. It gives us the rule for (1, 2)
taking values of x and using them to calculate values
for y. For example, (2, 1)
When x = 0, y = 3 – 0 = 3, When x = 1, y = 3 – 1 = 2 X' (3, 0) X
When x = 2, y = 3 – 2 = 1, When x = 3, y = 3 – 3 = 0 O (0, 0)
Now, the results can be shown in a table given below. Y'
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Equation, Inequality and Graph
x 0123
y=3–x 3 2 1 0
The values of x and y can be written in the form of coordinates, such as (x, y). Here, the
values of (x, y) are (0, 3), (1, 2), (2, 1), and (3, 0). If these coordinates are plotted on graph
and the points are joined, a straight line is obtained.
Thus, a linear equation always gives a straight line.
Worked-out examples
Example 1: From the equation y = 2x – 3, write the values of x and y, and complete
the table.
x
y
Also, draw a graph of the equation.
Solution:
Here, y = 2x – 3 is the given equation. A
Let x = 0, then y = 2× 0 – 3 = – 3 (2, 1)
Let x = 1, then y = 2 × 1 – 3= –1 X' O(0, 0) X
Let x = 2, then y = 2 × 2 – 3 = 1 (1, –1)
x012 (0, –3)
y –3 –1 1
B Y'
Now, plotting the coordinates (0, –3), (1, –1) and (2, 1), we get a straight line AB.
EXERCISE 11.6
General Section - Classwork
1. From the graphs of the straight lines, say and write the values of x and y in the
table.
a) Y b) Y
X' (0, 0) X X' (0, 0) X
Y' Y'
x x
y y
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Equation, Inequality and Graph
2. Let’s say and write the values of y and complete the tables.
a) y = x + 1 b) y = x – 3
x 0 1 2 3 x0 123
y y
c) y = 5 – x d) y = 2x + 1
x1 35 7 x0 – 1 1 –2
y y
Creative section
3. Let's copy and write the values of y. Plot the coordinates (x, y) and draw the graphs
of straight lines.
a) y = x + 2 b) y = x – 3 c) y = 2x – 1
x024 x013 x014
y
yy
4. Let's write the values of x of your own choice and find the values of y. Plot the
coordinates (x, y) and draw the graphs of straight lines. 3x – 1
2
a) y = x + 3 b) y = x – 2 c) y=
x x x0
y yy
5 1
5. Let's draw tables and write the values of x of your own choice. Then, find the
values of y. Plot the coordinates (x, y) and draw the graphs of straight lines.
a y = x + 3 b) y = x – 4
c) y = 2x + 1 d) y = 3x – 1
e) y = x + 1 f) y= 2x – 1
2 2
g) x + y = 5 h) x – y = 2
It’s your time - Project work
6. a) Let’s write linear equations of two variables of your own choice. Write any
four appropriate values of x and find the values of y in tables. Then, plot the
coordinates (x, y) and draw the graph of the straight lines.
b) Let's write the appropriate values of x of each pair of linear equations and find
the values of y in the tables. Plot the coordinates of each pair of equations and
draw two straight lines in each case. Find the coordinates of point of intersection
of each pair of straight lines. Do the coordinates of point of intersection satisfy
both the equations? Discuss in the class.
(i) x + y = 3 (ii) x + y = 4 (iii) x + y = 7
x – y = 1 x – y = 2 x–y=3
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Equation, Inequality and Graph
OBJECTIVE QUESTIONS
Let’s tick (√) the correct alternative.
1. The equation for the statement ‘the sum of x and 6 is 10’ is
(A) x + 6 = 10 (B) x + 10 = 6 (C) x – 10 = 6 (D) x – 6 = 10
2. The solution of the equation 2x + 9 = 5 is
(A) 4 (B) -4 (C) 2 (D) -2
3. What value of x satisfies the equation x + 25% of x = 25?
(A) 10 (B) 20 (C) 24 (D) 25
4. If x – y = 9 and the smaller number is 6, what is the greater one?
(A) 9 (B) 12 (C) 15 (D) 14
5. There are 200 students in a school. If the number of boys is 20 more than that of girls,
what is the number of girls?
(A) 100 (B) 110 (C) 90 (D) 180
6. 10 years ago, if Ram was p years old, how old will he be after 10 years?
(A) (p + 10) years (B) (p – 10) years (D) (p +20) years (D) (p – 20) years
7. The negation of the sign ‘=’ is
(A) = (B) > (C) ≠ (D) ≤
8. What is correct trichotomy statement for the statement ‘x + 1 is less than 6’ is
(A) x + 1 < 6 (B) x – 1 > 6 (C) x + 1 > 6 (D) x + 1 = 6
9. If a and b are any two whole number such that a<b and c is a negative integer, then
(A) ac > bc (B) ac < bc (C) ac ≤ bc (D) ac ≥ bc
10. The inequality represented by the graph in the number line is
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
(A) x<-2 (B) x> -2 (C) x≤ - 2 (D) x ≥ -2
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Unit 12 Coordinates
12.1 Coordinates – Looking back
Classwork - Exercise
1. Let's say and write the answers as quickly as possible.
a) The x and y-coordinates of the point (2, – 5) are ........... and ...........
b) If x-coordinate of a point is 3 and y-coordinate is – 2, the coordinates of the
point are ...................
c) The coordinates of the origin are ...................
d) The y-coordinate of a point on x-axis is ...................
e) The x-coordinate of a point on y-axis is ...................
f) The point (0, 5) lies on ................... axis
g) The point (4,0) lies on ................... axis.
h) The point (1,2) lies in the ......................................... quadrant.
i) The point (5,−3) lies in the ......................................... quadrant.
j) The point (−3,−4) lies in the ......................................... quadrant.
k) The point (−7,4) lies in the ......................................... quadrant.
2. From the given graph, let's say and write the coordinates of the points as
quickly as possible. Y
Coordinates of A are ........................... B A
Coordinates of B are ........................... OX
Coordinates of C are ........................... X'
Coordinates of D are ........................... C D
Y'
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Coordinates
In the given graph, point P is at a distance of 5 units along Y
to the direction of OX and 3 units up to the direction of OY
from the point O.
Here, the position of the point P on the graph is 5 units along 4
and 4 units up which is written as (5, 4).
Thus, the ordered pair of numbers which are used to indicate O 5 X
the position of a point on a grid is known as the coordinates
of the point.
Here, 5 is called x-coordinate or abscissa.
4 is called y-coordinate or ordinate.
The system of describing position of a point by using coordinates was invented by a
French Mathematician called René Descartes in the 17th century.
12.2 Coordinate axes and quadrants
In the figure alongside, XOX’ and YOY’ are Y
horizontal and vertical number lines respectively.
They are intersecting each other perpendicularly 2nd 1st
at the point O. These number lines are called the Quadrant Quadrant
coordinate axes.
X' O X
Here, XOX’ is called the x-axis 3rd
YOY’ is called the y-axis Quadrant 4th
The point O is called the origin. Quadrant
Furthermore, the intersection of x-axis and y-axis Y'
divide the coordinate plane into 4 regions. These regions are called the quadrants.
The region XOY is called the first quadrant.
The region X’OY is called the second quadrant.
The region X’OY’ is called the third quadrant.
The region XOY’ is called the fourth quadrant.
12.3 Finding points in all four quadrants
The co-ordinates of a point is always written as Y
the ordered pair of (x-coordinate, y-coordinate),
i.e. (x, y). Now, let’s learn to find out the B (-2, 4)
coordinates of points in different quadrants.
A (4, 2)
(i) In the first quadrant, the coordinates of a X' 2nd Quadrant 1st Quadrant X
point is (x, y). (-x, y) (x, y)
For example, the coordinates of A is (4, 2) 3rd QuadrantO 4th Quadrant
(-x, -y) (x, -y)
C (-1, -2)
D (4, -3)
(ii) In the second quadrant, the coordinates of a
point is (– x, y). Y'
For example, the coordinates of B is (– 2, 4)
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Coordinates
(iii) In the third quadrant, the coordinates of a point is (– x, – y).
For example, the coordinates of C is (– 1, – 2)
(iv) In the fourth quadrant, the coordinates of point is (x, – y).
For example, the coordinates of D is (4, – 3)
(v) The coordinates of the origin is (0, 0).
(vi) The coordinates of any point on the x–axis is (x, 0) or (– x, 0).
(vii) The coordinates of any point on the y-axis is (0, y) or (0, – y).
12.4 Plotting points in all four quadrants
Graph paper (or square grid) is used to plot the given point in different quadrants.
On the graph paper, we should first draw the coordinate axes XOX’ and YOY’.
Study the following illustrations and learn to plot the points in different quadrants.
plotting (3, 2) plotting (– 3, 4) plotting (– 3, – 2) plotting (2, – 4)
Y Y Y Y
Q(-3,4)
P(3,2) 4
X 3 2 -3 X' X -2 -3 O X' X O2 X'
O X' X O R(-3,-2) -4
S(2,-4)
Y' Y' Y' Y'
(3, 2) lies in (– 3, 4) lies in (– 3, – 2) lies in (2, – 4) lies in
the 1st quadrant. the 2nd quadrant. the 3rd quadrant. the 4th quadrant.
EXERCISE 12.1
General Section -Classwork
1. Let's say and write the answers in the blank spaces.
a) A point is 4 units right from the origin along x–axis and 3 units up along
y-axis, then its coordinates are …………………
b) A point is 5 units left from the origin along x–axis and 2 units down along
y-axis, then its coordinates are …………………
c) The coordinates of the origin are …………………
d) If the abscissa of a point is –6 and its ordinate is 4, then its coordinates
are …………………
e) The coordinates of a point (5, – 2), lies in the ………………… quadrant.
f) The coordinates of a point (– 3, – 8), lies in the ………………… quadrant.
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Coordinates
g) The coordinates of a point are (4, 0), lies on ………………… axis.
h) The coordinates of a point (0, – 7), lies on ………………… axis.
2. Let's tell and write the coordinates of the points from the graph as quickly as
possible.
Y Coordinates of P are ....................................
Coordinates of Q are ....................................
X' O X Coordinates of R are ....................................
Coordinates of S are ....................................
Y' Y
O
3. Plot these points in the graph as quickly as X' X
possible.
A (1, 3) , B (–3, 4)
C (–2, –4), D (5, – 2)
E (2, 0) , F (−5, 0)
G (0, 5) , H (0, −3)
Creative Section - A
4. a) Define coordinates of a point. Y'
b) What are quadrants? Express the ordered pairs of x and y-coordinates in
four different quadrants.
c) What are abscissa and ordinate? If the ordinate of a point is m and it's
abscissa in n, write the coordinates of the point.
d) What is origin? What are the coordinates of the origin?
5. Write down the coordinates of the points of intersection of each pair of straight
line segments. Y
5
a) Y b) Y c)
5 5 4
44
33 3
22 2
1
1 1 1 2 3 4 5 X X' –5 –4 –3 –2 –1–O1 1 2 3 4 5X
X' –5 –4 –3 –2 –1–O1 1 2 3 4 5 X X' –5 –4 –3 –2 –1–O1
–2
–2 –2
–3
–3 –3
–4
–4 –4
–5
–5 –5 Y'
Y' Y'
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d) Y e) Y f) Y
5 5 5
44 4
33 3
22 2
1
1 1 1 2 3 4 5 X X' –5 –4 –3 –2 –1–O1 1 2 3 4 5X
X' –5 –4 –3 –2 –1–O1 1 2 3 4 5 X X' –5 –4 –3 –2 –1–O1
–2 –2 –2
–3 –3 –3
–4 –4 –4
–5
–5 –5 Y'
Y' Y'
6. Let's plot the following points in graph papers.
a) (2, 4), (– 2, 4), (– 2, – 4), (2, – 4) b) (5, 3), (– 5, 3), (– 5, – 3), (5, – 3)
c) (6, 1), (– 6, 1), (– 6, – 1), (6, – 1) d) (7, 0), (0, 7), (– 7, 0), (0, – 7)
7. Let's plot the following points in graph papers and join them in order. Name
the geometrical shapes so formed.
a) A (2, 4), B (– 2, 3), C (0, 0)
b) P (– 5, – 2), Q (2, – 4), R (– 4, 6)
c) E (2, 5), F (– 3, 4), G (– 1, 6), H (3, – 5)
d) A (4, – 2), B (– 4, – 2), C (– 2, – 4), D (5, 6)
e) P (2, 3), Q (– 3, 3), R (– 3, – 2), S (2, – 2)
f) W (4, 3), X (– 4, 3), Y (–4, – 2), Z (4, – 2)
g) D (– 4, 0), E (0, 0), F (3, 3), G (– 1, 3)
h) K (– 4, – 5), L (– 2, 0), M (– 4, 2), N (– 6, 0)
Creative Section - B
8. Let's plot the points given in (i) and (ii) on the same axes of reference. Join them
separately by using a ruler. Find the coordinates of their points of intersection.
a) (i) (1, 0), (3, 2), (5,4) (ii) (5, 0), (3, 2), (1, 4)
b) (i) (2, – 4), (1, – 2), (– 2, 3) (ii) (4, 1), (1, – 2), (– 2, – 5)
c) (i) (1, 3), (4, 9), (– 3, – 5) (ii) (2, 2), (5, – 1), (– 2, 6)
d) (i) (2, 0), (5, 3), (– 1, – 3) (ii) (2, – 2), (3, 0), (4, 2)
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Coordinates
9. a) P (– 3, – 1), Q (– 1, – 1), R (1, 1) are three of the four vertices of a parallelogram
PQRS and S is the opposite vertex of Q. Plot these vertices in graph paper
and find:
(i) the coordinates of the vertex S.
(ii) the coordinates of the point of intersection of the diagonals PR and QS.
b) A (– 3, – 5), B (2, – 5), C (2, 2) are three of the four vertices of a rectangle
ABCD and D is the vertex opposite to B. Plot these vertices in graph paper
and find:
(i) the coordinates of the vertex D (ii) the area of the rectangle ABCD.
c) E (2, 3), F (–3, 3), and G (–3, –2) are three of four vertices of a square EFGH
where H is the opposite vertex of F. Plot these vertices in a graph paper and
find:
(i) the coordinates of the vertex H. (ii) the area of the square EFGH.
It's your time - Project work!
10. a) Let's mark a points respectively in the first, second, third and the fourth
quadrants of the coordinates plane in a graph paper given below. Then,
write the coordinates of the point in each quadrant.
b) Let's draw a rectangle in a graph paper. Write the coordinates of the vertices
of the rectangle .What is the area of your rectangle?
c) Let's draw a square in a graph paper write the coordinates of the vertices of
the square .What is the area of your square?
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Unit 13 Transformation
13.1 Reflection of geometrical figures
Let's study the following illustration and investigate the idea of reflection.
Mirror image Object Object Mirror image Object Mirror image
In the above illustrations, the objects are reflected by the mirror and their images are
formed inside the mirror. A
In the similar way, the reflection of a B Geometrical figure
geometrical figure means formation of M
C Line of reflection
M’
the image of the figure after reflecting B' C'
the figure about the line of reflection. Mirror image
A'
Let's have a look in the following geometrical figures and learn to find the image of
corresponding objects (points, line segments, angles, figures, etc) while reflecting
on the line XX'.
P KA
Q RL M
Q' R' L' M'
P' K' A'
In the above figure, XX' is the line of reflection. It is also called mirror line or axis
of reflection.
Here, image of point A is A', P is P', Q is Q', R is R', K is K', L is L', and M is M'.
Similarly, image of line segment, PQ is P'Q', QR is Q'R' ect. Likewise, image of DPQR
is DP'Q'R', image of ∠KLM is ∠K'L'M', and so on.
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Again, let’s learn to draw the image of a geometrical figure reflected about a line of
reflection.
In the figure, ABC is a triangle. MM’ is the line of reflection.
(i) Draw perpendiculars AP, BQ and CR from each vertex of ∆ABC on the line of
reflection. A
(ii) Produce AP, BQ and CR.
(iii) Measure the length of PA by using compasses B C M’
(iv) and cut off PA' = PA. Similarly, cut off QB' = QB
and RC' = RC. MQ PR
B’ C’
Join A', B' and C' by using a ruler.
Thus, ∆A'B'C' is the image of ∆ABC so formed. A’
Properties of reflection
(i) The geometrical figure and its image are at the equal distance from the line of
reflection.
(ii) The areas of the geometrical figure and its image are equal.
(iii) The appearance of the image of a figure is opposite to the figure.
13.2 Reflection of geometrical figures using coordinates
Here, we learn to find the coordinates of the images of geometrical figures formed
due to the reflection by x-axis and y-axis separately. In this case, x-axis and y-axis
are called the axis of reflection.
(i) x-axis as the axis of reflection
Let's study the following illustrations and learn to find the coordinates of the
image of a point in different quadrants when x-axis is the axis of reflection.
The point is in The point is in The point is in The point is in
the 1st quadrant. the 2nd quadrant. the 3rd quadrant. the 4th quadrant.
Y Y Y Y
P(2,3) P(-2,3) P(-2,-3) P(2,3)
XO X' X O X' X O X' X O X'
P'(2,-3) P'(-2,-3) P'(-2,-3) P'(2,-3)
Y' Y' Y' Y'
∴P (x, y) → P' (x, –y) ∴ P (– x, y) → P'(–x, –y) ∴ P (– x, – y) → P' (–x, y) ∴ P (x, –y) → P' (x, y)
x-coordinate x-coordinate x-coordinate remains x-coordinate
remains the same. remains the same. the same. remains the same.
Sign of y-coordinate Sign of y-coordinate Sign of y-coordinate Sign of y-coordinate
is changed. is changed. is changed. is changed.
197Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Coordinates
From the above illustrations, it is clear that, when a geometrical figure is reflected
about x-axis as the axis of reflection, the x-coordinate of the image remains the same
and the sign of y-coordinate of the image is changed.
(ii) y-axis as the axis of reflection
Study the following illustrations and learn to find the coordinates of the image
of a point in different quadrants when y-axis is the axis of reflection.
The point is in The point is in The point is in The point is in
the 1st quadrant. the 2nd quadrant. the 3rd quadrant. the 4th quadrant.
Y Y Y Y
P(-2,3) P'(2,3)
P'(-3,2) P(3,2)
X' O X X' O X X' O X X' O X
Y' Y' P(-3,-2) P'(3,-2) P'(-2,-3) P(2,-3)
∴P (x, y) → P' (– x, y) ∴ P (– x, y) → P'(x, y) Y' Y'
y-coordinate y-coordinate ∴ P (– x, – y) → P' (x, – y)
∴ P (x, – y) → P' (– x, – y)
remains the same. remains the same. y-coordinate y-coordinate
Sign of x-coordinate Sign of x-coordinate remains the same.
Sign of x-coordinate remains the same.
is changed. is changed. Sign of x-coordinate
is changed.
is changed.
From the above illustrations, it is clear that when a geometrical figure is reflected about
y-axis as the axis of reflection, the sign of x-coordinate of the image is changed but the
y-coordinate of the image remains the same.
EXERCISE 13.1
General Section - Classwork
1. Let's observe the graph given below and name the image point of following
objects under reflection on the line MN.
G
A CD Object Image
B I HE Point A Point P
MQ
N Point B .................
UV R Point W .................
ST Point E .................
PW
Vedanta Excel in Mathematics - Book 7 198 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
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Vedanta Excel in Mathematics Book 7 Final (2079)
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