Oasis Math 9

∴ x = – 1
11

(d) 4x+1 + 4x = 80

4x × 41+ 4x = 80

or, 4x (4 + 1) = 80

or, 4x × 5 = 80

or, 4x = 80
or, 5
or, 4x = 16

4x = 42

∴ x = 2

Example: 2

Solve: 4x + 1 =4 1
4x 1 4
4x 1
Solution: 4x + = 4 4

Let, 4x = a

Then, a+ 1 = 17
or, a 4

a2 + 1 = 147
a

or, 4a2 + 4 = 17a

or, 4a2 – 17a + 4 = 0

or, 4a2 – (16 + 1) a + 4 = 0

or, 4a2 – 16a – a + 4 = 0

or, 4a(a – 4) – 1(a – 4) = 0

or, (a– 4) (4a – 1) = 0

Either, a – 4 = 0 ......(i)

or, 4a – 1 = 0 ..... (ii)
From (i), a = 4
or, 4x = 41

or, x = 1

From (ii), 4a – 1 = 0

or, a = 1
4
or, 4x = 4–1

or, x = – 1

∴ x = ± 1

Oasis School Mathematics-9 145

Example: 3

If x = 3 1 + 3– 1 , prove that 3x³ – 9x – 10 = 0.
3 3

Solution:

Here, x = 3 1 + 3– 1
or. 3 3

( )or, x3³ = 3 1 + 3– 1 3
3 3

( ) ( ) ( )or, x3 = 3 1 3+ 3– 1 3+ 3. 3 1 . 3– 1 3 1 + 3– 1
3 3 3 3 3 3

or, x3 = 3 + 3 –1 + 3. 3º . x

or, x3 = 3 + 1 + 3x
or, x3 3
or, 3x3 10 + 9x Remember !
= 3
(a + b)3 = a3 + b3 + 3ab (a + b)
= 9x +10

or, 3x3 – 9x – 10 = 0 Proved.

Exercise 7.2

1. (a) If ax = a5, find the value of x.
(b) If ax = (b+1)x, write the relation between a and b.
(c) If ax = 1, what is the value of x?
(d) If 5x+y = 5z, write the relation among x, y and z.

2. Solve the following equations:

(a) 2x = 32 (b) 1 4x = 64 (c) 5x+2 = 1
(d) 42x × 2 = (32)½ 2
1
(e) 2(2x–4) = 8(x +5) (f) (64) 3 = 43x–5

( ) ( ) 1 1
(g) 3 x + 5= 33 (h) 2x × 42x × 83x = 1 (i) 3x+1 = 32x × 27
3

(j) 5x = 0.04 (k) 102x = 0.001

3. Solve: (a) 2x + 4 + 2x = 34 (b) 4x+2 = 22x+1 + 14 (c) 3x + 3x+2 = 10
3
(d) 3x+1 – 3x = 6 (e) 4x – 4x –1 = 192 (f) 3x+3 + 3x = 28

(g) 4x – 4x–1 = 48 (h) 52x + 52x + 1 = 150 (i) 3x + 3x+1 = 108

146 Oasis School Mathematics-9

4. Solve: (a) 23x–5 × ax–2 = 2x–2 × a1–x (b) 2x+3 × 3x+4 = 18 (c) 22x–3 × 5x–1 = 200

5. Solve:

(a) 3x + 1 = 10 (b) 2x + 2–x = 4 1 (c) 4x + 1 = 16 1
3x 3 4 4x 16
2
(d) 4x – 3 × 2x + 2 = 0 (e) 9x – 4 × 3x + 3 = 0 (f) 2x + 2x =3

6. (a) If ax = b, by = c and cz = a, prove that xyz = 1.

(b) If am. an = (am)n, prove that: m(n – 2) + n(m – 2) = 0

(c) If ax = by = cz and b2 = ac, prove that: 1 + 1 = y2 .
x z
1 –1

7. (a) If x = 3 2 + 3 2 , prove that 3x2 – 16 = 0.

(b) If x = 1 + 2– 1

23 3, prove that 2x3 – 6x – 5 = 0.

(c) If x = 1 – a– 1 show that: x3 + 3x = a – a1 .

a3 3,





Answer

Consult your teacher 2. (a) 5 (b) – 3 (c) – 2 (d) 3 (e) –19 (f) 2
1. 4 8 (g) 3 (h) 1

(g) – 43 (h) 0 (i) 4 (j) –2 (k) – 3
9 2

3. (a)1 (b) 0 (c) – 1 (d) 1 (e) 4 (f) 0
(c) 3
(i) 3 4. (a) 3 (b) – 2 5. (a) 1, – 1 (b) 2, – 2 (c) 2, – 2
(d) 0, 1 2 (f) 0, 1

(e) 0, 1

Oasis School Mathematics-9 147

Unit Ratio and
8 Proportion

8.1 Ratio

A ratio is a fraction obtained by dividing (or comparing) a quantity by another
quantity of the same kind. That means, a ratio gives a comparison between two
quantities of the same unit, i.e. a ratio measures what part or how many times the
first quantity is of the second.

(for e.g., if a and b are two quantities of the same kind, then the ratio of a to b is

denoted as a h: bisofrathbae.rRiasm10is:3100oyrea1r:s old and his father is 30 years. The ratio of ages
of Ram and 3 i.e. Ram's father is 3 times older than Ram.

Here, 10 is the ratio of age of Ram and his father. And 30 or 3 : 1 is the ratio of ages
30 10
of Ram's father and Ram

In a : b, a is called as Antecedent and b is called Consequent.

Note: Ratio is the comparison by division. A ratio is simply a number with no unit.
It means a ratio cannot be formed when two quantities are of different units.

Compound ratio:

cAkonmroawptionouawnsdhciorcamhtipioso=fuonbarmd×erdcadt=iaofb.ateFdcromr euxlatimplpylien,gif(obar athnedpdcroadreuctwt oof)ratwtioos,otrhmenore ratios is


Duplicate and sub–duplicate ratios:

The compounded ratio of two equal ratios is called duplicate ratio. The duplicate

ratio of a is a × a = a2
b b b b2
And, the square root of any given ratio is called sub-duplicate ratio.

For example. a2² a2 a
b2² b2 b
The sub-duplicate ratio of is =

Triplicate and sub–triplicate ratios:

A compound ratio formed by three equal ratios is called triplicate ratio. For exam-
a a a a a3
ple, the triplicate ratio of b = b × b × b = b3

And the cube root of triplicate ratio is called sub–triplicate ratio, for e.g. the sub

triplicate ratio of a3 = 3 a3 = ba.
b3 b3

148 Oasis School Mathematics-9

Worked Out Examples

Example: 1 Example: 2

Find the ratio between 160g and 1.5 kg. Find the compound ratio of 2 , 4 and 6 .
3 5 7
Solution:
Solution:

The ratio between 160g and 1.5 kg. The compound ratio of 2 , 4 and 6
3 5 7
160g 2 4 6
= 1500g = 3 × 5 × 7

= 8 = 8 : 75 = 16
75 35

Example: 3

If a:b = 2 , find the value of 2a + 3b
3 3a + 5b
Alternative method:
Solution:
2a 3b
Here, a = 2 i.e. a = b =k (suppose) 2a + 3b = b + b
b 3 2 3 3a + 5b +
3a 5b
then a = 2k and b = 3 k b b

Now, 32aa ++ 53bb = 32..22kk + 3.3k = 2×32+ 3
+ 5.3 k 3×32+ 5

= 64kk++195kk = 4 +3
3 +5
6
= kk((64++195)) = 13 3
21 4+9
13 3 13
= 3 = 3 × 21 = 21
6+15

3

Example: 4

If 2x + 3y = 18 , find x : y .
3x + 5y 29
Solution:

Here, 2x + 3y = 18
or, 3x + 5y 29

29(2x + 3y) = 18(3x + 5y)

or, 58x + 87y = 54x + 90y

or, 58x – 54x = 90y – 87y

or, 4x = 3y
or,
∴ x = 3
y 4

x : y = 3:4

Oasis School Mathematics-9 149

Example: 5

Divide 104 into two numbers which are in the ratio 6:7

Solution:

Let, two numbers in the ratio 6:7 be 6x and 7x

Here, 6x + 7x = 104

or, 13x = 104
or,
x = 104
13

x = 8

So, the required numbers are, 6 × 8 = 48 and 7 × 8 = 56

Example: 6

Two men's ages are in the ratio 2:3. After 9 years, the ratio of their ages will be 3:4. Find
their present ages.

Solution:

Let the ages of two men in the ratio 2:3 be 2x and 3x years.

Then, 2x + 9 = 3
or, 3x + 9 4

8x + 36 = 9x + 27

or, 36 – 27 = 9x – 8x

∴ x = 9

So, the required ages of two men are 2 × 9 = 18 years and 3 × 9 = 27 years.

Exercise 8.1

1. (a) On which condition, ratio of two quantities can be taken?
(b) In the ratio a : b, what are 'a' and 'b' called?
(c) What is the duplicate ratio of a : b?
(d) What is the triplicate ratio of p : q?
(e) What is the sub-duplicate ratio of a2 : b2?
(f) What is the sub-triplicate ratio of a3 : b3?
(g) What is the compound ratio of a : b and c : d?

2. Find the compound ratio of :

(a) 2 and 4 (b) 1 , 2 and 4 (c) x , a and p
3 5 2 3 5 y b q

150 Oasis School Mathematics-9

3. Find the duplicate and triplicate ratio of:

(a) p (b) 2 (c) 1
q 3 2

4. (a) Find the sub duplicate ratio of: (i) 4 (ii) 25
9 625

(b) Find the sub triplicate ratio of: (i) 27x3 (ii) 64a3b3
y3 27

5. If x = 4 , find the value of: (a) 2x – 53yy (b) 4x + 6y
y 5 4x – 3x + 8y

6. (a) If (5x + 2y) : (7x + 3y) = 9 : 13, find x : y.

(b) If 3x – 5y = 1 , find x: y.
3x + 5y 4

7. (a) If (7x – 3y) : (x + 4y) = 5 :14, evaluate (3x + 2y) : 2y2

(b) If 4x + 3y = 12, find the value of 5x – 2y .
2x – y x + y

8. If a : b : c =2 : 3 : 4, find the value of,

(a) 2a – 3b + 4c (b) 3a – 4b + 5c
a + b+ c a + b+ c

9. (a) Two numbers are in the ratio of 5:8 and their sum is 91. Find the two numbers.

(b) Two numbers are in the ratio of 7:5 and their difference is 6. Find the numbers.

10. (a) The ratio between two quantities is 2:3. If the first quantity is 34 kg, find the
other quantity.

(b) The ratio of two numbers is 3:4. If the antecedent is 24, find the consequent.

(c) Two numbers are in the ratio of 4:5. If the smaller number is 12, find the greater
number.

11. (a) Divide 130 in the ratio of 2 : 3 : 5.

(b) Divide 120 in the ratio of 2:3:7.

12. In a mixture of 28 litres, the ratio of milk and water is 3:1. If 4 litres of water is added
to the mixture, find the ratio of water and milk in the mixture.

13. (a) The ratio of the ages of two men is 2:3 and in 9 years the ratio of their ages will
be 3:4. Find their ages after 9 years.

(b) The ratio of the ages of two men is 3:4. After 4 years, the ratio of their ages will
be 7:9. Find their present ages.

(c) The ratio of the ages of two men is 3:4. 7 years ago, the ratio of their ages was
2:3. Find their present ages.

14. (a) Two numbers are in the ratio of 7:10. If 6 is added to each of them, they would
be in the ratio of 8:11. Find the numbers.

Oasis School Mathematics-9 151

(b) Two numbers are in the ratio of 3:4. If 10 is subtracted from both numerator
and denominator, the new ratio is 13:18. Find the numbers.

15. A sum of money is divided between A and B in the ratio of 2:5. If A gets Rs. 150 less
than B, how much will B get?

16. Two numbers are in the ratio of 3:2 and the difference of their squares is 125. Find
the numbers.

Answer 

1. Consult your teacher.

2. a) 8 b) 4 c) xap 3. a) p2 , p3 b) 4 , 8 c) 1 , 1 ( ) ( )4. (a) i)22ii) 52
15 15 ybq q2 q3 9 27 4 8 3 25

( ) ( ) 3x 3 ii) 4ab 3 5. (a) 79, (b) 23 6. (a) 1 : 2 (b) 25 : 9 (a) 2 : 1 (b) 1
(b) i) y y 26 7.

8. (a) 11 (b) 14 : 9 9. (a) 35, 56 (b) 21,15 10. (a) 51 kg. (b) 32 (c) 15
9
11. (a) 26, 39, 65 (b) 20, 30, 70 12. 11:21 13. (a) 27 years, 36 years.

(b) 24 years, 32 years (c) 21 years, 28 years 14. (a) 42, 60 (b) 75, 100 15. Rs. 250 16. 15, 10

8.2 Proportion

Take any four numbers 4, 8, 12, 24. Take the ratio of first two and last two ratio of first

two = 4 = 1 and ratio of last two = 12 = 1 . Thus 4, 8, 12, 24, are in proportion.
8 2 24 2

Hence, a, b, c and d are quantities, then they are said to be in proportion if the ratio
of first two quantities is equal to ratio of other two quantities.

i.e. if a = c then a, b, c and d are in proportion.
b d

Here, first term 'a' and last term 'd' are called extremes where as second term 'b' and
third term 'c' are called means.

In proportion, the product of extremes = the product of means.

In a = dc , if a, b, c and d are taken in order, then 'd' is generally called the fourth
b

proportional.

Continued proportion
Four or more quantities are said to be in continued proportion, if the ratio of the first

and the second is equal to the ratio of the second and the third and so on. i.e.

if. a = b = c ……………… then it is said to be continued proportion.
b c d

And, if a = b then b2 = ac
b c

152 Oasis School Mathematics-9

Some properties of proportion

1. If a = dc , then b = d (This property is called invertendo. )
b a c
ad b
⇒ ad = bc or, bc = ad or, ac = bc or, a = d
ac c

2. If a = dc , then a = db. (This property is called alternando.)
b c

Here, = a = c
b d
ad a b
⇒ ad = bc or, cd = cbdc, or, c = d

3. If a = dc , then a + b = c +d (This property is called componendo)
b b d

Here, a = c
b d
a c
or, b + 1 = d + 1 (Adding 1 on both sides)

∴ a+b = c + d
b d
a c a –b c –d
4. If b = d , then b = d ( This property is called dividendo)

Here a = c
b d
a c
or, b – 1 = d – 1 (Subtracting 1 on both sides)

∴ a–b = c –d
b d
a c a + b c+ d
5. If b = d , then a – b = c– d (This property is called componendo and dividendo)

Here, a = c
b d
a + b c+d
or, b = d ........ (i) (Using componendo property of proportion)

and, a = c
b d
a–b c–d
or, b = d ........ (ii) (Using dividendo property of proportion)

Dividing (i) by (2) we get.

a + b = c + d
a – b c – d

6. If a = c = e = … then each ratio is equal to
b d f

a + c + e + ... (Such property is called addendo)
b + d + f + ...

Oasis School Mathematics-9 153

Worked Out Examples

Example: 1

If 3, 8, 15, x are in proportion, find x.

Solution:

Since 3, 8, 15, x are in proportion,

3 = 15
or, 8 x

3x = 120

or, x = 120
∴ 3

x = 40

Example: 2

What constant number must be subtracted from 4, 6, 10 and 18 to make them in
proportion.

Solution:

Let, x must be subtracted to make them in proportion.

Then, 4 – x, 6 – x, 10 –x and 18 – x are in proportion.

Now, 4–x = 10 – x
6–x 18 – x

or, (4 – x) (18 – x) = (6 – x) (10 – x)

or, 72 – 4x – 18x + x2 = 60 – 6x – 10 x + x2

or, 72 – 22x = 60 – 16x

or, 12 = 6x

or, x = 12
6

∴ x = 2

Hence, 2 must be subtracted from 4, 6, 10 and 18 to make them in proportion.

Example: 3

If a = c , prove that a – b = c – d .
b d a + b c + d

Solution: a = c = k,
Let, b d

154 Oasis School Mathematics-9

∴ a = bk, c = dk
Now,
L.H.S. = a – b R.H.S. = c–d
a + b c+d

= bk – b = dk – d
bk + b dk + d

= b (k – 1) = d (k – 1)
b (k + 1) d (k + 1)

= (k – 1) = (k – 1)
(k + 1) (k + 1)

∴ LHS = RHS

Example: 4

If a, b, c are in continued proportion, prove that: a² b² c² 1 + 1 + 1 = a³ + b³ + c³
a³ b³ c³

Solution:

Let, a = b = k,
b c
b
c = k ⇒ b = ck,

a = k ⇒ a = bk = ck.k = ck2
b

L.H.S. = a2 b2 c2 1 + 1 + 1
a3 b3 c3

= (ck2)2. (ck)2 . c2 1 + 1 + 1
(ck2)3 (ck)3 c3

= c2.k4. c2.k2. c2 1 + 1 + 1
c3k6 c3.k3 c3

= c6.k6. 1 + k3 + k6
c3.k6

= c3(k6 + k3 + 1)

R.H.S. = a3 + b3 + c3

= (ck2)3 + (ck)3 + c3

= c3k6 + c3 k3³ + c3

= c3 ( k6 + k3 + 1)

∴ L.H.S. = R.H.S.

Example: 5

If a = b = c , prove that: (a² + b² + c²) (b² + c² + d²) = (ab + bc + cd)².
b c d

Solution: a b c
b c d
Let, = = = k,

Oasis School Mathematics-9 155

c = k ⇒ c = dk
d

b = k ⇒ b = ck = dk.k = dk2
c

a = k ⇒ a = bk = ck.k. = dk.k.k = dk3
b

L.H.S. = (a2 + b2 + c2) (b2 + c2 + d2)

= {(dk3)2 +(dk2)2+ (dk2)2} {(dk2)2+ (dk)2 + d2}

= (d2k6 + d2k4 + d2k2) (d2k4 + d2k2 + d2)

= d2k2(k4 + k2 +1). d2 (k4 + k2 + 1)

= d4k­2 (k4 +k2 +1)2

R.H.S. = (ab + bc + cd)2

= [dk3 . dk2 + dk2. dk + dk.d]²

= [d2k5 + d2k3 + d2k]²

= [d2k (k4 + k2 + 1]2

= d4k2 (k4 + k2 + 1)2

∴ L.H.S = R.H.S Proved.

Exercise 8.2

1. Identify whether the given terms are in proportion or not?

(a) 2, 3, 6, 39 (b) 10, 20, 30, 40 (c) ab, a2b², xab, xa2 b (d) 5a2, 10ab, 15ab2, 30b3

2. Find the fourth proportional to :

(a) 2, 3, 4 (b) 4x, 5y, 7z, (c) x², y², z² (d) 3, 7, 12

3. (a) If 8, 24, 30, x are in proportion, find x.

(b) If 4, 9, x and 18 are in proportion, find the value of x.

(c) If 2, 6 and 18 are in proportion, find the fourth proportional.

4. (a) Find the mean proportional to 12, 27.

(b) If 8, x and 18 are in continued proportion, find the value of x.

(c) If a, 25 and 125 are in continued proportion, find the value of a.

156 Oasis School Mathematics-9

5. (a) What constant number must be added to each of 2, 7, 17, 37 to make them in
proportion?

(b) What constant number must be subtracted from each of 8, 10, 17, 22 to make
them in proportion?

6. If a = c , prove that :
b d

(a) a –a b = c c d (b) a2 + ac + c2 = b2 + bd + d2
– a2 – ac + c2 b² – bd + d2

(c) (a2+ c2) (b2 + d2) = (ab + cd)2 (d) a2 + b2 = a+b 2
c2 + d2 c+d

(e) pa + qc = a2 + c2 (f) a2 + b2 = ab
pb + qd b2 + d2 c2+ d2 cd

7. If a = c = e , prove that :
b d f

(a) a2 + c2 + e2 = e (b) (ab+cd+ef)2 = (a2+ c2+ e2) (b2 + d2 + f2)
b2 + d2 + f2 f

(c) ba22 ++ c2 + e2 = badc (d) (a+c+e) (b+d+f) = ab + cd + ef
d2 + f2

(e) ba33 + c3 + e3 = ace (f) a2 + c2 + e2 + ac + ce + ae = ac
+ d3 + f3 bdf b2 +d2 + f2 + bd + bf + df bd

8. (a) If 3a² + 2b² = 3c² + 2d² , prove that a = dc .
3a² – 2b² 3c² – 2d² b

(b) If a2 + c2 = a2 – c2 , prove that a = dc .
b2 + d2 b2 – d2 b

9. If a = b , prove that:
b c

(a) a2 + ab = b2 + bc (b) a = a2 + ab + b2
a2² b2 c b2 + bc + c2

(c) abc ( a + b + c)3 = (ab + bc + ca)3 (d) a2 + b2 + c2 =a+b+c
a–b+c

(e) a+b+c = (a + b + c)2 (f) (a + b)2 = a2 + b2
a–b+c a2 + b2 + c2 (b + c)2 b2 + c2

(g) a1bc (ab + bc + ca)3 = (a +b + c)3 (h) a2 + b2 + c2 = (a – b + c) (a + b+ c)

10. If a = b = c , prove that:
b c d

(a) ba33 ++ b3 + c3 = a
c3 + d3 d

(b) (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2

(c) ab – bc + cd = (a – b + c) (b – c + d)

Oasis School Mathematics-9 157

(d) (b + c) (b + d) = (c + a) (c + d)

(e) (a – d)2 = (b – c)2 + (c – a)2 + (d – b)2

(f) a –c b + a– c = a – d
b c

11. (a) If y a z = b = x c y , prove that:
– z–x –

(i) a (y + z) + b(x + z) + c (x + y) = 0

(ii) a + b + c = 0

(b) If x = c + y – b = a + z – c, prove that (a – b)z + ( b – c)x + (c – a)y = 0.
b + c– a a b

12. If a+b = b+c = c + a , prove that each ratio is equal to 2.
c a b

13. If x : y : : y : z prove that:

(a) x2 y12 z2 (x3 + y3 + z3) = 1 + 1 + 1
x3 y3 z3
(y – z)2
(b) x – 2y + z = z

14. If x = y = z , prove that:
a b c

(a) x³ + y² + z³ = (x + y + z)3
a² b² c² (a + b + c)2

(b) x3 – y3 – z2 = (x – y – z)3
a2 b2 c2 (a – b – c)2

15. If a = 2b = 3c, prove that: a a c + a a b = 7 .
– – 2



Answer (b) 35yz (c) y2z2 (d) 28
4x x2 (c) 5


1. Consult your teacher 2. (a) 6,

3. (a) 90 (b) 8 (c) 54 4. (a) 18 (b) 12

5. (a) 3 (b) 2

158 Oasis School Mathematics-9

Unit Simultaneous Equations

9 in Two Variables

9.1 Warm-up Activities

• What is the degree of the equation x + 3y = 5?
• What are the variables of the equation 3x–y = 5 ?
• Guess four values of x and y which satisfy first equation.
• Guess four values of x and y which satisfy the second equation?
• What is that value called?

9.2 Simultaneous linear equations in two variables x and y.

An equation of the form ax + by + c = 0, where a, b, c are real numbers (a ≠ 0, b ≠ 0)
is called a linear equation of two variables x and y. 2x + 3y – 5 = 0, x – y + 1 = 0 are
the examples of linear equations.

And, two linear equations of two variables x and y are said to be simultaneous
linear equations if each of these equations is satisfied by the same pair of values of
x and y. In other word two equations which have only one pair of solution which
satisfies both the equations are called simultaneous equations.

e.g. consider the system of linear equations.
2x – 3y = 1 ………….. (i) and x + 2y = 4 ………….. (ii)
The equations (i) and (ii) are satisfied by the values x = 2 and y = 1.
Hence, these equations (i) and (ii) are called simultaneous equations of two varia-

bles x and y. There are many methods of solving equations. In this chapter, we will
discuss only three methods. They are:
(a) Substitution method
(b) Elimination method
(c) Graphical method

9.3 Substitution Method

In this method, value of one variable is obtained in terms of another variable and
that expression is substituted in another equation to get the value of the variable.

Oasis School Mathematics-9 159

Worked Out Examples

Example: 1

Solve by substitution method: x + 2y = 7, 2x + 3y = 12.

Solution: x + 2y = 7 ................ (i)


2x + 3y = 12 ................ (ii)
From equation (i),

x = 7 – 2y ................ (iii)
Substituting, x = 7 – 2y in equation (ii)

we get,

2(7 – 2y) + 3y = 12

or, 14 – 4y + 3y = 12

or, – y = 12 – 14

or, –y = –2 or, y= 2

Now, substituting y = 2 in (iii), we get,

x+2×2 = 7

or, x+4 = 7

or, x = 7 – 4 = 3

∴ x = 3 and y = 2

Example: 2

Solve : 4x + 3y = 20 and 6x – 3y = 0

Solution: 4x + 3y = 20 ................ (i)

6x – 3y = 0 ................ (ii)
From equation (ii),
6x – 3y = 0

or, 6x = 3y
x = 63x
or, y = 2x.

Substituting the value of y in (i)

4x + 3 × 2x = 20 or, 4×2+3y = 20

or, 4x + 6x = 20 or, 8 + 3y = 20

or, 10x = 20 or, 3y = 20–8

20 or, 3y = 12
x = 10 = 2
12
Substituting the value of x in (i) or, y = 3 =4

4x + 3y = 20 ∴ x = 2 and y = 4.

160 Oasis School Mathematics-9

Exercise 9.1

1. Solve (by substitution method):

(a) y = 2x and x + y = 3 (b) y = x+ 5 and 2x + y = 8

(c) y = 2x + 1 and 2y – 3x = 4 (d) y = x – 1 and x + 2y = 10

(e) y = 2x + 4 and 3x + 2y = 7 (f) y = 3x – 8 and 5x + 2y = – 4
3 4

2. Solve (by substitution method) :

(a) x + y = 8 and 2x – y = 7 (b) x + y = 5 and x – 3y = –3
(d) 2x + 3y = 7 and x – y = 1
(c) 7x – y = 5 and 5x + 4y = 13 (f) x + 2y = 9 and 3x – y = 13

(e) x + 3y = 13 and x + y = 7

3. Solve (by substitution method) :

(a) x + y = 4 and 3x + y = 11
5 2
1 1
(b) x = 3 y + 11 and y = 9 – 2 x

(c) 3(y – 2) – 2(x + 3) = 1 and 2(y – 3) + 3(x + 2) = 0



Answer (c) 2, 5 (d) 4, 3 (e) 1, 2 (f) 0, – 2
(c) 1, 2 (d) 2, 1 (e) 4, 3 (f) 5, 2
1. (a) 1, 2 (b) 1, 6 (c) – 2, 3

2. (a) 5, 3 (b) 3, 2
3. (a) 2, 10 (b) 12, 3

9.4 Elimination Method

In this method, we have to follow the following steps:
• Multiply the given equations by suitable number to make the coefficient of one
of the variables equal.
• Add or subtract two equations to eliminate the variable having equal co–efficient.
• Value of another variable is obtained.
• Substitute this value to one of the equations to get the value of another variable.

Oasis School Mathematics-9 161

Worked Out Examples

Example: 1

Solve : 2x + 2y = 7 and 3x – 2y = 13

Solution: Here,

2x + 2y = 7 .............. (i)

3x – 2y = 13 .............. (ii)

Adding equation (i) and equation (ii), we get

5 x = 20

or, x = 20 =4
5
Substituting x = 4 in equation (i), we get

2 × 4 + 2y = 7

or, 2y = 7 – 8

or, 2y = – 1

or, y = – 1
2
1
∴ x = 4 and y = – 2

Example: 2
Solve : 5x + 3y = 11 and 5x + 4y = 13
Solution: Here,
5x + 3y = 11 .............. (i)
5x + 4y = 13 .............. (ii)

Here, subtracting equation (ii) from equation (i)

5x + 3y = 11 Substituting, y = 2 in equation (i) we get

5x + 4y = 13 5x + 3 × 2 = 11
– – – or, 5x + 6 = 11
–y = –2

or y = 2 or, 5x = 11 – 6

or, 5x = 5

Hence, x = 1 and y = 2 ∴ x = 1

162 Oasis School Mathematics-9

Example: 3

Solve: 7x – y = 13 and 5y – 3x = – 1
Solution:

7x – y = 13.............. (i)

5y – 3x = – 1 .............. (ii)

Multiplying equation (i) by 5 and equation (ii) by 1 and adding equation (i) and (ii), we get,

35x – 5y = 65 Substituting x = 2 in equation (i) we get,

or, – 3x + 5y = – 1 7 × 2 – y = 13

or, 32x = 64 or, 14 – 13 = y

∴ x = 6324 or, 14 – 13 = y

= 2 or, 1 = y

Example: 4 ∴ x = 2 and y = 1

Given x – 6 = 6(y – 6) and 3(x + 4) = 8(y + 4)

Solution:

Given equations are

x – 6 = 6(y – 6)

or, x – 6 = 6y – 36

or, x – 6y = – 36 + 6

or, x – 6y = –30 .................... (i)

And, 3(x + 4) = 8(y + 4)

or, 3x + 12 = 8y + 32

or, 3x – 8y = 32 – 12

or, 3x – 8y = 20 ................ (ii)

Multiplying equation (i) by 3 and subtracting (ii) from (i)

3x – 18y = – 90

or, 3x – 8y = – 20

–10y = – 110

or, y = – 110 = 11
10

Substituting the value of y in equation (i)

x – 6y = – 30

or, x – 6 × 11 = – 30

or, x – 66 = – 30

Oasis School Mathematics-9 163

or, x = – 30 + 66

= 36

Hence, x = 36

y = 11

Example: 5

Solve : 2 + 5 = 1 and 3 + 2 = 19
x y x y 20

Solution: Here, 2 + 5 = 1 .............. (i)
x y

3 + 2 = 2109 .............. (ii)
x y
Multiplying equation (i) by 2 and equation (ii) by 5 and then subtracting (ii) from (i)

4 + 1y0 = 2 Substituting x = 4 in equation (i), we get,
x
1–x5 + 1y0– 19 2 + 5 = 1
–4 4 y
=
1 + 5 = 1
2 y
– 1x1 11
= – 4 or, 5 = 1 – 1
y 2
1 1
x = 4 5 = 1
y 2
or, x = 4 y = 10

Hence, x = 4 and y = 10
Example: 6

Solve: 14 + 3 =5 and 21 – 1 =2
x+y x–y x+y x–y

Solution: Here, 14 + 3 = 5 .............. (i)
x+y x–y

21 – 1 = 2 .............. (ii)
x+y x–y

Here, multiplying (i) by 1 and (ii) by 3 and then adding (i) and (ii)

14 + 3 = 5
x+y x–y

63 – 3 = 6
x+y x–y

77 = 11
x+y

or, x + y = 7 .............. (iii)

164 Oasis School Mathematics-9

Substituting x + y = 7 in (i) we get,

14 + 3 = 5
7 x–y
or,
2+ 3 = 5
or, x–y
or,
3 = 3
x–y
x – y = 1 .............. (iv)

Now, adding equation (iii) and (iv),

x + y = 7
x–y = 1

2x = 8
x = 4
or,

Now, substituting x = 4 in equation (iii),

we get,

4+y = 7

or, y = 3

Hence, x = 4 and y = 3.

Exercise 9.2

Solve the following system of equations by using elimination method:

1. (a) x + y = – 2 and x – y = 4 (b) 2x + y = 6 and 2x – 3y = 14

(c) 4x + 3y = 6 and x – 3y = 9 (d) 4x – 3y = –1 and 4x + y = 3

2. (a) x + 2y = 7 and 2x – y = 4. (b) x + 2y = – 1 and 2x – 3y = 12

(c) 2x + y = 35 and 3x + 4y = 65 (d) 3x + y = 100 and x – 4y = 120

(e) 2x – 3y = 7 and 7x + 2y = 12 (f) 2x + 3y = 1 and 5x + 2y = – 3

3. (a) 7(x + 3) – 2 (y + 2) = 14 and (x + 2) + (y + 3) = 11

(b) 3(3y – 1) = x + 42 and (5x +4) – 4 (y – 7) = 7

(c) 10x + y = 4 (x + y) and 10x + y + 27 =10y + x

4. (a) 1 + 1 = 5 and 2 + 3 = 13 (b) 4 – 3 = 7 and 3 + 4 = 24
x y x y x y x y

(c) y5 + 7 = 31 and 5 + 7 = 29 (d) 2 + 18 = 3 and 4 – 9 =1
x x y x y x y

Oasis School Mathematics-9 165

(e) x6 – 4 = 4 and 3 + 6 = – 2 (f) 3x + 4y = 8xy and x + 2y = 3xy
y x y

5. (a) x – y = 2 and x – y =–1 (b) x – y = 2 and x + y = 2
3 4 9 2 3 4 9 4

(c) 23x – y = 2 and x + 2y = 6 (d) 2y – 3 = 5 and y – 3x – 1 = 8
5 3 5 5–x 5

6. (a) x+7 = 2x – y = 2y + x (b) y+3 = 8–x = 3(x + y)
3 2 4 5 4 8

7. (a) 14 – x 3 y = 8 and 21 + 1 = 5
x+y – 3 x+y x–y 4

(b) x 5 y + 2 =3 and 15 – y 4 x =7
+ x–y x+y –

Answer (b) 4, – 2 (c) 3, –2 (d) 1 , 1 2. (a) 3, 2 (b) 3, – 2 (c) 15, 5
(f) – 1, 1 2 (c) 3, 6


1. (a) 1, –3

(d) 40, – 20 (e) 2, -1 3. (a) 1, 5 (b) – 3, 5
2
12, 1 (b)41, 1 13, 1 1
4. (a) 3 3 (c) 2 (d) 2, 9 (e) 3, -2 (f) 2, 2 5. (a) 9, 4 (b) 9, 4 (c) 6, 10

(d) 2, 9 6. (a) 8, 6 (b) –4, 12 7. (a) 5,7 (b) 3, 2

9.5 Graphical Method of Solving Simultaneous Equations

While solving two simultaneous equations by this method, we have to find same
pair of points in each of the equations. The pair of points in each equation should be
plotted in the graph to get two lines. The point of intersection of two lines gives the
solution of two simultaneous equations.

Worked Out Examples

Solve (Graphically)

Example: 1 Y

3x + 4y = 24 and x + y = 7. (0, 7)

Solution: (0, 6)
Here, 3x + 4y = 24 (i)
(4,3)
or, 4y = –3x + 24

or, y = –3x + 24 X' O(0, 0) (7, 0) x+(y8=7,30x+)7y=2X4
4 Y'

x084

y603

Let's plot these points on the graph paper.

166 Oasis School Mathematics-9

Again , x + y = 7 ................(ii)
y = –x + 7
or,

x072
y705
Let's plot these points on the graph paper.
From the graph, the above pairs of straight lines meet at (4, 3).
Hence, required solution is (4, 3)

∴ x = 4 and y = 3

Example: 2
y = 2x + 1, 3x + 2y = 23.

Solution: Y

(1, 10)
Here, y = 2x + 1 ...............(i) y=2x + 1

x 0 2 –2 (3, 7)3x+2y=23
y 1 5 –3 (2, 5)

Let's plot these points on the graph paper. (0, 1)
O
Again, 3x + 2y = 23 .....................(ii) X' X
Y'
or, 2y = 23 –3x X

or, y = 23–3x
2

x135

y 10 7 4

Let's plot these points on the graph paper.

From the graph, above pair of straight lines meet at the point (3, 7).

Hence, required solution is (3, 7).

∴ x = 3 and y = 7 Y

Example: 3 (0,7) (1,6) y = 2x –5
2x – 5 = –x + 7 (2, 5)

Solution: Let, 2x – 5 = – x + 7 = y (4, 3)

Here, y = 2x – 5 ................... (i) X'

x012 (2, – 1) y=–x+7
y –5 –3 –1 (1, – 3)
Let's plot these points on the graph paper. (0,–5)
Y'
Again, y = – x + 7 ................... (ii)

Oasis School Mathematics-9 167

x012
y765

Let's plot these points on the graph paper.

The above pair of straight lines meet at the point (4,3). Hence, required solution
is (4,3).

∴ x = 4 and y = 3.

Exercise 9.3

Solve the following system of equations graphically.

1. (a) x + y = 7, x – y = 3

(b) 2x – y = 3, 3x + 2y = 1

(c) 2x – 3y = –9, 2x + y = 11

(d) 3x – 2y = 8, 5x + y = 9

(e) 2x – 3y = 6, 2x + y +10 = 0

2. (a) y = –2x + 4, x = –3y – 3 (b) y = x, 3x – 2y = 2

(c) x – 4y – 5 = 0, x – y + 1 = 0 (d) 2x + y = 8, y – x = 2

3. (a) x4 + y = 12, 3y – 2x = 13 (b) x–5 = 10 – 2x
2 2 3 3 2 3

(c) 4 – 3x = 4x – 11
2 3

4. (a) 5x – 3y = 11, 2x –3y = –1

(b) x = 4, y = x – 3

(c) y = 2, 2x + y = 6

5. (a) Cost of 1 pen and 2 copies is Rs. 80 and the cost of 2 pens and 1 copy is Rs. 100.
Make the equation of this verbal problem and solve the equations graphically
to get the cost of 1 pen and 1 copy.

(b) The sum of two numbers is 9 and their difference is 1. Find the numbers using
graphical method.



Answer (b) x = 1, y = –1 (c) x = 3, y = 5 (d) x = 2, y = –1
(c) x = – 3, y = – 2
1. (a) x = 5, y = 2 (c) x = 2
(b) 4, 5
(e) x = –3 , y = –4 2. (a) x = 3, y = –2 (b) x =2, y = 2

(d) x = 2, y = 4 3. (a) x = – 2, y = 2 (b) x = 5,

4. (a) x = 4, y = 3 (b) x = 4, y = 1 (c) x= 2, y = 2 5. (a) Rs. 40, Rs. 20

168 Oasis School Mathematics-9

Unit

10 Quadratic Equations

10.1 Warm-up Activities

• What are the degree of the following equations?
x2 – 9 = 0, x2–5x+6=0, x2–29=7, (x+1) (x+2) = 0
• What are these equations called?
• Guess two values of x which satisfy the given equations?
• Compare the above equations with ax2+bx+c=0 and find the values of x.
• Factorise the above equations.

10.2 Quadratic Equations

In the above given examples, the degree of variable x is 2. So, these
equations are quadratic equations. Hence, the quadratic equation is
defined as which contains the squares of the unknown quantities. The
the equation general form of quadratic equation is ax2 + bx + c = 0 where a, b, c are real
numbers, and a ≠ 0.

(a) Solution of quadratic equation by the method of factorisation
The general form of quadratic equation is, ax2 + bx + c = 0, where a, b and c are real

numbers. The method for solving such equation is based on the following property.
If a × b = 0, then either a = 0 or b = 0. This principle is called zero – factor property.
The steps for solving a quadratic equation by the method of factorisation are as

follows:
• Express the given equation in the general form ax2 + bx + c = 0 (by

simplification if necessary).
• Factorise the left side.
• Equate either of the factor equal to zero.
• Solve them and find two values of x.

Worked Out Examples

Example: 1
Solve : 4x² = 25
Solution:

Oasis School Mathematics-9 169

Here, 4x2 = 25

or, 4x2 – 25 = 0

or, (2x)2 – (5)2 = 0

or, (2x + 5) (2x – 5) = 0

Either, 2x + 5 = 0 .......... (i)

or, 2x – 5 = 0 ............(ii)

From equation (i), we get, Similarly, from equation (ii) we get,

2x + 5 = 0 2x – 5 = 0

or, 2x = – 5 or, 2x = 5
or, or,
x = – 5 ∴ x = 5
2 2

Example: 2 x = ± 5
2
Solve : 6x² + x = 12

Solution:

Here, 6x2 + x = 12

or, 6x2 + x – 12 = 0

or, 6x2 + (9 – 8) x – 12 = 0

or, 6x2 + 9x – 8x – 12 = 0

or, 3x (2x + 3) – 4(2x + 3) = 0

or, (2x + 3) (3x – 4) = 0

Either, 2x + 3 = 0 .......... (i) Similarly,


or, 3x – 4 = 0 .......... (ii) From equation (ii), we get,

From equation (i), we get, 3x – 4 = 0

2x + 3 = 0 or, 3x = 4
or,
or, 2x = – 3 ∴ x = 4
or, 3
Example: 3 x = – 3 3 4
2 x = – 2 or, 3

Solve : x 3 1 + 1 = 5
+ x–1 4
Solution:
Here, x 3 1 + 1 = 5
+ x–1 4

or, 3(x – 1) + (x + 1) 5
(x + 1) (x – 1) = 4

170 Oasis School Mathematics-9

or, 3x – 3 + x + 1 = 5
x2 – 1 4
or,
or, 4x – 2 = 5
x2 – 1 4
5x2 – 5 = 16x – 8

or, 5x2 – 16x + 3 = 0

5x2 – 15x – x + 3 = 0

or, 5x (x – 3) – 1 (x – 3) = 0

or, (x – 3) (5x – 1) = 0

Either, x – 3 = 0 .......... (i) or, 5x – 1 = 0 .......... (ii)

From equation (i), we get, Similarly, from equation (ii), we get,

x – 3 = 0 5x – 1 = 0

or, x = 3 or, 5x = 1
∴
x=3 and 1 or, x = 1
5 5

(b) Solution of quadratic equation by completing the perfect square

The method of completing the square can be applied to all quadratic equations.
The process of this method is given below:
• Transpose the constant term to the right–hand side of the equations:
• Divide both sides of the equations by coefficient of x2.
• Make the left side of the equation a perfect square.

• Take the square root of each side.

• Solve the resulting simple equations.

Example: 4

Solve : x² – 5x + 6 = 0

Solution:

Here, x2 – 5x + 6 = 0

or, x2 – 5x = – 6

or, x2 – 2.x. 5 + 5 2 = –6+ 52
2 2 2


or, x – 5 2 = 25 –6
2 4

or, x – 5 2 = 25 – 24
2 4

Oasis School Mathematics-9 171

or, x – 5 2 = 1
2 4

or, x – 5 2 ± 1 2
2 2
=

or, x – 5 = ± 1
2 2

Taking positive sign, Taking negative sign,

x = 52 + 1 x = 5 – 1
2 2 2

= 26 = 4
2
= 3 = 2

∴ x = 3 or 2

Example: 5

Solve : 3x² – x – 2 = 0

Solution:

Here, 3x2 – x – 2 = 0

Dividing both sides by 3

3x2 – x – 2 = 0
3 3 3

or, x2 – x – 2 = 0
3 3

or, x2 – 2.x. 1 + 1 2 – 2 = 12
6 6 3 6

or, x – 1 2 = 1 + 2
6 36 3


or, x – 1 2 = 25
6 36


or, x – 1 2 = ± 5 2
6 6


or, x – 16 = ± 5
6

∴ x = ± 5 + 1
6 6

172 Oasis School Mathematics-9

Taking positive sign Taking negative sign

or, x = 5 + 1 or, x = 1 – 5
6 6 6 6

= 66 = – 2
= 1 3

∴x = 1 or – 2
3

(c) Solution of quadratic equation by using formula

Every quadratic equation can be expressed in the form of ax2 + bx + c = 0. To find the
values of x that satisfy the equation, we use the method of completing the square,

ax2 + bx + c = 0

or, ax2 + bx = – c

or, x² + b x = – c [Dividing both sides by a.]
a a

or, x2 + 2. x . b + b2 = – c + b 2 [Adding b 2
2a 2a a 2a 2a
to both sides]

or, x + b 2 = b2 – c
or, 2a 4a2 a
( )x
+ b 2 = b2 – 4ac 2
2a 4a2

or, x + b 2 = ± b2 – 4ac [Taking square root on both sides]
or, 2a 4a2
∴

x + b = ± b2 – 4ac
2a 2a

x = ± b2 – 4ac – b
2a 2a

or, x = – b ± b2 – 4ac
2a

From the above example, we have two solutions for unknown quantity x which are
given as:

–b+ b2 – 4ac and –b- b2 – 4ac
2a 2a

where a, b and c are real number and a ≠ 0.

This general formula is used in solving a quadratic equation. It can be consequently
applied when the left side cannot be factorised.

Oasis School Mathematics-9 173

Example: 6

Solve: x² + 2x – 15 = 0.

Solution:

Here, given quadratic equation is x2 + 2x – 15 = 0

comparing this equation with ax2 + bx + c = 0

a = 1, b = 2 and c = –15

We have, x = – b ± b2 – 4ac
2a

x = – 2 ± 22 – 4 × 1 × (– 15)
2×1

= –2± 4 + 60
2

= – 2 ± 64 = –2±8
2 2

Now, taking +ve sign,

x = – 2+ 8 = 6 =3
2 2

Similarly, taking –ve sign,

x = – 2- 8 = –10 = – 5
2 2

∴ x = 3 and – 5

Example: 7

Solve: 4x² – 3x = 6
Solution:

Here, 4x2 – 3x = 6

4x2 – 3x – 6 = 0

Now, comparing the given equation with ax2 + bx + c = 0, we get a = 4, b = – 3 and c = – 6

We have, x = – b ± b2 – 4ac
∴ x = 2a

– (– 3) ± (–3)2 – 4 × 4 × – 6
2×4

= 3± 9 + 96
8

= 3 ± 105
8
Now, taking + ve sign, we get,

x = 3 + 105
8

174 Oasis School Mathematics-9

Similarly, taking –ve sign, we get,

x = 3- 105
8

∴ x = 3 + 105 or, x= 3 - 105
8 8

Example: 8

Solve : x + x–1 = 2 1
x–1 x 2
Solution:

Here, the given equation is, x + x –1 = 2 1
x–1 x 2

or, x2 + (x – 1)2 = 5
x(x – 1) 2

or, x2 + x2 – 2x + 1 = 5
or, x2 – x 2

2 (2x2 – 2x + 1) = 5 (x2 – x)

or, 4x2 – 4x + 2 = 5x2 – 5x

or, 4x2 – 4x + 2 – 5x2 + 5x = 0

or, – x2 + x + 2 = 0

or, –(x2 – x – 2) = 0

or, x2 – x – 2 = – 0

Now, comparing it with ax2 + bx + c = 0, we get, a = 1, b = – 1 and c = – 2

Using the formula, x = –b± b2 – 4ac
2a

∴ x = – (– 1) ± (–1)2 – 4 × 1 × – 2
2×1

= 1 ± 1+8
2

= 1±3
2

Now, taking + ve sign we get,

x = 1 + 3 = 4 = 2
2 2

Similarly, taking –ve sign we get,

x = 1-3 = –2 = – 1
2 2

Hence, x = 2 and – 1.

Oasis School Mathematics-9 175

Exercise 10.1

1. (a) If (x + 1) (x – 1) = 0, find the values of x.
(b) What are the values of x in the quadratic equation ax2 + bx + c = 0.

(c) If x (x–2) = 0, what are the values of x?

2. Solve:

(a) (x – 1) (x + 2) = 0 (b) (x – 5) (2x + 3) = 0

(c) (3x – 1) (x + 4) = 0 (d) (5 – 2x) (3x – 7) = 0

(e) (5x – 1) (2x +7) = 0

3. Solve: (b) 4x2 + 3x = 0 (c) x2 – 7x = 0
(a) x2 – 2x = 0

(d) x2 – 9 = 0 (e) 9x2 – 16 = 0 (f) x2 = 4x
(g) 5x2 = 125

4. Solve (by factorisation method):

(a) x2 – 10x – 24 = 0 (b) x2 – 5x – 6 = 0 (c) 5x2 – 2x – 3 = 0

(d) 3y2 + 7y – 6 = 0 (e) x2 + 50 = 7 – 2x2 + 10
25 15

(f) 5x + 7 = 3x + 2 (g) x + 1 – x –1 = 7
x–1 x – 1 x +1 12

(h) x x 1 + x + 1 = 2 1 (i) 4 – x 5 2 = 3
+ x 12 x–1 + x

(j) x x 1 + x + 1 = 13 (k) x + 3 – 2x – 3 = x – 3
+ x 6 x + 2 x–1 2 – x

5. Solve (by completing the square method):

(a) x2 – 10x + 21 = 0 (b) x2 + 6x – 40 = 0

(c) x2 – 5x – 36 = 0 (d) x2 – 4x + 1 = 0

(e) 2x – 3 = x+2 (f) x+5 = x–2
3x – 7 x+3 5x – 2 3x – 8

(g) x 4 1 – x 5 2 = 3 (h) x + 2 – x–2 = 4 4
– + x x – 2 x+2 5

6. Solve (by using formula): (b) 3x2 – 10x – 13 = 0
(a) x2 - 11x + 30 = 0 (d) (4x - 3)2 = 16x
(c) 7x2 + 2 = 9x

176 Oasis School Mathematics-9

(e) 2x2 – 1 = 2 – 3x2 (f) x x 1 + x + 1 = 5
6 15 + x 2

(g) x + 3 + x–3 = 2x – 3 (h) x 3 1 + x 2 3 = 2
x + 2 x–2 x–1 – – x

7. Solve (using any method):
(a) 6 x2 - 4x - 2 6 = 0

(b) (2x + 3) ( 3 - x) = 2x – 6 (c) x 1 1 + 2 = 2 3 x
– x –
5
x– 2
(d) x–4 = 4 – 10 (e) 1 2 + x 3 6 = 1
x 3x + + x

(f) x + x+2 =4 (g) x–1 + x–3 = 3 1
x+2 x x–2 x– 4 3



Answer

1. Consult your teacher.

2. (a) 1, –2 (b) 5, – 3 (c) 31, – 4 (d) 25, 7 (e)51, –7 3. (a) 0, 2 (b) 0, – 3
2 3 2 4

(c) 0, 7 (d) 3, –3 (e) 43, -4 (f) 0, 4 (g) 5, – 5 4. (a) 12, –2 (b) 6, –1
3

(c) 1, – 3 (d) – 3, 2 (e) 5, –5 (f) 3, – 1 (g) 7, –1 (h) – 4, 3
5 3 7

(i) 3, –1 (j) 2, –3 (k) 0, 4 5. (a) 7, 3 (b) 4,– 10 (c) 9, –4
3

(d) 2 ± 3 (e) 5, –1 (f) 4, 121 (g) 3, – 1 (h) 3, – 4
2 3

6. (a) 5, 6 (b) 13 , – 1 (c) 1, 2 (d) 94, 1 (e) ± 3 (f) 1, –2 (g) 4, 0 (h) 2, – 1
3 7 4 4

7. (a) 6, –6 (b) 3, –5 (c) 4 , 1 (d) 5 , 16 (e) 2, – 6 (f) –1 ± 3 (g) 5, 5
3 2 3 2 2 3 7 2

Oasis School Mathematics-9 177

Assessment Test Paper

4 3
2
9

a3
27b3

3 a6b–3c6 4 a–4b4c8 625 1
4
1296

3x – 5y 10
3x + 5y 1 3x
4y

bc xbc xac xba
c a b
xb
xc xa
(a + b + c) ( b + c + d)
a bc bc cd
b cd

11 23
xy xy

x+2 x–2 5
x–2 x+2 6
178 Oasis School Mathematics-9

Geometry

Contents 45Estimated Teaching Hours

• Basic concept

• Triangle

• Congruence of triangles

• Parallelograms

• Mid-point theorem and its converse

• Construction of different types of quadrilaterals

• Similarity

• Circle

Expected Learning Outcomes

At the end of this unit, students will be able to develop the
following competencies:

• To use the properties of pair of angles, pair of angles made by the
transversal with parallel lines

• To prove and verify experimentally various properties of triangle
• apply these properties to solve the problems related to them

• To make the pair of triangles congruent using various axioms and
make the corresponding sides and angles equal

• To identify the various types of quadrilateral

• To prove the various properties of parallelograms theoretically

• To prove the mid point theorem and apply it to prove various
problems related to it

• To construct the different types of quadrilaterals

• To make triangles similar and make corresponding sides
proportional

• To prove the theorem on circle related to the chords and apply it to
solve the problems related to them

Teaching Materials
• Geometrical instruments like ruler, compass, protractor, set square,

flash card, chart paper, A4 size paper, etc.

Oasis School Mathematics-9 179

Unit

11 Triangles

11.1 Warm-up Activities O A
B
Discuss the following questions in your class.
C
From the given figure, name a pair of adjacent angles. C
• Name their common vertex and common arm.
• Define adjacent angles.

From the given figure:
• Name two angles which form a linear pair.
• What is their sum? Define linear pair.

From the given figure: A OB
• Name two pair of alternate angles. A E
• Name two pair of interior angles.
• Name two pair of interior angles. GB
Under which condition:
(i) Alternate angles are equal. CH D
(ii) Corresponding angles are equal. F
(iii) Sum of co-interior angles is 1800.

Try to find the answer to these questions.
• How many lines can be drawn from a given point?
• How many lines can be drawn from two given points?
• How many mid-points does a line segment have?
• Is the distance between two parallel lines always same?
• How many bisectors does an angle have?

Do you know!
• Greek Mathematician Euclid is known as Father of Geometry
• The Geometry on the plane figure is known as Euclidean Geometry
• Geometrical assumptions without any proof are called postulates

180 Oasis School Mathematics-9

11.2 Triangles A
'c' 'b'
We have already discussed about triangles in previous classes. B 'a' C
Let's review it.
Given figure is a triangle ABC. It can be written as ∆ABC.
Its three angles are ∠A, ∠B and ∠C.
Its three sides are AB, BC and AC.
Side opposite to vertex A is denoted by 'a'.
Side opposite to vertex B is denoted by 'b'.
Side opposite to vertex C is denoted by 'c' .
∴ BC = a, AC = b and AB = c

Theorem-11.1 A
A
The sum of angles of a triangle is 180º.

Experimental Verification:

Draw two triangles ABC of different shapes B C
C
and size. B Fig. (ii)

To verify : ∠A + ∠B + ∠C = 180º Fig. (i)

Observations:

Figure ∠A ∠ B ∠ C ∠ A + ∠ B + ∠ C Remarks
(i)
….º ….º ….º 180º ∠ A + ∠ B + ∠ C = 180º

(ii) ….º ….º ….º 180º ∠ A + ∠ B + ∠ C = 180º

Conclusion : Hence, the sum of the angles of a triangle is 180º

Theoretical Proof: MA N
Given : ABC is a triangle

To Prove : ∠ABC + ∠BAC + ∠ACB = 180º

Construction: Draw a line MAN || BC. BC
Proof:

S.N. Statements Reasons

1. ∠ MAB = ∠ ABC 1. Alternate angles formed on MN || BC.

2. ∠NAC = ∠ACB 2. Same as above reason.

3. ∠MAB + ∠BAC + ∠CAN = 180º 3. Being MAN a straight line.

4. ∠ABC + ∠BAC + ∠ACB = 180º 4. From (1), (2) and (3).
Hence, the sum of three angles of a triangle is 180º.



Oasis School Mathematics-9 181

Corollary: Sum of two interior angles of a triangle is equal to opposite exterior angle.
OR

Exterior angle of a triangle formed by producing any one side of a triangle is equal
to sum of non-adjacent interior angles.
A

Given : In ∆ABC, side BC is produced to D.

To Prove : ∠ BAC + ∠ABC = ∠ACD B CD
Proof.

S.N. Statements Reasons
1. ∠ABC + ∠BAC + ∠ ACB = 180º 1. Sum of angles of a triangle.
2. ∠ACB + ∠ACD = 180º 2. Being linear pairs.
3. ∠ABC + ∠BAC + ∠ACB 3. From statement (1) and (2).

= ∠ACB + ∠ACD

4. ∠BAC + ∠ABC = ∠ACD 4. From (3) eliminating common ∠ACB from
both sides.

Hence, sum of two interior angles of a triangle is equal to opposite exterior angles.

Try: What is the sum of three exterior angles of a triangle.

Worked Out Examples

Example: 1

Find the value of 'x' in the given figure. A
Solution: x

Produce AD to F of BC. 110º
D
Here, ∠ADE = ∠AFC = 110º. B 50º E
Again, ∠ABC + ∠BAF = ∠AFC F C
A
[∵ Sum of the angles of a triangle is equal to opposite exterior angle] E
or, 50º + x = 110º x C

or, x = 110º – 50º B 50º 110º C
∴ x = 60º D

F

Example: 2 A 32º

Find the value of unknown angles from the given figure. bD
Solution: c aa
In the given figure, AB ||DE.
Now, In ∆DBC, 50º
BE

182 Oasis School Mathematics-9

∠DBC + ∠DCB + ∠BDC = 180º [Sum of three angles of a triangle]

or, 50º + 32º + 2a = 180º

or, 82º + 2a = 180º

or, 2a = 180º – 82º

or, 2a = 98º

or, a = 49º

Again , ∠ABD = ∠BDE [Alternate angles]

or, c = a

or, c = 49º

Again, ∠DAB = ∠ CDE [Corresponding angles]

or, b = a

or, b = 49º

∴ a = 49º, b = 49º, c = 49º

Example: 3

In the given figure, BE and CE are the bisectors of ∠ABC and ∠ACD respectively. Prove
that: ∠BAC = 2∠BEC.

Solution: A E
Given: In the given figure, C D
BE and CE are the bisectors of ∠ABC and ∠ACD respectively. B
To prove: ∠BAC = 2∠BEC.

Proof.

S.N. Statements Reasons
1. 2∠EBC = ∠ABC 1. Being EB, the bisector of ∠ABC.
2. 2∠ECD = ∠ACD
3. ∠ABC + ∠BAC = ∠ACD 2. Being EC, the bisector of ∠ACD.

4. 2∠EBC + ∠BAC = 2∠ECD 3. Sum of two angles of a triangle is equal to
5. ∠BAC = 2 (∠ECD–∠EBC) opposite exterior angle.
6. ∠EBC + ∠BEC = ∠LCD
4. From statement 1, 2 and 3.
7. ∠BEC = ∠ECD –∠EBC
8. ∠BAC = 2∠BEC. 5. From statement 4.

6. Sum of two angles of a triangle is equal to
opposite exterior angle

7. From statement 6.

8. From statements 5 and 7.

Hence proved.

Oasis School Mathematics-9 183

Exercise 11.1

1. (a) What is the sum of three angles of a triangle?

(b) What is the relation among A

∠BAC, ∠ABC and ∠ACD? B CD

(c) In the given figure, what is the relation c

among a, b and c? ab

2. Find the value of x, y and z from the given diagrams.

(a) A (b) A (c) (d) X
x
x P 56º R S
xy

B 45º 65º C B 56º xC 49º T 112º Z y 56º Y
Q

T C
A
(e) 120º P (f) C (g) 3x
5x C
A E 40º B 4x
80º y
Q x R 113º S 65º D D E

x

B

3. Find the value of xº, yº and zº from the given diagrams.

(a) A 70º 80º B (b) A (c) B

35º MD xº EN
B zº
yº C D 60º
E yº H
D xº zº E 70º XB 130º Y
C xº
C
F
A
C AB
(d) (e) 12 0º
C
80º F x 130º D
BE
25º F
x
EE
D
G
R SA B

(f) P Q x (g) xP
100º

40º

CH D

T

F

184 Oasis School Mathematics-9

4. Find the values of unknown angles from the given figure.

(a) P (b) T A

(c) 85º

25º y 55º P 140º R Bx
zS x
Q 30º40º x 53º zº T 60º 45º yº F D
RS y 30º
30º
QU E C

(d) A (e) A (f) P

x

D xº E My N
F 100º 40º
Q R35º
By xC B
B 40º 50º E D
C

5. (a) In the given figure, ∠B = 500, OA bisects ∠BAC and OC 50º

bisects ∠BCA. Find the value of ∠AOC. A O
C

A

(b) In the given figure, OB and OC are the bisectors of ∠PBC and B 600
∠QCB respectively, find the value of ∠BOC.
C

PO Q
A

6. (a) In ∆ABC, side BC of ∆ABC is produced to both sides to D

and E. Prove that ∠ABD + ∠ACE = 180º + ∠A.

D E
B C
D

(b) In the given figure, BA is produced to D, CB is A
produced to E and AC is produced to F, prove that B
∠DAC + ∠ABE + ∠BCF = 360º.
E C
 F

Answer

1. Consult your teacher

2. (a) x = 70º (b) x = 34º (c) x = 75º, y = 105º (d) x = 56º, y = 68º

(e) x = 53º (f) x = 35º, y = 60º (g) x = 30º

3. (a) x = 70º, y = 30º, z = 80º (b) x = 110º, y = 75º, z = 75º (c) x = 70º(d) x = 55º

(e) x = 70º (f) x = 120º (g) x = 90º

4. (a) x = 72º , y = 32º, z = 23º (b) x = 80º, y = 40º, z = 80º c) x = 65º, y = 20º

(d) x = 115º, y = 65º (e) x = 60º, y = 20º (f) x = 50º, y = 95º

5. (a) 115º (b) 60º

Oasis School Mathematics-9 185

11.3 Theorem on the relation between three sides and angles of a triangle

Theorem-11.2

The sum of two sides of a triangle is greater than the third side.

Experimental verification:

Draw two triangles of different shapes and sizes with the help of pencil and scale.

A A

B C BC
Figure (i) Figure (ii)

To verify : AB + BC > AC, AB + AC > BC

AC + BC > AB

Measure all the sides and tabulate in the given table.

Observation:

Figure AB BC AC AB +BC AB + AC AC + BC Remarks

(i) ....cm ....cm ....cm ....cm ....cm ....cm AB + BC > AC

AB + AC > BC

AC + BC > AB

(ii) ....cm ....cm ....cm ....cm ....cm ....cm

Conclusion: Hence, the sum of two sides of a triangle is greater than its third side.

Theorem-11.3

The side opposite to the greatest angle of a triangle is longest and the side

opposite to smallest angle is the shortest.

Experimental Verification:

Draw two triangles of different shape and size.

B
A

BC AC
Figure (ii)
Figure (i)

To verify : Side opposite to the greatest angle is the longest and side opposite to the

smallest angle is the shortest one.

Measure all the angles and sides. Tabulate in the given table.

186 Oasis School Mathematics-9

Fig. No. ∠A BC ∠B AC ∠C AB Remarks
(i) .......0 .......0 .......0 .......0 .......0 .......0
(ii) .......0 .......0 .......0 .......0 .......0 .......0

Conclusion: The side opposite to the greatest angle is the longest and vice versa.

Converse: In a triangle, angle opposite to longest side is the greatest. Following the
same method we can verify the converse.

Activity :

Take match sticks of 2cm, 3cm, 4cm, 5cm, 6cm.
• Make a triangular shapes using the match stick of
(i) 2cm, 3cm and 4cm
(ii) 2cm, 3cm and 6cm
(iii) 4cm, 5cm and 6cm
Is it possible to makes triangle in each case? If not, find out the reasons behind it.



Worked Out Examples

Example: 1 B

In ∆ ABC, ∠A = 50º, ∠B = 57º. Find the longest and shortest sides. 57º

Solution:

Here, In ∆ABC, ∠A = 50º, ∠B = 57º 50º

Now, ∠A + ∠B + ∠C = 180º° (Sum of angles of a triangle) A C

50º + 57º + ∠C = 180º

or, ∠C = 180º – 107º

= 73º

Here, ∠C >∠B >∠A

So, the longest side is AB (opposite to the greatest angle) and the shortest side is BC
(opposite to the smallest angle).

Example: 2

In the given figure, if AB > AC and D is any point on side BC of ∆ABC A
C
Prove that AB > AD. B D
Given: In ∆ABC, AB> AC and D is a point on BC.
To prove : AB > AD.

Proof:

S.N. Statements Reasons

1 AB > AC 1. Given

2. ∠ACB > ∠ABC 2. Being angle opposite to longer side is the greater one

Oasis School Mathematics-9 187

3. ∠ADB > ∠ACD 3. In ∆ADC, exterior angle is greater than non adjacent
interior angles

4 ∠ADB > ∠ABD 4. From 2 and 3

5 AB > AD 5. Side opposite to greater angle is longer

Hence, proved

Exercise 11.2


1. (a) How to identify the longest side and the shortest side of a triangle if its all three

angles are given?
(b) How to identify the largest and the smallest angle of triangle if its all three sides

are given?

2. From the given figure, name the longest and shortest sides.

(a) A (b) A (c) P
105º xº
2x

B 30º C B xC Q 3x x R
2

3. Find the smallest and greatest angles from the given figures.

(a) A (b) A (c) X

6cm 10cm 7cm
5cm
5cm
6cm

B C B CY 10cm Z
7cm 8cm

4. (a) In ∆ ABC, ∠A = 80º, ∠B = 60º. Arrange the sides in the ascending order of
magnitude.

(b) In ∆ PQR, PQ = 8.5 cm, QR = 6.5 cm and PR = 7 cm. Arrange the sides in the
descending order of magnitude.

5. (a) Is it possible to construct a triangle having sides 4.5 cm, 8 cm and 2.5 cm? Justify
your answer.

(b) Is it possible to construct a triangle having sides 5 cm, 6 cm and 7 cm? Justify
your answer.
A

6. In the given figure AB > AC, BD and DC are the bisectors D
of ∠B and ∠C respectively. Prove that: BD > DC.

BC

7. In ∆ABC, ∠A + ∠B = 140° and ∠B + ∠C = 85°. Write the sides of
the triangle in ascending order according to their lengths.

188 Oasis School Mathematics-9

8. In the figure, BO and CO are the bisectors of ∠B A
and ∠C of ∆ABC. O

Prove that : AB + BC + ∠AC= 2(OA+OB+OC). B DC



Answer

1. Consult your teacher 2. (a) Longest side is BC, Shortest side is AB
(b) Longest side is AC, Shortest side is AB (c) Longest side is PR, Shortest side is PQ
3. (a) Greatest angle is ∠A, Smallest angle is∠C (b) Greatest angle is ∠B, Smallest angle is ∠C
(c) Greatest angle is ∠X, Smallest angle is ∠Z
4. (a) AB < AC < BC b) ∠R > ∠Q > ∠P 5. Consult your teacher. 6. AB < AC < BC

11.4 Congruent Triangles AD

Let's observe ∆ABC and ∆DEF.

Both are similar in shape.

Both are equal in size. B CE F

Then they are equal in area.

Hence, ∆ABC and ∆PQR are congruent triangles.

Condition of congruency: AD

I. ASA Axiom

In ∆ABC and ∆DEF

∠A = ∠D (A), AB = DE (S), ∠B = ∠E (A). B CE F

Angle – Side – Angle of ∆ABC are equal to Angle – Side –Angle in order of ∆DEF.

∆ABC is congruent to ∆DEF.

Mathematically, it is written as ∆ABC ≅ ∆DEF.

This axiom is known as ASA axiom.

How to make corresponding sides and angles equal?
Side opposite to ∠A is BC

Side opposite to ∠D is EF I understand !
aaasrnirdeegeclseocsororroperpesppspopoosonistnidetdeitniontgogeaqenqsugiuadlelaeslas.nsiagdnleedss
BC = EF [sides opposite to ∠A and ∠D]

Similarly, AC = DF [sides
opposite to ∠B and ∠E]

∠C = ∠F [Angles opposite
to AB and DE]

Oasis School Mathematics-9 189

II. SAS Axiom

In ∆ABC and ∆DEF,

AB = DE(S), ∠B = ∠E (A) and BC = EF (S)

Side – Angle – Side of ∆ ABC are equal to Side – Angle – Side of ∆DEF.

∆ABC ≅ ∆DEF by SAS axiom. D
How to make corresponding sides and angles equal? A

Angle opposite to AB is ∠C

Angle opposite to DE is ∠F B CE F

∴∠C = ∠F [Corresponding angles of congruent triangles]

Guess the next corresponding angles.

Side opposite to ∠B is AC

Side opposite to ∠E is DF

∴ AC = DF (Corresponding sides of congruent triangles)

III. S.S.S. Axiom

In ∆ABC and ∆DEF, AB = DE (S) AC = DF (S) and BC = EF (S)

Side – Side – Side of ∆ABC are equal to A D

Side – Side – Side of ∆DEF.

∴∆ ABC ≅ ∆DEF by S.S.S. axiom.

B CE F

∠C = ∠F, ∠A = ∠D and ∠B = ∠E are corresponding angles of congruent triangles

ABC and DEF.

IV. R.H.S. Axiom AD

In right angled ∆ABC and ∆DEF

∠ABC = ∠DEF (Right angles)

AC = DF (Hypotenuse), BC = EF (Sides) B CE F

Right angle – Hypotenuse – Side of ∆ABC are equal to Right angle – Hypotenuse –
Side of ∆DEF

∴ ∆ ABC ≅ ∆DEF by R.H.S. axiom.

Their corresponding sides and angles are

AB = DE ∠A = ∠D ∠C = ∠F AD

V. AAS Axiom B CE F
In ∆ ABC and ∆DEF
∠A = ∠ D (A) ∠B = ∠E (A) BC = EF (S )

190 Oasis School Mathematics-9

Angle – Angle – Side of ∆ABC are equal to
Angle – Angle – Side of ∆DEF.
∴ ∆ABC ≅ ∆DEF by A.A.S. axiom.
Their corresponding sides and angles are AB = DE, AC = DF, and ∠C = ∠F

Note: A.A.S. axiom also written as S.A.A. axiom.

Worked Out Examples

Example: 1
Make the following pair of triangles congruent and identify the corresponding sides
and angles of the triangle.

AP

B CQ R

Given : In ∆ABC and ∆PQR, AB = PQ, BC = QR, ∠B = ∠Q.
To prove : ∆ABC ≅ ∆PQR
To find : Corresponding sides and angles.

Proof.

Statements Reasons
1. In ∆ABC and ∆PQR 1.
(i) AB = PQ (S)       (i) Given
(ii) ∠ABC = ∠PQR (A) (ii) Given

(iii) BC = QR (S) (iii) Given
2. ∆ABC ≅ ∆PQR 2. By S.A.S. axiom
3. ∠C = ∠R, ∠A = ∠P 3. Angle opposite to the equal sides of

4. AC = PR congruent triangles
4. Sides opposite to the equal angles of

congruent triangles

Exam ple: 2 A Hence, proved.
B
Make the following pair of triangles congruent and PQ
make their corresponding sides and angles equal.
Given : In ∆ ABC and ∆ PQR CR
AB = PQ, ∠B = ∠Q = 90º, AC = PR

Oasis School Mathematics-9 191

To prove : ∆ABC ≅ ∆PQR and to make corresponding sides and angles equal.
Proof.

Statements Reasons
1. In ∆ABC and∆PQR 1.

(i) ∠AB C= ∠PQR (R) (i) Given

(ii) AC = PR (H) (ii) Given

(iii) AB = PQ (S) (iii) Given

2. ∆ABC ≅ ∆PQR 2. By R.H.S. axiom

3. BC = QR 3. Corresponding sides of congruent triangles

4. ∠BAC = ∠QPR and ∠ACB = ∠PRQ 4. Corresponding angles of congruent triangles
Hence, proved.

Exercise 11.3

1. (a) Write all axioms under which two given triangles are congruent.

(b) Are two congruent triangles equal in area? P

(c) In the given figure, ∆PQR = ∆PSQ + ∆PSR, what is this axiom R
called?

QS

(d) How to take corresponding sides and corresponding angles

of congruent triangles?

2. Make the following pairs of triangles congruent and also make their corresponding

sides and angles equal. MP

AP

(a) (b)

B CQ RN OQ R
A CF DP RZ X

(c) (d) Y

B PY EQ
Z
(e)

QR X

192 Oasis School Mathematics-9

3. Make the following pairs of triangles congruent and hence, find the size of x, y

and z. AD

(a) A P (b) x

25º

B 55º y cm x CQ z 8 cm 65º R B 55º z y 100º F
CE



(c) P y cm E (d) C R
F 35º
55º

z 7 cm x cm

Q xR D Ay B P 65º Q
6.5 cm 4.5 cm
z cm

4. (a) In the given figure, make ∆ABD ≅ ∆ACD. A C
B D
A

(b) In the given figure, make ∆ABD ≅ ∆ADC.

B C
D P

A

(c) In the given figure, make ∆ABP ≅ ∆CDP. C

B

A B D

(d) In the given figure, make ∆ACD ≅ ∆ABD. D
C D
A
(e) In the given figure, make ∆ABC ≅ ∆DBC.

BC

A C
D
5. (a) In the given figure, AO = OD and BO = OC. Prove that,

(i) ∆ABO ≅ ∆COD B O

(ii) AB = CD


Oasis School Mathematics-9 193

(b) In the given figure, AB = AC and BD = DC, prove that, A

(i) ∆ABD ≅ ∆ADC

(ii) ∠ADB = ∠ADC BC
D

(c) In the given figure, AB = DC and AB || DC. Prove that, A B
C
(i) ∆ABC ≅ ∆ADC

(ii) AD = BC D



Answer

1. Consult your teacher 2. Consult your teacher.

3. (a) x = 65º, y = 8 cm, z = 55 º (b) x = 25º, y = 55º, z = 100º

(c) x = 35º, y = 6.5 cm, z = 55º (d) x = 7 cm, y = 65º, z = 4.5 cm

4. Consult your teacher. 5. Consult your teacher.

11.5 Types of Triangles C

I. According to side: 3cm
4cm
(a) Scalene triangle : If all the three sides of a triangle are 2cm 3cm 2cm
A
different in length, then it is a scalene triangle. C C
4cm C
In the figure alongside, AB, BC and AC are of different length.

∴ ∆ABC is a scalene triangle

(b) Isosceles triangle: If any two sides of a triangle are equal, then 3cm 2cm
it is an isosceles triangle.

In the figure alongside, AB = AC = 3cm

∴ ∆ABC is an isosceles triangle. B

(c) Equilateral triangle: If all the sides of a triangle are equal, A
then it is an equilateral triangle.
2cm
In the figure, AB = BC = AC.

∴ ∆ABC is an equilateral triangle. B C

Note: Each angle of an equilateral triangle is 60°.

Altitude and Median of a Triangle A

Altitude: In the given ∆ABC, AD ⊥ BC, BE ⊥ AC and CF ⊥ AB F E

AD, BE and CF are altitudes of ∆ABC. BDC


194 Oasis School Mathematics-9