Math 8

18.2 Some special product and formulae (Continue)

Formula 4

(a + b)3 = a³ + 3a²b + 3ab² + b³

Proof: We have,

(a + b)3 = (a + b) (a + b)2 = (a + b) (a² + 2ab + b²)

= a(a² + 2ab + b²) + b(a² + 2ab + b²)

= a³ + 2a²b + ab² + a²b + 2ab² + b³

= a³ + 3a²b + 3ab² + b³

∴ (a + b)³ = a³ + 3a²b + 3ab² + b³

Again, (a + b)3 = a³ + 3a²b + 3ab² + b³

= a³ + 3ab (a+b) + b³

= a³+ b³+ 3ab (a + b)

or, ( a + b)3 – 3ab(a + b) = a³ + b³

∴ a³ + b³ = (a + b)³ – 3ab(a + b)

Formula 5

(a – b)3 = a³ – 3a²b + 3ab² –b³

Proof: We have,

(a – b)3 = (a – b) (a – b) (a – b)

= (a – b) (a – b)2

= (a – b) (a² – 2ab + b²)

= a(a² – 2ab + b²) – b(a² – 2ab + b²)

= a³ – 2a²b + ab² – a²b + 2ab² – b³

= a³ – 2a²b – a²b + 2ab² + ab² – b³

(a – b)3 = a³ – 3a²b + 3ab² – b³ Remember !
∴ (a – b)³ = a³ – 3a²b + 3ab² – b³
(a + b)3= a3 + b3 + 3ab (a + b)
Cor: We have, = a3 + 3a2b + 3ab2 + b2
(a – b)3 = a3 – b3 – 3ab (a – b)
(a – b)3 = a³ – 3a²b + 3ab² – b³ = a3 – 3a2b + 3ab2 – b3

(a – b)3 = a³ – b³– 3ab (a–b)

or, = (a–b)³ +3ab (a – b) = a³ – b³

∴ a³ – b³ = (a – b)³ + 3ab(a – b)

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Geometrical Interpretation of (a + b)3 a-b
a
Take a cubical object having the length of each side (a+b)cm.
Cut the cube into 8 parts as shown in the figure.

b b
b

a b
a

Here,

Volume of first part = a × a × a = a3

Volume of second part = a × b × b = ab2

Volume of third part = a × b × b = ab2

Volume of fourth part = a × b × b = ab2

Volume of fifth part = a × a × b = a2b

Volume of sixth part = a × a × b = a2b

Volume of seventh part = a × a × b = a2b

Volume of eighth part = b × b × b = b3

∴ Volume of given object = a3 + ab2 + ab2 + ab2 + a2b + a2b+ a2b + b3

(a + b)3 = a3 + 3a2b + 3ab2 + b3

∴ (a + b)3 = a3 + 3a2b + 3ab2 + b3

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Geometrical Interpretation of (a – b)3 b a a-b
bb
Take a cubical object having the length of each side a cm. a-b
From each side, cut b cm as shown in the figure. a
Volume of the whole solid = a3
-b
Here we can get 8 parts as shown in the figure.

-b -b
-b
-b

-b

-b

-b -b
-b

Volume of first part = (a–b)3

Volume of second part = b2(a-b)

Volume of third part = b2(a-b)

Volume of fourth part = b2(a-b)

Volume of fifth part = (a-b)2.b

Volume of sixth part = (a-b)2.b

Volume of seventh part = (a-b)2.b

Volume of eighth part = b3

Here,

∴ a3 = (a – b)3 + (a – b)b³ + (a – b)b2 + (a – b)b2 + (a – b)2b + (a – b)2b + (a – b)2b + b3

a3 = (a – b)3 + 3 (a – b)b2 + 3 (a – b)2b + b3

a3 = (a – b)3 + 3 (ab2 – b3) + 3(a2 – 2ab + b2) b + b3

a3 = (a – b)3 + 3ab2 – 3b3 + 3a2b – 6ab2 + 3b3 + b3

a3 = (a – b)3 – 3ab2 + 3a2b + b3

∴ (a – b)3 = a3 – 3a2b + 3ab2 – b3

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Worked Out Examples

Example: 1 1
3b
Find the cubes of : 2a –

Solution:

We know that (a – b)3 = a³ – 3a²b + 3ab² – b³

So, the required cube of (2a – 31b) = (2a – 31b)3 (31b)2 (31b)³

= (2a)3 – 3 × (2a)2 × 1 + 3 × 2a × –
3b
1 1 1
= 8a³ – 3 × 4a² × 3b + 3 × 2a × 9b2 – 27b3
Example: 2
= 8a³ – 4ba2 + 2a – 271b3
3b2

If a + b = 8 and ab = 5, then find the value of a³ + b³.

Solution:

Now, we know that

a³ + b³ = (a + b)3 – 3ab (a + b)

= (8)3 – 3 × 5(8)

= 512 – 120 = 392

∴ a³ + b³ = 392

Example: 3

If a – 1 = 7, find the value of a³ – 1 .
a a3

Solution:

Here, a– 1 = 7
a

Now, cubing on both sides, we get

( )a –1 3 = (7)3 [∵ (a – b)3 = a³ – 3ab(a – b) – b³]
a

( )or, 1 1 1
(a)3 – 3a × a a– a – a3 = 343

or, a³ – 3(7) – 1 = 343 [∵ a– 1 = 7]
a3 a

or, a³ – 21 – 1 = 343
a3

∴ a³ – 1 = 343 + 21 = 364
a3

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Example: 4

Simplify: (5x + 2y)3 + 3(5x + 2y)2 (3x – y) + 3(5x + 2y) (3x – y)2 + (3x – y)3

Solution:

Let 5x + 2y = a and 3x – y = b,

then (5x + 2y)3 + 3(5x + 2y)2 (3x – y) + 3(5x + 2y) (3x – y)2 + (3x – y)3

= a³ + 3× (a)2 × b + 3 × a × b² + (b)3

= a³ + 3a²b + 3ab² + b³

= (a + b)3 [∴a³ + 3a²b + 3ab² + b³ = (a + b)3]

Now, replacing the value for a = 5x + 2y and b = 3x – y, we get

= {(5x + 2y) + (3x – y)}3

= {5x + 2y + 3x – y}3
= (8x + y)3

Example: 5

If x + 1 = k, prove that x3 + 1 = k3 – 3k.
x x3

Solution:

L.H.S. = x3 + 1
x3

= (x + 1 )3 – 3.x. 1 (x + 1 ) [ ∵ a³ + b³ = (a + b)³ – 3ab (a + b)]
x x x

= k3 – 3(k) [∵ x + 1 = k]
x

= k3 – 3k = R.H.S.

∴ Hence proved.

Example: 6

Evaluate: 99³

Solution: Here, 993 = (100 – 1)3

= (100)³ – (1)³ – 3 × 100× 1 (100 –1) [ ∵ (a – b)³ = a³ – b³ – 3ab (a – b)]

= 1000000 – 1 – 300 × 99

= 1000000 – 1 – 29700

= 970299

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Exercise 18.2

1. Find the cubes of the following binomials:

(a) (4a + 5b) (b) (3x – 4y) (c) y + 1 (d) 2x – 1
y 3y

2. (a) If a + b = 9 and ab = 4, find the value of a³ + b³.

(b) Find the value of a³ + b³, if a + b = 8 and ab = 3.

(c) Find the value of x3 –y3, if x – y = 6 and xy = 4.

(d) Find the value of a³ – b³, if a – b = 3, ab = 10.

(e) Find the value of x3 – 1 , if x – 1 = 4.
x3 x

(f) Find the value of a³ + 1 , if a + 1 = 9.
a3 a

3. (a) If x + 1 = m, prove that x3 + 1 = m3 – 3m.
x x3

(b) If 3x – 2y = 5, prove that 27x3 – 8y3 – 90xy = 125.

(c) If a + 1 = 3, prove that a³ + 1 = 0.
a a3

(d) If p – 2q = 4, prove that p3 – 8q3 –24pq = 64.

(e) If x – y = 7, prove that x3 – y3 – 21xy = 343.
4. Simplify:
(a) (p + q)3 – 3(p + q)2 (p – q) + 3(p + q) (p – q)2 – (p – q)3
(b) (2x + 7y)3 + 3(2x + 7y)2 (x – 3y) + 3(2x + 7y) (x – 3y)2 + (x – 3y)3
(c) (3m + 4n)3 + 3(3m + 4n)2 (m + 3n) + 3(3m + 4n) (m + 3n)2 + (m + 3n)3

5. Find the value of the following without actual multiplication.

(a) (97)3 (b) (106)3 (c) (1001)3

6. Simplify the following: (b) (5m +2n)3 – (5m – 2n)3
(a) (3a + b)3 + (3a – b)3

Answer

1. (a) 64a3 + 240a2b + 300ab2 + 125b3 (b) 27x3-108x2y+144xy2-64y3

(c) y3+3y+3/y+1/y3 (d) 8x3-4x2/y+2x/3y2-1/27y3

2. (a) 621 (b) 440 (c) 288 (d) 117 (e) 76 (f) 702

4. (a) 8q3 (b) (3x+4y)3 (c) (4m+7n)3 5. (a) 912673 (b) 1191016 (c) 1003003001

6. (a) 54a3 + 18ab2 (b) 300m2n+16n3

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Unit

19 Factorisation

19.1 Factorisation–Review

Look and learn the following.

5 × 3 = 15, 5 and 3 are the factors of 15.

Similarly, 6×x = 6x, 6 and x are the factors of 6x.

(a+b) (a–b) = a²–b², (a+b) and (a–b) are the factors of a²–b².

Thus, the product of 6 and x is 6x, the term 6x is called a multiple of 6 and x,
whereas 6 and x are the factors of 6x.

The process of finding two or more expressions whose product is equal to the given
expression is called factorisation.

For example: Factorisation of 2a + 4ab = 2a(1 + 2b)

Method of factorisation of algebraic expression:
Factorisation of a polynomial having a monomial factors

Steps: Find by inspection, what is common throughout in the given expression,
then find the quotient of the given expression by this common factor.

For example: Factorise: 2x2y + 6xy2 + 4xy

Solution:
By inspection, we find that each term of given expression 2x2y + 6xy2 + 4xy is exactly

divisible by 2xy. So, 2xy is a common factor.

Here, 2x2y + 6xy2 + 4xy

= 2xy (x + 3y + 2)

Note: If first term is '–', then it is usual to take '–' as common.


When the common factor is a polynomial.

In this case, we take out the common factor and use distributive property. Look and
learn

For example :

Factorise: (a) x (a + b) + y(a + b) + z(a + b)

Solution:

Here, x(a + b) + y(a + b) + z(a + b)=(a + b) (x + y + z)

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(b) ax + by + ay + bx

Solution:

Here, ax + by + ay + bx = ax + ay + bx + by

= a(x + y) + b(x + y) = (x + y) (a + b)

Worked Out Examples

Example: 1 Example: 2

Factorise : 2a²b + 3ab² Factorise: x(a – b) + y (a – b) + z (a – b)
Solution: Solution:
2a²b + 3ab² x(a – b) + y(a – b) + z (a – b)
= ab(2a + 3b) = (a – b) (x + y + z)

Example: 3 Example: 4

Factorise: 2x + 2y + xy + y² Simplify : 7 × 16 + 7 × 14
Solution: Solution:
2x + 2y + xy + y² 7 × 16 + 7 × 14
= 2(x + y) + y (x + y) = 7 (16 + 14)
= (x + y) (2 + y) = 7 × 30 = 210

Exercise 19.1

1. Factorise

(a) 2x + 6y (b) 15x²y + 25xy2 (c) 3a² – 15ab

(d) 6x4y – 18x3y2 (e) – 10 x³y²z³ – 15 x²yz (f) – 21x4a²c³ + 14 x³ac²

2. Factorise

(a) 4a + 3ab + 5a² (b) 3x²y – 6xy + 9xy² (c) 4x³y7 – 2xy6 + 6y9

(d) –abx – acy – adz (e) 4x2y – 6xy2 + 10xy

3. Factorise (b) a2(a + 1) + 1(a + 1)
(a) a(x + y) + b(x + y)

(c) y(a – b) – x(a – b) (d) x(a – b) + y(b – a) (e) x(x – 2y) + y(2y – x)

4. Factorise
(a) (a – 2) (x+ 4) + (a – 2) (x + 5) (b) (x – y)2 + (x – y)

(c) (a – b) – (a – b)2 (d) p(a – b) + q(a – b) + r(a – b)

(e) x(a + b + c) + y(a + b + c) + z(a + b + c)

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5. Factorise

(a) ax + ay + px + py (b) ax + bx + ma + mb

(c) ab – ac – b + c (d) a² – (x + y) a + xy (e) xy² – y(x – z) – z

6. Simplify using factorisation method.

(a) 6×25 + 6 × 13 (b) 29 × 86 + 29 ×14

(c) 100 × 33 + 100 × 67 (d) 279 × 300 – 279 × 200 (e) 15×15 – 15 × 12 + 15 × 13



Answer

1. (a) 2(x+3y) (b) 5xy(3x+5y) (c) 3a(a–5b) (d) 6x3y (x–3y) (e) –5x2yz(2xyz2+3)

(f) –7x3ac2(3xac–2) 2. (a) a(4+3b+5a) (b) 3xy (x–2+3y) (c) 2y6(2x3y–x+3y3)

(d) –a(bx+cy+dz) (e) 2xy (2x–3y+5)

3. (a) (x+y) (a+b) (b) (a+1) (a2+1) (c) (a–b) (y–x) (d) (a–b) (x–y) (e) (x–2y) (x–y)

4. (a) (a–2) (2x+9) (b) (x–y) (x–y+1) (c) (a–b) (1–a+b) (d) (a–b) (p+q+r)

(e) (a+b+c) (x+y+z) 5. (a) (x+y) (a+p) (b) (a+b) (x+m) (c) (b–c)(a–1)

(d) (a–x) (a–y) (e) (y–1) (xy+z) 6. (a) 228 (b) 2900 (c) 10000 (d) 27900 (e) 240

19.2 Factorisation of difference of two squares

As we know that (a + b) (a – b)
= a² + ab – ab – b²
= a² – b²
Factors of a² – b² are (a + b) and (a – b)
Hence, a² – b² = (a + b) (a – b)

Geometrical Interpretation of a² – b²

• Take a square sheet of paper having a side 'a' units.

Its area is a² a

a

• From the square sheet take 'b' units from each side as shown in the figure.



ba

b
a

Area of that part = b²

Area of the shaded part = a² – b² .............. (i)

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• Cut the shaded part.

a

ba

b
a–b

• Again, cut it along the dotted line.

a

a–b

a
bb

a–b

• Join these two parts as shown in the figure.

ab

a–b a–b

ba

Then the new sheet is a rectangle having two sides (a + b) and (a – b)
ab

a–b a–b

b a

Area of shaded part

∴ a² – b² = (a + b) (a – b)

Factorisation of sum of two cubes
We already know that,
(a + b)³ = a³ + 3a²b + 3ab² + b³
or, (a + b)³ = a² + b³ + 3ab (a + b)
or, (a + b)³ – 3ab (a + b) = a³ + b³
or, (a + b) [(a + b)² – 3ab] = a³ + b³
or, (a + b) (a² + 2ab + b² – 3ab) = a³ + b³
or, (a + b) (a² – ab + b²) = a³ + b³
Hence, a³ + b³ = (a + b) (a² – ab + b²)

Factorisation of difference of two cube
We already know that,
(a – b)³ = a³ – 3a²b +3ab² – b³

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or, (a – b)³ + 3a²b – 3ab² = a³ – b³.
or, (a – b)³ + 3ab (a – b) = a³ – b³
or, (a – b) (a² – 2ab + b² + 3ab) = a³ – b³
or, (a – b) (a² + ab + b²) = a³ – b³
∴ a³ – b³ = (a – b) (a² + ab + b²)

Worked Out Examples

Example: 1 Example: 2

Factorise : x² – 16y² Factorise: 25x² – 64y²
Solution:
x² – 16y² Solution:
= (x)² – (4y)² 25x² – 64y²
= (5x)² – (8y)²

= (x + 4y) (x – 4y) [∵ a2–b2 = (a+b ) (a-b)] = (5x + 8y) (5x – 8y) [∵ a2–b2 = (a+b ) (a-b)]

Example: 3 Example: 4

Factorise: (i) (x + 3y)2 – (x – 3y)2
Solution: Factorise: 4a² – 9(b – c)²
(i) (x + 3y)2 – (x – 3y)2 Solution:
4a² – 9 (b – c)²

= {(x + 3y) + (x – 3y)} {(x + 3y) – (x – 3y)} = (2a)² – {3(b – c)}²

[∵ a² – b²= (a + b) (a – b)] = (2a)² – (3b – 3c)²

= (x + 3y+ x – 3y) (x + 3y – x + 3y) = {2a + (3b – 3c)} {2a–(3b–3c)} [∵ a2–b2 = (a+b ) (a-b)]

= 2x × 6y = 12xy = (2a + 3b – 3c) (2a – 3b + 3c)

Example: 5 Example: 6

Factorise : a³ + 27

Solution: Factorise: 8(x – y)³ – z³

(i) Here, Solution:
a³ + 27 = (a)3 + (3)3 8(x – y)³ – z³

= (a + 3) {(a)2 – a× 3 =+ ([32)(2x} – y)]³ – z³

= (a + 3) (a² – 3a +9).= (2x – 2y)³ – z³

= {(2x – 2y) – z} {(2x – 2y)² + (2x – 2y) z + z²}

= (2x – 2y – z) (4x² – 2. 2x. 2y + 4y² + 2xz – 2yz + z²)

= (2x – 2y – z) (4x² – 8xy + 4y² + 2xz – 2yz + z²)

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Exercise 19.2

1. Factorise each of the following:

(a) a² – 16 (b) 4x² – 9y² (c) 9y² – 16z² (d) 25p² – 144q²
(e) a² – b1²
(f) a² – 1 (g) 2x5² – y²
2. (a) 4 – (x – y)² 9 4b² 49

(b) (x + y)² – 9z² (c) (x + y)² – (x – y)²

(d) 9z² – 4(x – y)² (e) (4x – y)² – 9z²

(f) (2a + 3b)² – (2a – 3b)² (g) 25(a + b)² – 16(a – b)²

3. (a) x³ – 4x (b) 2x³ – 72x (c) 5 – 125y² (d) a³b³ – 16ab (e) 2 a ³ b ² – 8 a
z²
4. Factorise each of the following:

(a) x³ + 8 (b) x³ + 27 (c) 8x³ + 125y³ (d) m³ + 27n³

(e) 8x³ – 1 (f) 27m³ – 8n³ (g) 64a³ – 125b³

5. Factorise:

(a) x³ + 1 (b) a³ – 1 (c) a³ + b³ (d) x³ – 125 (e) 2674x³ +1
x³ b³ 27 64 27 y³

6. Factorise:

(a) (x + y)3 + (x – y)³ (b) (a + b)³ – (a – b)³ (c) (a – 2b)³ – 8c³

(d) 64 – (2a – 3b)³ (e) (2x + 3y)³ – (2x – 3y)³

7. Factorise: (a) a³ + b³ + a + b (b) a³ – b³ – a + b

Answer

1. (a) (a+4) (a-4) (b) (2x+3y) (2x-3y) (c) (3y+4z) (3y–4z) (d) (5p+12q) (5p–12q)

(e) (a+b1) (a–b1) (f) (a3+ 1 ) (3a– 1 ) (g) (x5 +7y ) (x5 – y )
2b 2b 7

2. (a) (2+x–y) (2–x+y) (b) (x+y+3z) (x+y–3z) (c) 4xy

(d) (3z+2x–2y) (3z–2x+2y) (e) (4x–y+3z) (4x–y–3z) (f) 24ab (g) (9a+b) (a+9b)

3. (a) x(x+2) (x-2) (b) 2x(x+6) (x-6) (c) 5(1+5y) (1–5y) (d) ab (ab+4z) (ab–z4) (e) 2a (ab+2) (ab–2)

4. (a) (x+2) (x2–2x+4) (b) (x+3) (x2–3x+9) (c) (2x+5y) (4x2–10xy+25y2)

(d) (m+3n) (m2–3mn+9n2) (e) (2x–1) (4x2+2x+1) (f) (3m–2n) (9m2+6mn+4n2)

(g) (4a–5b) (16a2+20ab+25b2)

( )5.(a)x + 1  x2 −1+ 1  ( )(b) a − 1  a 2 + a + 1  ( )(c)a + b  a2 − ab + b2 
x x2 b  b b2  3 4  9 12 16 

( )(d) x − 5  x2 + 5x + 25  (e) 3x + 1  9x2 − 3x + 1 
 3 y   9 3y y2  4  16 4
 

6. (a) 2xy (x2+3y2) (b) 2b (3a2+b2) (c) (a–2b–2c) (a2–4ab+4b2+2ac–4bc+4c2) (d) (4–2a+3b)
(16+8a–12b+4a2–12ab+9b2) (e) 18y(4x2+3y2) 7. (a) (a+b) (a2–ab+b2+1) (b) (a–b) (a2+ab+b2–1)

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19.3 Factorisation of the trinomial of the form a2 ± 2ab + b2

We know that (a + b)² = a² + 2ab + b²
i.e. factors of expression a² + 2ab + b² are (a + b) and (a + b).
Similarly, (a – b)² = a² – 2ab + b²
i.e. factors of expression a² – 2ab + b² are (a – b) and (a – b)
Let's convert the expression 4x² + 12xy + 9y² in the form of (a + b)²
4x² + 12xy + 9y²
= (2x)² + 2. 2x. 3y + (3y)²
= (2x + 3y)²

Worked Out Examples Example: 2

Example: 1 Resolve into factors : 9x² + 24xy + 16y²
Solution:
Convert the trinomial 4a² – 20ab + 25b²
in the form of (a – b)². 9x² + 25xy + 16y²
Solution: = (3x)² + 2.3x. 4y + (4y)²
4a² – 20ab + 25b² = (3x + 4y)²
= (2a)² – 2.2a.5b + (5b)² = (3x + 4y) (3x + 4y)
= (2a – 5b)²
= (2a–5b ) (2a–5b)

Exercise 19.3

1. Write the missing term in each of the following to make perfect squares:

(a) x² + ..... + (2)² (b) 4a² + 24ab + ..... (c) ..... + 16 mn + 16n²

(d) x² – ..... + 16y² (e) 25a² – 60ab + .....

2. Convert the following in the from of (a + b)² or (a – b)².

(a) x² + 10x + 25 (b) 4x² – 4xy + y² (c) 9x² + 24xy + 16y²
(d) 49a² – 84ab + 36b² (e) 64m² – 112mn + 49n²
3. Factorise: (f) x² + 2 + 1
x²

(a) a² – 14a + 49 (b) 25 – 60y + 36y² (c) m² – 24m + 144
(d) 3b² – 6b + 3
(e) 32x² – 48xy + 18y² (f) x² – 1 + 1
4x²

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4. (a) What must be added to x² + 14x to make it a perfect square?
(b) What must be added to 9a² + 4b² to make it a perfect square?
Answer
1. (a) 4x (b) 36b2 (c) 4m2 (d) 8xy (e) 36b2 2. (a) a.(x+5)2 (b) (2x–y)2 (c) (3x+4y)2
(d) (7a-6b)2 (e) (8m–7n)2 (f) (x+1/x)2 3. (a) (a-7)2 (b) (5-6y)2 (c) (m-12)2
(d) 3(b-1)2 (e) 2(4x–3y)2 (f) (x–1/2x)2 4. (a) 49 (b) 12ab.

19.4 Factorisation of the trinomial of the form ax2 + bx + c.

Let's recall the multiplication of binomials, such as (x + 2) (x + 3)
Now, (x + 1) (x + 3)
= x² + x + 3x + 3
= x² + 4x + 3
= x² + (sum of the constants) x + product of the constant
= x2 + (1 + 3)x + 1 × 3
Hence, to factorise x2 + 5x + 6
We have to find two numbers whose sum is 5 and the product is 6.
i. e. x² + 5x + 6
= x² + (2 + 3) x + 6
= x² + 2x + 3x + 6
= x (x + 2) + 3(x + 2)
= (x + 2) (x + 3)

Steps:
• Arrange the terms in ascending order or descending order of power of

variable.
• Find the product (P) of the coefficient of first term and the last term of

given expression with their sign.
• Split the middle term of the given expression such that the sum (S) or

difference (D) of these two terms is equal to the middle term and their
product is equal to the product obtained in step 1.

• By forming the suitable groups, factorise the given trinomial.

258 Oasis School Mathematics-8

Worked Out Examples

Example: 1 Example: 2

Factorise: x² – 7x + 12 I have to find two Factorise: x² – x – 6 Look! I have found
Solution: numbers whose sum is Solution: 1 and 6,
= x² – (3 + 4) x + 12 = x² – (3 – 2) x – 6 3–2=1
7 and product is 12. = x² – 3x + 2x – 6 3×2=6
= x(x – 3) + 2(x – 3)
= x² – 3x – 4x + 12 = (x – 3) (x + 2)

= x(x – 3) – 4(x – 3)

= (x – 3) (x – 4)

Example: 3

Factorise: 3a² + a – 14 Example: 4

Solution: Product = 42 Factorise: 2(a+b)2 + 9(a + b) + 7
3a² + a – 14 Difference = 1
= 3a² + (7 – 6) a – 14 Numbers are 7 and 6 Solution:

Let (a + b) = x, then

= 3a² + 7a – 6a – 14 2(a + b)² +9(a + b)+7

= a(3a + 7) – 2(3a + 7) = 2x² + 9x + 7

= (3a + 7) (a – 2) = 2x² + (7+2)x +7

= 2x² + 7x + 2x + 7

= x(2x+7) + 1(2x+7)

= (2x + 7) (x + 1)

Now, substituting for x = (a + b) we get,

= {2(a+b)+7)} {(a+b)+1}

= (2a+2b+7) (a+b+1)

Exercise 19.4

1. Find the two numbers, whose (b) Sum = 5, product = 6
(a) Sum = 7, product = 12 (d) Difference = 5, product = 24
(c) Sum = 8, product = 7 (f) Difference = 2, product = 80
(e) Difference = 1, product = 72

Factorise (from 2 to 11): (b) x2 + 7x + 10 (c) x2 + 10x + 9
2. (a) x2 + 3x + 2 (e) x² + 17x + 72
(d) x² + 12x + 20

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3. (a) m² – 3m + 2 (b) a² – 7a + 12 (c) m2 – 8m + 15
(d) y2 – 11y + 30 (e) b² – 9b + 20

4. (a) a² + a – 2 (b) x² + 2x – 3

(c) a² + a – 20 (d) m2 + 7m – 18 (e) x2 + 3x – 18

5. (a) x2 – x – 12 (b) x2– 5x – 24

(c) y2– y – 72 (d) m² – 2m – 80 (e) b² – 5b – 150

6. (a) 2a² + 7a + 5 (b) 2x2 + 7x + 3

(c) 3a² + 5a + 2 (d) 4x² + 17x + 15

7. (a) 2x2 – 3x + 1 (b) 5a² – 19a + 12

(c) 3a² – 8ab + 5b² (d) 3x² – 36x + 33

8. (a) 2m2 + m – 3 (b) 6x2 +5x – 6

(c) 5a²+11a – 12 (d) 7x² + 4x – 11

9. (a) 3a² – 2ab – 8b² (b) 3x2 – 7xy –6y2

(c) 4x2 – 16x – 9 (d) 2b² – 5b – 12

10. (a) 1 – 3x – 28x2 (b) –7x2 + 3 – 4x

(c) 3x2 + 11x + 6 3 (e) (x + 1) (2x – 1) – 5

11. (a) (a – b)2 + 4(a – b) + 3 (b) (x + y)2 + 4(x + y) – 12

(c) 2(x + 2)² – 3 (x + 2) – 5 (d) (2x + y)2 + 4(2x + y) – 32

Answer

1. (a) 3 and 4 (b) 2 and 3 (c) 1 and 7 (d) 3 and 8 (e) 8 and 9 (f) 8 and 10
2. (a) (x+1) (x+2) (b) (x+5) (x+2) (c) (x+1) (x+9) (d) (x+2) (x+10) (e) (x+8) (x+9)
3. (a) (m–1) (m–2) (b) (a–3) (a–4) (c) (m–5) (m–3) (d) (y–5) (y–6) (e) (b–4) (b–5)
4. (a) (a+2) (a-1) (b) (x+3) (x-1) (c) (a+5) (a-4) (d) (m+9) (m-2) (e) (x+6) (x-3)
5. (a) (a) (x-4) (x+3) (b) (x-8) (x+3) (c) (y-9) (y+8) (d) (m-10) (m+8) (e) (b-15) (b+10)
6. (a) (2a+5) (a+1) (b) (2x+1) (x+3) (c) (a+1) (3a+2) (d) (x+3) (4x+5)
7. (a) (2x-1) (x-1) (b) (a-3) (5a-4) (c) (a-b) (3a-5b) (d) (x-11) (3x-3)
8. (a) (2m+3) (m-1) (b) (2x+3) (3x-2) (c) (a+3) (5a-4) (d) (x-1) (7x+11)
9. (a) (a-2b) (3a+4b) (b) (x-3y) (3x+2y) (c) (2x+1) (2x-9) (d) (b-4) (2b+3)
10. (a) (1+4x) (1-7x) (b) -(x+1) (7x-3) (c) (x+3 3 ) ( 3 x+2) (d) (x+2) (2x-3)
11. (a) (a-b+1) (a-b+3) (b) (x+y+6) (x+y-2) (c) (2x-3) (x+3) (d) (2x+y+8) (2x+y-4)

260 Oasis School Mathematics-8

Self Practice Materials

Complete the given tables:

(i) Sum Product Numbers (ii) Difference Product Numbers

7 12 3,4 2 15 5,3
9 14 3 28
11 28 12 45
21 108 9 70
12 27 7 44
10 21 11 180

S.N. Algebraic Middle-term Product Numbers Factors
(iii) Expression (Sum of the numbers) (x – 2) (x – 3)

i. x² – 5x + 6 –5 6 – 2, – 3

ii. x² + x – 6

iii. x² – 5x – 24

iv. x² – x – 6

v. y² – y – 12

vi. y² – 11y + 24

vii. x² – 7x + 12

viii. x² + 5x + 6

Additional Text

Factorisation of the trinomial in the form of a4 +a²b² +b4

We know that,

(a + b)² = a² + 2ab + b²

or, (a + b)² – 2ab = a² + b²

∴ a² + b²= (a + b)² – 2ab ……………….(i)

Again (a – b)² = a² – 2ab + b²

or, (a – b)2 + 2ab = a² + b²

∴ a² + b²= (a – b)2 + 2ab ……………….(ii)

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Example: Alternative method

Factorise: a4 + a²b² + b4
(i) a4 + a²b² + b4 = (a²)² + (b²)² + a²b²
Solution: = (a² + b²)² – 2a²b² + a²b²
= (a² + b²)² – a²b²
(i) Here, a4 + a²b² + b4 = (a² + b²)² – (ab)²
= (a²)² + 2 × a² × b² + (b²)² – a²b² = (a² + b² + ab) (a² + b² – ab)
= (a² + b²)² – (ab)² = (a2 + ab + b2) (a2-ab+b2)
= {(a² + b²) + ab}{(a² + b²) – ab}
= (a² + b² + ab) (a² + b² – ab)
= (a2 + ab + b2) (a2–ab + b2)

(ii) a4 – 6a² – 7 – 8x – x²
Solution:

Here, a4 – 6a² – 7 – 8x – x²
= a4 – 6a² + 9 –16 – 8x – x²
= (a4 – 6a² + 9) – (16 + 8x + x²)
= {(a²)² – 2.a².3 + (3)²} – {(4)² + 2.4.x + (x)²}
= (a² – 3)² – (4 + x)²
= {(a² – 3) + (4 + x)} {(a² – 3) – (4 + x)}
= (a² – 3 + 4 + x) (a² – 3 – 4 – x)
= (a² + 1 + x) (a² – 7 – x)
= (a2 + x+ 1) (a2–x–7)

Additional Exercise

Factorise each of the following:

1. x4 + x2y2 + y4 2. x4 + 4

3. x4 + 6x2 + 25 4. a4 + 7a²b² + 16b4

5. 9a4 – 24a²b² + 4b4 6. a² – 10a + 24 + 6b – 9b²

7. 16 – x2 + 6xy – 9y2

Answer

1. (x2–xy+y2) (x2+xy+y2) 2. (x2+2x+2) (x2–2x+2)

3. (x2–2x+5) (x2+2x+5) 4. (a2–ab+4b2) (a2+ab+4b2)

5. (3a2+2b2–6ab) (3a2 – 6ab+b2) (3a2+6ab+2b2) 6. (a+3b–6) (a–3b–4)

7. (4–x+3y) (4+x–3y)

262 Oasis School Mathematics-8

Unit

20 H.C.F. and L.C.M.

20.1 Highest Common Factor (H.C.F.)

Let's take any two algebraic terms a² and a³.
Possible factors of a² are a and a².
Possible factors of a³ are a, a² and a³.
Common factors of a² and a³ are 'a' and 'a2'
Among them, the highest common factor is a².
Hence, H.C.F. is the greatest number or expression which divides the given number

or expression exactly without remainder.

Note: H.C.F. is also written an G.C.F. or G.C.D.

Where G.C.F. = Greatest common factor

G.C.D. = Greatest common divisor

H.C.F. by factorisation method

To find the H.C.F. of an algebraic expression by factorisation method, apply the
following steps.

• Factorise the given algebraic expressions.

• Find the common factors from all expressions.

• Common factor or the product of all common factors.

Let's see an example.

Find the H.C.F. of x2 + 2x and x2 + x – 2.

Solution:

1st expression = x2 + 2x = x(x + 2)

2nd expression = x2 + x – 2

= x2 + 2x – x – 2 = x(x + 2) – 1(x + 2) = (x + 2) (x – 1)

∴ H.C.F. = (x+2)

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Worked Out Examples

Example: 1

15x³y7 and 18x7y³

Solution:

First expression = 15x³y7

= 3 × 5x³y7

Second expression = 18x7y³

= 2 × 3 × 3x7y³

∴ H.C.F. = 3x³y³

Example: 2

Find the H.C.F. of x2 – 4 and x2 – x – 6

Solution:

First expression = (x2 – 4)

= (x)2 – 22

= (x + 2) (x – 2)

Second expression = x2 – x – 6 = x2 – (3 – 2) x – 6 = x² – 3x + 2x – 6

= x(x – 3) + 2(x – 3) = (x – 3) (x + 2)

∴ H.C.F. = (x + 2)

Example: 3

Find the H.C.F. of (x + y)2, x2 – y2, 2x2 + xy – y2.

Solution:

First expression = (x + y)²

= (x + y) (x + y)

Second expression = x² – y² = (x + y) (x – y)

Third expression = 2x² + xy – y²

= 2x² + (2 – 1)xy – y²

= 2x² + 2xy – xy – y²

= 2x(x + y) –y (x + y)

= (x + y) (2x – y)

∴ H.C.F. = (x + y)

264 Oasis School Mathematics-8

Example: 4

Find the H.C.F. of a2 + 2a – 3, a2 – 3a + 2 and a3 – 1.

Solution:

First expression = a2 + 2a – 3

= a² + 3a – a – 3

= a(a + 3) – 1(a + 3)

= (a + 3) (a – 1)

Second expression = a² – 3a + 2

= a² – 2a – a + 2

= a(a – 2) – 1(a – 2)

= (a – 2) (a – 1)

Third expression = a³ – 1

= (a)³ – (1)³

= (a – 1) (a² + a + 1)

∴ H.C.F = (a – 1)

Exercise 20.1

Find the H.C.F. of the following by factorisation:

1. (a) a³b², a³b (b) x5y³, x³y4 (c) 8x²y, 4xy²

(d) 5x3y2z, 15x2yz3 (e) 16a³b4c, 24a²b³c² (f) 8a²b4c2, 12a³bc5 and 20a³b6

(g) 21m3n2p, 15m2n3p2 and 12m4n2p4

2. (a) x²(x + y) and xy (x + y) (b) (x – 1) (x + 2) and (x – 1) (x – 2)

(c) 3ab (a + b) (a – b), 6ab (a + b) (a + b) (d) a² + ab, a²c + abc

(e) x³y² +x²y³, x²y+xy²

3. (a) x² – y², (x – y)² (b) a³–b³, a²–b²

(c) a³ + b³, (a + b)³ (d) 2x² – 8, (x – 2)²

4. (a) (a² – b²) and (a² – 2ab + b²) (b) (x³ – y³) and (x² + xy + y²)

(c) (8x³+27y³) and (4x² – 9y²) (d) (a + b)³, a² + 2ab + b², a² – b²

(e) a³ – 8b³, a² – 4b², (a – 2b)²

5. (a) x² – 9 and x² – 5x + 6 (b) x² + x – 20 and x² – 7x + 12

(c) a² – a – 2 and a² + 2a + 1 (d) x² – 36, x² – 12x + 36
(e) x³ – 25x, x² – 6x + 5 (f) 2x² – 18, x² – 4x + 3, x² – 6x + 9

(g) x³ – xy², x² + 3xy + 2y², x² + 2xy + y² (h) 2(a² – 9), 4(a – 3)² and 6(a² – 6a + 9)

(i) x² – 3x + 2, x² – 5x + 6 and x² – 6x + 8


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Answer
1. (a) a3b (b) x3y3 (c) 4xy (d) 5x2yz (e) 8a2b3c (f) 4a2b (g) 3m2n2p
2. (a) x(x + y) (b) (x – 1) (c) 3ab(a + b) (d) a(a + b) (e) xy (x + y)
3. (a) (x–y) (b) (a–b) (c) (a+b) (d) (x–2)
4. (a) (a–b) (b) x2 + xy + y2 (c) (2x + 3y) (d) (a + b) (e) (a – 2b)
5. (a) (x–3) (b) (x – 4) (c) (a+1) (d) (x–6) (e) (x–5) (f) (x–3)
(g) (x+y) (h) 2(a–3) (i) (x–2)



20.2 Lowest Common Multiple (L.C.M.)

Let's consider any two algebraic terms a² and a³.
Multiplies of a² are a², a³, a4, a5 .............Multiplies of a³ are a³, a4, a5
Here, the common multiples are a³, a4, a5...................
Among these, lowest common multiple is a³.
∴ L.C.M. = a³.

L.C.M. by factorisation method

To find the L.C.M. of an algebraic expression, apply the following steps.
♦ Factorise the given algebraic expressions.
♦ Find the common factors.
♦ Multiply the common factors with all the remaining factors.
∴ The product is L.C.M.

Let's be clear with the help of the given example.

Example:

Find the L.C.M. of x² – 2x and x² – 3x + 2.

Here,

First expression = x² – 2x

= x(x – 2)

Second expression = x² –3x + 2

= x² – (2 + 1) x + 2

= x² – 2x – x + 2

= x(x – 2) – 1 (x – 2)

= (x – 2) (x – 1)

266 Oasis School Mathematics-8

Here, common factor = (x – 2)

Remaining factors are x and (x – 1)

∴ L.C.M. = (x – 2) × x (x – 1) = x (x – 1) (x – 2)

Note: L.C.M. of 3 expressions = common from all expressions

× common from two expressions

× remaining factors from all expressions.

Worked Out Examples

Example: 1

Find the L.C.M. of (a + b) (a – b) and (a + b)².

Solution:

First expression = (a + b) (a – b)

Second expression = (a + b)² = (a + b) (a + b)

∴ L.C.M. = Common factors × Remaining factors

= (a + b) × (a – b) × (a + b) = (a + b)² (a – b)

Example: 2

Find the L.C.M. of x² – 4, x³ + 8

Solution:

First expression = x² – 4

= (x)² – (2)²

= (x + 2) (x – 2)

Second expression = x² + 8

= (x)² + (2)³

= (x + 2) (x² – 2x + 4)

∴ L.C.M. = (x + 2) (x – 2) (x² – 2x + 4)

Common Remaining from Remaining from
first expression second expression

Example: 3

Find the L.C.M. of x³ + 1, x² + 3x + 2 and (x + 1)².

Solution:

Oasis School Mathematics-8 226677

Fist expression = x³ + 1

= (x)³ + 1³ L.C.M. = (x+1) (x2–x+1)
= (x + 1) (x² – x + 1) (x+2) (x+1)

Second expression = x² + 3x + 2 Common from Remaing Remaing
all expression in 2nd in 1st
expression expression
= x² + (2 + 1) x + 2

= x² + 2x + x + 2 Remaing
in 3rd

expression

= x(x + 2) + 1(x + 2)

= (x + 2) (x + 1)

Third expression = (x + 1)²

= (x + 1) (x + 1)

∴ L.C.M = (x + 1) (x² – x + 1) (x + 2) (x + 1)

= (x + 1)² (x + 2) (x² – x + 1)

Example: 4

Find the L.C.M. of :

x³ + 7x² + 12x, 2x³ + 13x² + 20x, x³ + 64

Solution:

First expression = x³ + 7x² + 12x

= x(x² + 7x + 12) Common in all = (x + 4)
= x [x² + (3 + 4) x + 12] Common in two = x
= x[x² + 3x + 4x + 12] Remaining = (x + 3) (2x + 5)
= x[x(x + 3) + 4 (x + 3)] and (x² –4x + 16)

= x(x + 3) (x + 4) ∴ L.C.M. =
(x + 4)x (x + 3) (2x+5)
Second expression = 2x³ + 13x² + 20x (x² – 4x + 16)

= x(2x² + 13x + 20]

= x(2x² + (8 + 5) x + 20]

= x[2x² + 8x + 5x + 20]

= x[2x (x + 4) + 5(x + 4)]

= x(x + 4) (2x + 5)

Third expression = x³ + 64

= x³ + 4³

= (x + 4) (x² – 4x + 16)

∴ L.C.M. = (x + 4). x (x + 3) (2x + 5) (x² – 4x + 16)

= x (x+3) (x+4) (2x+5) (x2–4x+16)

268 Oasis School Mathematics-8

Exercise 20.2



Find the L.C.M. of the following by factorisation:

1. (a) a²b³c and a4b5c4 (b) x2y3z and x3y2z4

(c) 16a²bc, 8a³b²c³ and 4ab4c² (d) 8a5b³c², 4a4b5c³ and 12a²b³c³

(e) 6x5y6z4, 3x7y²z and 9x10y4z²

2. (a) (x–y) (x–y) and (x+y) (x+y) (x–y) (b) (x + 3) (x + 4) and (x + 3) (x+1)

(c) (2x + 1) (2x – 1) and (x – 1) (2x – 1)

(d) (a + 2b) (a – b) (a – b) (2a – b) and (1 + b) (a – b)

(e) (x + 3) (x + 1), (x + 1) (x + 2) and (x + 2) (x + 3)

3. (b) (x – y)² and (x² – y²) (b) (a³ – b³) and (a – b)²

(c) x² + xy and xy + y² (d) x² – y² and x³ – y³

(e) a² – 4b², (a + 2b)² and a² + 2ab

4. (a) x² + 6x + 9 and x² + 5x + 6 (b) 2x² + 7x + 6 and x² + 7x + 10
(c) x² + 2xy and x³ – 4xy² (d) x² – 9 and x³ – 7x² + 12x
(e) a³ + 8, a³ + 9a² + 14a

5. (a) a² + 5a +6, a² + 9a + 20 and a² + 7a + 12
(b) 2a² – 8, a³ + 8 and a² – 2a – 8
(c) a4 – 8a, 2a² – 5a + 2, and 3a² – 8a + 4
(d) 2x² – 18, x² – 2x – 3 and x² + x – 12



Answer

1. (a) a4b5c4 (b) x3y3z4 (c) 16a3b4c3 (d) 24a5b5c3 (e) 18x10y6z4­
2. (a) (x+y)2 (x-y)2 (b) (x+1) (x+3) (x+4) (c) (2x-1) (2x+1) (x-1)
(d) (a+2b) (2a-b) (a-b)2 (1+b) (e) (x+3) (x+2) (x+1)
3. (a) (x-y)2 (x+y) (b) (a-b)2 (a2+ab+b2) (c) xy (x+y) (d) (x+y) (x-y) (x2+xy+y2)
(e) a(a+2b)2 (a-2b) 4. (a) (x+3)2 (x+2) (b) (x+2) (x+5) (2x+3)
(c) x(x+2y) (x-2y) (d) x(x+3) (x-3) (x-4) (e) a (a+2) (a+7) (a2–2a+4)
5. (a) (a+2) (a+3) (a+4) (a+5) (b) 2(a+2) (a-2) (a2–2a+4) (a–4)
(c) a(a-2) (2a-1) (3a-2) (a2+2a+4) (d) 2(x-3) (x+3) (x+1) (x+4)

Oasis School Mathematics-8 226699

Unit Rational
21 Expressions

21.1 Rational Expressions - Review

In a fraction, if the numerator and the denominator consist of algebraic expressions,
then the fraction is called the algebraic fraction (or rational algebraic expression).

For examples : 3x , 3x + 4y etc. are algebraic fractions or rational algebraic
4y 3x – 4y

expression.



Note: If the denominator of the rational expression is zero, it has no meaning.

i.e. in a if b = 0 it has no meaning.
b

Reduction of rational expression into its lowest term

To reduce the rational expression into its lowest term, apply the following steps.

Steps

• Resolve the numerator and denominator in their factors.
• Cancel the common factors.

Note: +x = x , –x = x , +x =– x , –x =– x
+y y –y y –y y y y


Worked Out Examples



Example: 1

For what value of x is the rational expression 2x – 1 undefined?
x – 2
Solution:

2x – 1 is undefined, if its denominator is 0.
x – 2

i.e. x – 2 = 0

or, x = 2.

Example: 2

Reduce –10x³y4 to its lowest terms.
20x²y5

Solution: –x
2y
Here, –10x³y4 = –10 × x3–2 = [∵ am = am–n]
20x²y5 20 × y5–4 an

270 Oasis School Mathematics-8

Example: 3

Reduce x² + x – 2 to its lowest terms.
x² – 4

Solution:

x² + x – 2
x² – 4

= x² + (2 – 1) x – 2
(x + 2) (x – 2)

= x² + 2x – x – 2
x² – 2²

= x(x + 2) – 1(x + 2)
(x + 2) (x – 2)

= (x + 2) (x – 1) = x–1
(x + 2) (x – 2) x–2

Exercise 21.1

1. For what value of x are the following rational expressions undefined ?

(a) 1 1 (b) x 2 5 (c) x+5 (d) x+4 (e) 5x+2
x– + x–4 3x – 2 5x – 2

2. Reduce the following algebraic fractions to the lowest term:

(a) a4b²c (b) 15p³q4 (c) – 6x14y9
a²bc³ – 10pq² 24x10y5

(d) 30x4y5z6 (e) 15a7b²c7
45x7yz9 5a9bc4

3. Reduce to the lowest terms:

(a) 2x + 4 (b) 2x – 2y (c) a² + ab
x² – 4 2(x – y)² a² – b2

(d) 2m² + 4mn (e) 2x² + 8x
m² – 4n² x² –16

4. Simplify:

(a) x² – 9 (b) a³ – 8 (c) –3–x
x² – 6x + 9 a² – 4 x² – x – 12

(d) x² + 5x + 6 (e) a² – 5a + 6 (f) a³ + b³
x² + 4x + 4 a² – 3a + 2 (a – b)² + 4ab

(g) x³ – y³ – 3xy (x – y) (h) 4 – (a + b)² (i) a² – 5a – 14
(a + 2)² – b² 2a³ – 13a² – 7a
x³ – y³



Oasis School Mathematics-8 227711

Answer

1. (a) 1 (b) -5 (c) 4 (d) 2/3 (e) 2/5

2. (a) a2b (b) -3p2q2 (c) -x4y4 (d) 2y4 (e) 3bc3
c2 2 4 3x3z3 a2

3. (a) 2 (b) 1 (c) a (d) 2m (e) 2x
x-2 x-y a-b m–2n x-4

4. (a) x+3 (b) a2+2a+4 (c) –(x1-4) (d) x+3 (e) a-3 (f) a2–ab+b2
x-3 a+2 x+2 a-1 a+b

(g) x2–2xy+y2 (h) 2–a–b (i) a+2
x2+xy+y2 a+2–b a(2a+1)



21.2 Multiplication and Division of Rational Expressions

Multiplication of Rational Expressions
While multiplying the rational expressions, apply the following steps.

Steps

• Multiply numerator with numerator and denominator with denominator.
• Reduce it into its lowest term.

For example : Multiply : 3a²b × 8ac²
4ab 6bc
Solution:

3a²b × 8ac²
4ab 6bc

= 3 × 8a²b × ac²
4 × 6 × ab × bc

= a³bc²
ab²c

= a²c
b

Division of Rational Expressions

Division of rational expression means multiplication of first rational expression
with the reciprocal of second.

For example,

Simplify : 4x² – 81y² ÷ 2x – 9y
1 – 4a² a – 2a²

Solution:

272 Oasis School Mathematics-8

4x² – 81y² ÷ 2x – 9y
1 – 4a² a – 2a²


= (2x)² – (9y)² × a – 2a2
(1)² – (2a)² 2x – 9y

= (2x + 9y) (2x – 9y) × a(1 – 2a)
(1 – 2a) (1 + 2a) (2x – 9y)

= a(2x + 9y)
(1 + 2a)

Worked Out Examples

Example: 1

Multiply: 3a²b4 by 8a²b² Alternate method
6a²b³ 20a²b

Solution: 3a2b4 × 8a2b2
6a²b³ 20a2b
Here, 3a²b4 by 8a²b² 2
6a²b³ 20a²b
= 3×8×a2×a2×b2×b2
= 3a2b4 × 8a2b² 3 6×4×5a2×a2×b3×b
6a²b³ × 20a²b
= b6 = b2
3 × 8 × a2 × a2 × b4 × b² 5b4 5
= 6 × 20 × a² × a² × b³ ×b

= a4 × b6
5 × a4 × b4

= 1 × a4–4 × b6–4
5

= 1 × b² = b2 [ ∵ a0 = 1]
5 5

Example: 2

Simplify : a² + 4a – 12 ÷ a² – 2a – 3
a² – 5a + 6 a² + 2a + 1

Solution:

a² + 4a – 12 × a² + 2a + 1
a² – 5a + 6 a² – 2a – 3

= a² + (6 – 2) a – 12 × a² + 2.a.1 + (1)²
a² – (3 + 2) a + b a² – (3 – 1) a – 3

= a² + 6a – 2a – 12 × (a + 1)²
a² – 3a – 2a + 6 a² – 3a + a – 3

= a(a + 6) –2(a + 6) × (a + 1)²
a(a – 3) – 2 (a – 3) a(a – 3) + 1 (a – 3)

Oasis School Mathematics-8 227733

= (a + 6) (a – 2) × (a + 1)
(a – 3) (a – 2) (a + 1) (a – 3)

= a+6 × a+1 = (a + 6) (a + 1)
a–3 a–3 (a – 3)²

= a2 + 7a + 6
a2 – 6a + 9

Example: 3

Simplify : y × (a + b)² ÷ ax + bx
a² – b² a–b (a – b)²

Solution:

y × (a + b)² × (a – b)2
a² – b² a–b ax+bx

= a² y b² × (a + b)² × (a – b)2
– a–b x(a+b)

y (a + b) (a + b) (a –b) (a – b) y
(a + b) (a – b) (a – b) x(a + b) x
= × × =

Exercise 21.2

1. Simplify

(a) a³b² × a³c (b) 8x5y³ × 15a³b
a²c² b 9a4b² 5x4y³

(c) x² × y² × z² (d) 3ab³ × abc × 4c4d4
y xz zx 4ac² c³d³ 3a²b

(e) 18x³y4 × 25x²yz × 3a6b4c
15a5b³ 6a²bc² 5x4y5

2. Simplify:

(a) xy ÷ x²y³ (b) 6ab² ÷ 18a³b
ab a²b³ 21b³c 63a²c²

(c) x²a² ÷ x²y³ (d) 8x5y² ÷ 6x4y²
b4y² a³b² 9a4b² 15a³b

(e) 7x²y³ × 12ab ÷ 3a²b² (f) x × 3z ÷ 9x
8a5b4 21xy 4x4y5 2y 4c 8ac

3. Simplify:

(a) a+b × a² – b² (b) x × x² – y²
(a – b)² (a + b)² x–y y

(c) a² + 5a + 6 × a² + a – 6 (d) x² + 5x + 6 × x² – 2x – 3
a² – 4 a³ – 9a x² – 1 x² – 9

(e) 4x² – 9y² × a²b + ab² (f) x² + 3x + 2 × x + 3 × x² – 4
a² – b² 4x – 6y x² + 4x + 3 (x – 2)² x+2

274 Oasis School Mathematics-8

(g) a² + ab + b² × a + b ÷ a³ – b³
ab (a – b) a² – b²

4. Simplify:

(a) x² + 3x + 2 ÷ x² – 4 (b) 4x² – 100 ÷ 4x – 20
x² + 4x + 3 x² – 9 x² + 2x – 15 3–x

(c) 2x² – 5x + 2 ÷ 4x – 2 (d) a² + 8a + 15 ÷ 3b + 2a + ab + 6
3x² – 5x – 2 3x² + x ab – a + 5b – 5 ab – a + 2b – 2

5. Simplify:
a³ – b³
(a) a³ + b³ ÷ a+b × (a + b)²
a–b (a – b)²

(b) a² – 8a – 9 × a² – 25 ÷ a² + 4a – 5
a² – 17a + 72 a² – 1 a² – 9a + 8

(c) x² – x – 20 × x² +3x+2 ÷ x+1
x² – 25 x² +6x+8 x² + 5x

Answer

1. (a) a4b (b) 8x (c) y (d) b3d (e) 3xz
c 3ab a ac

2. (a) ab2 (b) c (c) a5 (d) 20x (e) 2x5y7 (f) za
xy2 b2 b2y5 9ab 3a6b5 3y

3. (a) 1 (b) x(x+y) (c) aa(a+-33) (d) x+2 (e) ab(2x+3y) (f) x+2 (g) (a+b)2
a–b y x–1 2(a-b) x–2 ab(a-b)

4. (a) x-3 (b) -1 (c) x(x-2) (d) a+2 5. (a) a2+ab+b2 (b) a-5 (c) x
x-2 x-1 b+2 a2–ab+b2 a-1

21.3 Addition and Subtraction of Rational Expressions

(or Algebraic Fractions): (Review)

Like and Unlike Fractions: If the fractions have the same denominator, then they

are called like fractions and if the denominators of the fractions are not same, then

they are called unlike fractions.

For example : a and c are like fractions and d and e are unlike fractions.
b b e ƒ

To reduce fractions to their lowest common denominator.
Rule:
• If possible factorise the numerator and denominator.

• Cancel out the factors that are common to both numerator and denominator.

• Find the L.C.M. of the given denominators.
• Simplify the given terms.

Oasis School Mathematics-8 227755

For example: Express with the lowest common denominator of xaz, b and c .
xy yz

Solution:

Here, L.C.M. of xz, xy and yz = xyz.

Now, a = a × y = ay
xz xz y xyz

b = b × z = bz and
xy xy z xyz

c = c × x = cx
yz yz x xyz


When the fractions have same denominator.

Working steps

(i) Add or subtract the numerators.

(ii) Write the result of numerators over the common denominator of the like
fractions.

(iii) Reduce the resulting fraction to the lowest term.


Worked Out Examples

Example: 1

3a + 2b + 5c
x x x

Solution:

3a + 2b + 5c
x x x

= 3a + 2b + 5c
x

Example: 2

x+3 – x–3
x–3 x+3

Solution:

x+3 – x–3
x–3 x+3

= (x + 3)² – (x – 3)²
(x – 3) (x + 3)

= (x² + 6x + 9) – (x² – 6x+9)
x² – 9

= x² + 6x + 9 – x² + 6x – 9 = 12x
x² – 9 x² – 9

276 Oasis School Mathematics-8

Example: 3

a a b – b – 2ab
– a+b a² – b²

Solution:

a – b – 2ab
a–b a+b a² – b²

= a – b – 2ab
a–b a+b a² – b²

= a(a(a+–bb))–(ab(+a – b) – 2ab
b) a² – b²

= a² + ab – ab + b² – 2ab
a² – b² a² – b²

= aa²² –+ bb²² – 2ab
a² – b²

= a² +a²b–² – 2ab
b²

= (a +(ab)–(ba)²– b)

= (a – b) (a – b)
(a +b) (a – b)

= aa –+ bb

Example: 4

x+y – x–y – 4xy
x–y x+y x² + y²

Solution:

x+y – x–y – 4xy
x–y x+y x² + y²

= (x(x+ –y)y²)–(x(x+–yy))² – 4xy
x² + y²

= (x² + 2xy + y²) – (x² –2xy + y²) – 4xy
x² – y² x² + y²

= x² + 2xy + y² – x²+2xy – y² – 4xy
x² – y² x² + y²

= x²4x–yy² – 4xy
x² + y²

= 4xy(x(x² ²+–yy²)²)–(4xx²y+ (x² – y²)
y²)

= 4x3y + 4xy3 – 4x3y + 4xy³
(x²)² – (y²)²

= 8xy3
x4 – y4

Oasis School Mathematics-8 227777

Example: 5

a–1 + a–2 + a–5
a² – 3a + 2 a² – 5a + 6 a² – 8a + 15

Solution:

a–1 + a–2 + a–5
a² – 3a + 2 a² – 5a + 6 a² – 8a + 15

= a² – a–1 a + 2 + a–2 + a–5
(2 + 1) a² – (3 + 2) a + 6 a² – (5 + 3) a + 15

= a² – 2aa––1a + 2 + a–2 + a–5
a² – 3a – 2 a + 6 a² – 5a – 3a + 15

= a – 1 + a(a – a – 2 – 3) + a–5
a(a – 2) – 1(a – 2) 3) – 2(a a(a – 5) – 3(a – 5)

= (a – a1)–(a1 – 2) + a–2 + a–5
(a – 2) (a – 3) (a – 5) (a – 3)

= a 1– 2 + 1 + 1
a–3 a–3

= 1(a – 3) + 1 (a – 2) + 1(a – 2)
(a – 2) (a – 3)

= a –3 + a–2 + a– 2
(a – 2) (a – 3)

= (a 3a –7 3)
– 2) (a –



Exercise 21.3

Simplify the following:

1. (a) 2x + 3x (b) 3a + a (c) a+b + a–b
5 5 b b 2ab 2ab

(d) 5x – 5 (e) x – 3y – 3x – y (f) a² + 4a + 4
x–1 x–1 a+b a+b a+2 a+2

(g) a² – 16 (h) x+4 + x+8 (i) 3a² + 15a + 12
a–4 a–4 x² – 36 x² – 36 a+4 a+4

(j) x + 2x + 3 + 3
x² + 5x + 6 x² + 5x + 6 x² + 5x + 6

(k) 2x – 3x + 4y + 2x + 5y
x–y x–y x–y

2. (a) 1 + 1 (b) 1 + 1 y (c) x+y – x–y
x–2 x+2 2x+y 2x – x–y x+y

(d) a – a² (e) 1 + 2b (f) 1 + 1
a–5 a² – 25 a–b b² – a² x² – 1 1–x

(g) 1 – 1 (h) y(xx– y) – y y) (i) a²b – ab²
(a – b) (c – a) (a – c) (b – c) x(x – a–b a+b

278 Oasis School Mathematics-8

3. (a) 1 + 1 + 4 4 (b) 1 + 1 + 2y
a+2 a–2 – a² x+y x–y y² – x²

(c) 1 + 1 + a 9 (d) 1 + 1 + 1 2
a–3 2(a + 3) a² – 1–x 1+x – x²

(e) 2 + 2x – x² + 3 (f) x+y – x–y + 2xy
x+1 x–1 x² – 1 x–y x+y x² – y²

(g) x–y – x – x2
x x+y y(x+y)

4. (a) 1 + 3 + 2
(a – 3) (a – 2) (a – 2) (4 – a) (a – 3) (a – 4)

(b) 1 + 1 + 1
(1 –x) (x – 2) (x – 2) (x – 3) (x – 3) (x – 1)

(c) x + y + z
(x – y) (x – z) (y – z) (y – x) (z – x) (z – y)

5. (a) 1 + 1 + 1
a² + 8a + 15 a² + 4a + 3 a² + 6a + 5

(b) a–2 + a–4 + a–3
a² – 5a + 6 a² – 6a + 8 a² – 6a + 9

(c) x–1 + x–2 + x–5
x² – 3x + 2 x² – 5x + 6 x² – 8x + 15



Answer

1. (a) x (b) 4a (c) 1 (d) 5 (e) -2(x + y) (f) (a+2) (g) (a+4)
b b a+b

(h) 2 (i) 3(a +1) (j) 3 (k) x+y
x-6 x+3 x-y

2. (a) 2x (b) 4x (c) 4xy (d) 5a (e) 1 (f) x
x2 - 4 4x2 - y2 x2 - y2 a2 - 25 a+b 1- x2

(g) (a - -1 (h) x+y (i) ab(a2 +b2 )
b)(b - c) xy a2 - b2

3. (a) 2 (b) 2 (c) 5a +3 (d) 4 (e) x+5 (f) 6xy (g) - (x 2 - xy + y 2 )
a+2 (x + y) 2(a2 - 9) 1- x2 x+1 x2 - y2 xy

4. (a) 1 - 4) (b) x - 3) (c) 0
(a - 2)(a - 3)(a (x -1)(x - 2)(x

5. (a) 3 (b) 3a - 7 (c) 3x - 7
(a +1)(a + 5) (a - 2)(a - 3) (x - 2)(x - 3)

Oasis School Mathematics-8 227799

Unit

22 Indices

22.1 Indices (Review)

Let us consider an algebraic term axb. Here, 'a' is coefficient, 'x' is the base and 'b' is
the index (or power) of the base. An index of a base means how many times the base
is multiplied. The plural form of index is called indices.

Laws of Indices
If m and n are two positive integers, then

(i) am × an = am+n [Product law]
(ii) am ÷ an = am–n
(iv) (am)n = amn [Quotient law]

[Power law]
Note:

(ab)m = ambm ( ) ( )am –2 = bn 2
bn am
( ) a m = am
b bm


(v) aº = 1 [Law of zero index]
[Root law of index]
(vi) n am m

= a n

Worked Out Examples

Example: 1

Find the products in their exponential form. 2 × 2 × 2
(i) 3 × 3 × 3 ×3 (ii) (–2) × (–2) × (–2) × (–2) × (–2) (iii) 5 5 5
Solution:

(i) 3 × 3 × 3 × 3 = 34

(ii) (–2) × (–2) × (–2) × (–2) × (–2) = (–2)5

( ) 2 × 2 2 = 23
(iii) 5 5 × 5 5

280 Oasis School Mathematics-8

Example: 2

Find the value of the following:

(i) 32 × 53 3 ( )(iii) 8 1
3
(ii) 36 2
125

Solution:

(i) Here, 32 × 53 = 3 × 3 × 5 × 5 × 5 = 9 × 125 = 1125

3 = 6 × 6 × 6 = 216

(ii) Here, 36 2 = 62 × 3/2 = 63

( ) ( )(iii) 1 1 1 [ ( ) ]∵
8 23 23 × 3 = 52 a n an
Here, 125 3 = 53 3 = b bn
53 × 1 =
3

Example: 3

Simplify by using laws of indices. The answer should contain only positive indices.

(i) 24 ÷ 26 (ii) a6 × a4 (iii) ax–y × ay–z × az–x
a²

Solution:

(i) Here, 24 ÷ 26 = 24 = 24–6 = 2–2 [∵ am = am–n]
26 an

= 1 = 1 [∵ a–m = 1 ]
2² 4 am

(ii) Here, a6 × a4 = a6+4 [∵ am × an = am+n]
a² a²

= a10 = a10–2 = a8. [∵ am = am–n]
a² an

(iii) Here, ax–y × ay–z × az–x = a x–y+y–z+z–x [∵ am × an = am+n]

= ax–x–y+y–z+z = a0 = 1

Example: 4

( ) ( )Evaluate:5×  8 1 ÷ 9 −1 
9  27 3 4 2 



( ) ( )Solution:5  8 1 9 −1  ( ) ( ) ( )=5 2 3× 1 3 2x −1 
9  27 3 4 2  9  3 3 2 2 
×  ÷ ×  ÷ 


( )= 5 ×  2 ÷ 3 −1 
9  3 2 
 

( )= 5 ×  2 ÷ 2 1
9  3 
 3 

= 5 ×  2 × 3  = 5
9  3 2  9

Oasis School Mathematics-8 228811

Exercise 22.1

1. Express the product in their exponent form.

(a) 6 × 6 × 6 × 6 (b) (–b) × (–b) × (–b) × (–b) × (–b)

( ) ( ) ( ) (c) 3× 3 × 3 (d) x × x × x
5 5 5 4 4 4

2. Express the following exponent form into the product form.

(a) (2x)3 (– )(b) 2 2 ( )(c) x 3
3 a


3. (a) What power of x yields unit value?

(b) What index of m will yield 1 ?
m²
1
(c) What is the power of 6 so that its value becomes 36 ?

(d) If ax = 1, then what is the value of x?

(e) If x = 2, what is the value of 2x?

(f) If x = 2, what is the value of 4 ?
2x

4. Simplify by using laws of indices. Answer should contain only positive indices.

(a) 34 × 35 (b) a6 × a–5 (c) 45 ÷ 43

(d) c6 ÷ c3 (e) (43)4× (44)2 (f) (54)3 × (52)3

( ) ( ) p 5 p 2 ( ) ( )(h) 32 3 4 (– ) (– ) (– )(i) 13 1 4 15
(g) q q 5 5 3 3 3
× × × ×

(j) a² × a3 (k) (4p)3 × (4p)² (l) 4² × 9² × 49
a4 (4p)4 2³ × 7 × 3²

(m) 8p2× 5p3 (n) 4r5 × 4r4 (o) 9x2y2 × 7x3y4

5. Simplify: (b) ax²–y²× ay²–z² × az²–x² (c) (–4x²y)³ × (3x–3y)²
(a) bx-y × by-z× bz-x

(d) (–2ab²c–1)³× (3a²b³c)–1 (e) (2x²y³z-1)²× (3x²y³z)–3

( )
(f) (3x²y³z–1) × (4x–2y3z2)–2 (g) a³b–5 –5
a²b–3

6. (a) If x = 1, y = 2 and z = 3, find the value of each of the following:

(i) x–2 (ii) 27xºy6z–3 (iii) (x5y)–4 (iv) xyyzzx

(b) If a = 2, b = 0, c = 1, d = –1, find the value of

(i) a–1.b–2 (ii) b³c3d2 (iii) ca. ab.db (iv) ab.cd.da

282 Oasis School Mathematics-8

7. Evaluate:

(a) 82/3 (b) 323/5 (c) 272/3 (d) (27)–2/3

(e) (–125)–2/3 ( )(f) 243 2/5 ( )(g) 1 – 1 ( )(h) 5 –5
32 216 3 4


(i) (625)–1/4 ( )(j) 1 0.25 (k) (40.5)2 ( ) ( )(l) 91 32 3
16 16 243
2× 5

(m) (20)³ × (30)² (n) 3 26 × 39 × 512
(40)² × (10)³ 23 × 33 × 125

8. Simplify:

(a) 3x+1 – 3x (b) 2x+1 – 2x (c) 5x+1 – 5x (d) ax+2 – ax+1
2 × 3x 3 × 2x 4 × 5x ax+1 –ax

9. Simplify:

(a) 128x6y–3 (b) (8x6y–3)1/3 (c) (4xy)² (d) 81x6y–5z²
64x–3y³ 2xy (2x–3y–3)³ (3x²y³z)³

10. Simplify:

(a) ax+y × ay–z (b) x3a+3b × x3a+b
ax+2y a2x+z x2a–2b x2a+b

(c) x × xa(b + c) –b(c+a) × x-c(a + b) (d) (ya+b)a–b × (yb–c)b+c × (yc+a)­ c-a

( ) ( ) ( )
(e) za a²+ab+b² zb b²+bc+c² zc c²+ca+a² (f) xa²–b² × xb²–2a²
zb zc za x2b²–a² xb²
. .

(g) (xa+b)a–b × (xb+c)b–c
(xa+c)a–c

Answer

1. (a) 64 (b) (-b)5 (c) (3/5)3 (d) (x/4)3 2. (a) 2x×2x×2x (b) (-2/3)×(-2/3)×(-2/3)
(c) x/a×x/a×x/a 3. (a) 0 (b) (-2) (c) -2 (d) 0 (e) 1 (f) 1
4. (a) 39 (b) a (c) 42 (d) c3 (e) 420 (f) 518 (g) (p/q)7 (h) (3/5)6 (i) (-1/3)12­
(j) a (k) 4p (l) 126 (m) 40p5 (n) 16r9 (o) 63x5y6
5. (a) 1 (b) 1 (c) -192y5 (d) -8/3 ab3/c4 (e) 4/27x2y3z5 (f) 3x6/16y3z5 (g) b10/a5
6. (a) (i) 1 (ii) 64 (iii) 1/16 (iv) 24 (b) (i) 0 (ii) 0 (iii) 1 (iv) 1

7. (a) 4 (b) 8 (c) 9 (d) 1 (e) 1 (f) 9 (g) 6
9 25 4

(h) 1024 (i) 1 ( j) 1 (k) 4 (l) 2 (m) 9 (n) 750
3125 5 2 9 2

8. (a) 1 (b) 1/ 3 (c) 1 (d) a

9. (a) 2x9 (b) x (c) 2x11y11 (d) 3
y6 y2 y14 z

1 1 1
(a) a 2(1x+z) (b) x2a+5b x 3 b2
10. (c) x 2 bc (d) 1 (e) 1 (f) (g) 1
a2(x+3)

Oasis School Mathematics-8 228833

Unit Equation, Inequality

23 and Graph

23.1 Linear Equations in One Variable

Let us consider some statements such as
• 5 added to a number x gives 12
i.e. 5 + x = 12.
• 10 subtracted from a number x results 8
i.e. x – 10 = 8
• Twice the number 'x' added to 5 gives 15

i.e. 2x + 5 = 15.

All these are statements where two expressions are equal for a particular value of a
variable. Hence, all three above expressions are equations.

A statement where two expressions are equal for a particular value of a variable is
called an equation.

Rules for solving

Linear equations of one variable

• We can add the same number on either side of the equation.

i.e. x–3=4

⇒ x – 3 + 3 = 4 + 3

• We can subtract the same number from either side of the equation.

i.e. x + 5 = 12

⇒ x + 5 – 5 = 12 – 5

• We can multiply both sides of the equation by the constant number.

i.e. x = 6
3
x
⇒ 3 ×3=6×3

• We can divide both sides of the equation by the constant number.

i.e. 5x = 10

⇒ 5x = 10
5 5

284 Oasis School Mathematics-8

Note: Any term of the equation can be taken to the other side by changing its
sign. This process is called transposition.

e.g. 2x + 5 = 8–x
⇒ 2x + x = 8–5

Worked Out Examples

Example 1:

Solve : 2x–3 = 5

Solution:

or 2x–3 = 5

or, 2x = 5+3
or,
∴ x = 8
2

x = 4.

Example: 2

Solve: 2x + 1 = 11
3x + 5 20

Solution:

2x + 1 = 11
3x + 5 20

or, 20(2x +1) = 11 (3x +5) (Cross multiplying)

or, 40x +20 = 33x + 55

or, 40x – 33x = 55 – 20

or, 7x = 35
or,
∴ x = 35
7
x = 5

Example 3:

Solve : 9x – 7 = 3x – 4
3x+5 x+6

Solution:

9x – 7 = 3x – 4
3x+5 x+6
or, 44x = 22
or, (9x-7) (x+6) = (3x–4) (3x+5) or,
or, 22
or, 9x2+54x–7x–42 = 9x2+15x–12x–20 x = 44

or, 9x2+47x–9x2–3x = –20+42 x = 12

Oasis School Mathematics-8 228855

Exercise 23.1

Solve the following:

1. (a) x – 2 = 3 (b) y – 2 = 10 (c) z – 1 = 3

2. (a) x + 1 = 6 (b) y + 2 = 4 (c) z + 6 = 2

3. (a) 2x = 4 (b) 3x = -18 (c) 5x = 20

4. (a) x = 3 (b) x = -2 (c) x =3
2 4 4

5. (a) 2x–5 = 15 (b) 3x–6=24 (c) 4x–2 = 18

6. (a) 2x+6 = 26 (b) 3x+5 = 20 (c) 4x + 5 = 25

7. (a) x = 7 (b) 3x = 15 (c) 2x = 8
2 2 4 3 3

8. (a) 4x–3 = 2x + 5 (b) 5x–3 = 4x+6 (c) 4x – 15 = 25 – 6x

9. (a) x–2 = x+6 (b) x–8 = x–12 (c) 2x – 8 = 8
2 4 2 3 x+5 5

(d) 2x+1 = 3 (e) 2x+5 = 3
3x-1 2 3x+5 4

10. (a) x+2 – x–3 = 1 (b) 4x – 3 + 2x + 1 =1
4 3 2 2 6

(c) 5x-4 – x–3 = x+6 (d) 3x-1 – x – 3 = 3 (e) x+12 – x = 13
8 5 4 5 4 6 12 2

11. (a) x + x = 66 (b) x–3 – x = 2x–5
10 3

(c) 2(4x-1) – 5x-1 = x+4 (d) 2(3x+1)–3(x+3) = 1
3 3 3 (x+3)+(3x+1) 2

12. (a) (x+2) (x+3)–(x-1)(x+4) = 14

(b) (x+1) (x+2) – (x-1) (x+3) = 7

(c) (2x+1) (2x+3) – (2x–1) (2x–3) = 16



Answer
1. (a) 5 (b) 12 (c) 4 2. (a) 5 (b) 2 (c) -4 3. (a) 2 (b) -6 (c) 4
4. (a) 6 (b) -8 (c) 12 5. (a) 10 (b) 10 (c) 5 6. (a) 10 (b) 5 (c) 5
7. (a) 7 (b) 20 (c) 4 8. (a) 4 (b) 9 (c) 4 9. (a) 10 (b) 0 (c) 40 (d) 1 (e) 5
10. (a) 12 (b) 1 (c) 8 (d) 7 (e) 54 11. (a) 60 (b) 3/2 (c) 5/2 (d) 9
12. (a) 2 (b) 2 (c) 1

286 Oasis School Mathematics-8

23.2 Applications of Linear Equation

We can convert the verbal problems into the linear equation of one variable. While
solving such problems, apply the following steps.

Steps

• Read the given problem carefully to find out what is given and what is to
find.

• Denote the unknown quantity by the letter x or by some other letter such as
a, b, y, z, etc.

• Make the equation using unknown quantity.
• Solve the equation to find the value of x.
• Verify the solution, if necessary.

Worked Out Examples

Example: 1

A number increased by 5 is 12. Find the number.

Solution: Let the number be x;

From the given condition,

x + 5 = 12

or, x = 12–5

or, x = 7

∴ Required number = 7.

Example: 2

When a number is multiplied by 4 and decreased by 7, the result is 65. Find the number.
Solution: Let the number be x

From the given condition, 4x – 7 = 65


or, 4x = 65 + 7

or, 4x = 72

or, x = 72
or, 4

x = 18.

∴ Required number = 18.

Oasis School Mathematics-8 228877

Example: 3

If 12 is added to three times of a number, the result is 42, find the number.

Solution:

Let the required number be x.

From the given condition,

12 + 3x = 42

or, 3x = 42–12

or, 3x = 30

or, x = 30
3
∴ Required number = 10.

Example: 4

A is now 12 years old and B is 2 years old. In how many years, will A be three time as
old as B?

Solution: Let's suppose after x years, A will be three times as old as B.

After x years,

A's age = (12+x) years, B 's age = (2 +x ) years

From the given condition, ( 12 + x) = 3(2 + x)

or, 12 + x = 6 + 3x

or, 3x–x =12 – 6

or, 2x = 6
or,
x = 6
2

x = 3

∴ After 3 years, A will be three times as old as B.

Example: 5

Length of a rectangle is 5m more than its breadth. If its perimeter is 50m, find its length
and breadth.

Solution:

Let, the breadth of the rectangle = x, then its length = x + 5

From the given condition,

Perimeter = 2(length + breadth)

or, 50 = 2 (x + 5 + x)

or, 50 = 2(2x + 5)

or, 50 = 4x + 10

288 Oasis School Mathematics-8

or, 50 – 10 = 4x

or, 4x = 40
or,
x = 40 m
4

x = 10m

∴ Length = x + 5 = (10 + 5)m = 15m, breadth = x = 10m.

Exercise 23.2

1. (a) If 5 is added to a number, the sum is 15. Find the number.

(b) A number increased by 20 is 45. Find the number.

(c) If a number is decreased by 10, the result is 25. Find the number.

(d) By how much 35 should be decreased to make 21?

(e) Twice the number decreased by 3 is 17. Find the number.

(f) If a number is multiplied by 4 and increased by 12, the result is 48. Find the
number.

(g) The sum of three consecutive natural numbers is 33. Find the numbers.

(h) Find three consecutive natural numbers whose sum is 63.

(i) Find three consecutive even numbers whose sum is 48.

(j) Find three consecutive odd numbers whose sum is 27.

2. (a) In 16 years, Anil's age will be twice his age 12 years ago. What is his present age ?

(b) A father's age is three times that of his son and in 10 years hence it will be twice
as greater as his son's. Find their ages.

(c) Ages of two brothers are in the ratio 2:3. 5 years hence the ratio of their ages will
be 5:7. Find their present ages.

(d) Ages of a son and his father are in the ratio 1:3. 10 years ago, ratio of their ages
was 1:5, find their present ages.

3. Find the value of x from the given figures.

(a) A (b) A D (c) 3x+2

(3x+1)cm (6x+1)cm (4x+10)cm x+4

B (5x+1)cm C BC Perimeter = 36cm

(10x-2)cm
Perimeter = 31cm



4. (a) The length of a rectangular field is 10 m longer than its breadth. If its perimeter

is 40 m, find its length and breadth.

(b) Two equal sides of a triangle are 3 cm more than three times the third side. Find
the sides of the triangle if its perimeter is 21 cm.



Oasis School Mathematics-8 228899

Answer

1. (a) 10 (b) 25 (c) 35 (d) 14 (e) 10 (f) 9 (g) 10, 11, 12 (h) 20, 21, 22 (i) 14,16,18
(j) 7, 9, 11 2. (a) 40 yrs (b) 10 yrs, 30 yrs. (c) 20 yrs, 30 yrs, (d) 20 yrs, 60 yrs.
3. (a) 2 (b) 2 (c) 3 4. (a) 15m, 5m (b) 8cm, 8cm, 5cm.

23.3 Graph of the Linear Equation

The equation of the form ax + by + c = 0 is called a linear equation. To draw the graph
of a linear equation, let's take an equation.

2x – y + 1 = 0 which is in the form of ax + by + c = 0.

It can be written as,

– y = –2x–1

or, y = 2x +1

Let's fill the given table with suitable variables.

x 0 1 2 When x = 0, y = 2 × 0 + 1 =1
y 1 3 5 When, x = 1, y = 2 × 1 + 1 =3
When, x = 2, y = 2 × 2+ 1 =5

Plot the points (0, 1), (1, 3) and (2, 5) on the graph and join them.

Y



(2, 5)

y=2x+1 (1, 3)

(0, 1) X
X' O

Y'

Example: Draw the graph of the line 3x +2y = 6, Y

Solution: O
Y'
Given equation of a straight line

3x +2y = 6 3x+2y=6 (2,0) (4,-3)
(0,3)
or, 2y = 6 – 3x X
or,
6 –3x X'
2
y =

x 0 2 4

y 3 0 -3

290 Oasis School Mathematics-8

23.4 Slope of a Straight Line R

The slope is usually measured as a ratio of vertical rise to the
horizontal distance.

In the given figure, slope of line PR is P θQ
Vertical rise (QR) Y
Q(10, 11)
Slope (m) = Horizontal distance (PQ)
In the graph, a straight line PQ

is passing through the points P(6,6) R(10,6)
P(6, 6) and Q(10, 11).

Now, draw a line (PR) through P parallel to x–axis

and the line (QR) through Q parallel to y–axis which X

intersect at R(10, 6). X' O

Here, PR = 10–6=4 Y'

QR = 11-6 = 5

∴ From P to Q, horizontal displacement = change in x-coordinates = 4

Vertical displacement = Change in y-coordinates = 5

change in y-coordinate = 5 Y B(x2, y2)
∴ Slope of a line (AB) = m = change in x-coordinate 4

Hence, the slope of straight line passing through X' A(x1, y1) X
A(x1, y1) and B(x2, y2), is

Slope of AB = m = y2–y1 Y'
x2–x1

Standard Equation of Straight Line

A linear equation is in the form of ax+by+c = 0

It can be written as by = –ax–c

or, y = -ax–c
b
ba) x + (– c
or, y = (– a )

W here, (– ba) represents the slope of straight line and (– ac) Remember!
y = mx + c.
represents the Y– intercept.

Hence this equation can be written as y = mx + c. slope Y–intercept

Oasis School Mathematics-8 229911

Note : The slope of a straight line may be positive, negative, zero or undefined.

YY

Positive slope Negative slope

X' O X X' O X

Y' Y'
Y Y

No slope Slope is undefined

X' O X X' O

Y' Y'

Worked Out Examples

Example: 1

Find the slope of a line joining the points (2, 1) and (1, 6).

Solution:

Here, (2, 1) = (x1, y1) and (1, 6) = (x2, y2).
Now, the required slope (m)
= y2–y1
x2–x1

= 6–1 = 5 = –5
1–2 -1

Example: 2

Find the value of k if the slope of the line passing through (3, k) and (–2, 4) is – 3 .
5

Solution:

Here, (3, k) = (x1, y1) and (–2, 4) = (x2, y2) and slope (m) = – 3 .
5

We have, slope (m) = y2–y1
x2–x1

292 Oasis School Mathematics-8

or, -3 = 4–k
or, 5 -2–3
or, 15 = 20 – 5k
-3 = 4–k or, –3 × –5 = 5(4 – k)
5 –5

or, 5k = 20 – 15 ∴ k= 5 =1
5

Example: 3

Find the slope and y–intercept of the line whose equation is 3x + 2y = 6.

Solution:

Here, 3x + 2y = 6

or, 2y = –3x + 6

or, y = -3x + 6
2 2

or, y = -3 x + 3
2

Now, comparing with y = mx + c, we get,

Slope (m) = -3 and y–intercept (c) = 3.
2

Example: 4

Draw a straight line passing through the points A(3, 2) and B(5, 4).

(i) Find the difference of y–coordinate and x–coordinate for the line AB.

(ii) Find the slope of AB. Y
Solution:
Now, plotting the points A(3, 2) and B(5, 4) 4 B(5, 4)
on the graph and joining them, line AB is obtained. 3 45 6

(i) Here, (3, 2) = (x1, y1) and (5, 4) = (x2, y2) X' A(3,2) X
2
1

O1 2 3

difference of y–coordinates = y2 – y1 = 4 – 2 = 2 Y'

difference of x–coordinates = x2 – x1 = 5 – 3 = 2

(ii) Slope of AB = y2–y1 = 2 = 1
x2–x1 2

Example: 5

Draw the graph of the equation y = 3x – 2 and find its slope taking any two points.
Solution:
Here, y = 3x – 2

Oasis School Mathematics-8 229933

Table of values: 1 2 Y
1 4
x0 7
y -2 6
5
By plotting (0, –2), (1, 1), and (2, 4) 4 (2, 4)
a straight line PQ is obtained. 3
2
Taking any two points (1, 1) and (2, 4) 1 (1,1)

(1, 1) = (x1, y1) X' O 123 456 X
and
∴ (2, 4) = (x2, y2). (0,-2)

Slope of PQ = y2–y1 Y'
x2–x1

= 4–1 = 3 = 3
2–1 1

Exercise 23.3

1. Find the slope of straight lines passing through the following pairs of points.

(a) A(2, 3), B(6, 5) (b) P(–5, 2), Q(3, –5) (c) M(2, 5), N(–6, 9)

(d) R(4, 7), S(6, 10) (e) C(–1, –4), D(5, 6) (f) K(2, 1), L(5, 7).

2. (a) Find the value of k, if the slope of the line passing through (5, k) and (9, 8) is 1 .
2

(b) Find the value of a, if the slope of the line passing through (a, 7) and (6, 10) is -23.

3. Find the slope and y–intercept of the lines whose equations are:

(a) y = 2x+9 (b) y = 2 x + 5 (c) y = –2x + 7
3 3

(d) 2x + 3y = 5 (e) y + 2x = 3

4. A straight line is passing through the points P(1, 2) and Q(2, 4).

(i) Draw the graph of the straight line.

(ii) Find the difference of x-coordinates and y-coordinates for the line PQ.

(iii) Find the slope of PQ.

5. Complete the following table.

(a) x 1 2 3 (b) x 0 1 2 (c) x 1 4 -2

y = 2x – 1 y = 3x – 1 y = 2x – 1
3

6. Draw the graphs of the following equations.

(a) y = 3x + 4 (b) 2y – 3x = –5 (c) x + 2y = 12

(d) 3x + 4y = 12 (e) 2x + 6y = 3

294 Oasis School Mathematics-8