# ML Aggarwal Solutions for Chapter 5 Simultaneous Linear Equation Class 9 Maths ICSE

**Exercise 5.1**

**Solve the following systems of simultaneous linear equations by the substitution method (1 to 4):**

**1. (i) x + y = 14**

**x – y = 4**

**(ii) s – t = 3**

**s/3 + t/2 = 6**

**(iii) 2x + 3y = 9**

**3x + 4y = 5**

**(iv) 3x – 5y = 4**

**9x – 2y = 7**

**Solution**

**(i)** x + y = 14

x – y = 4

It can be written as

x = 4 + y

By substituting the value in the above equation

4 + y + y = 14

By further calculation

2y = 14 – 4 = 10

Dividing by 2

y = 10/2 = 5

So we get

x = 4 + 5 = 9

Hence, x = 9 and y = 5.

**(ii)** s – t = 3

s/3 + t/2 = 6

By taking LCM

2s + 3t = 6×6 = 36

We know that

s – t = 3 **…(1)**

2s + 3t = 36 **…(2)**

So we get

s = 3 + t **…(3)**

By substituting the value of s in equation (2)

2(3 + t) + 3t = 36

By further calculation

6 + 2t + 3t = 36

So we get

5t = 36 – 6 = 30

By division

t = 30/5 = 6

Substituting t in equation (3)

s = 3 + 6 = 9

Hence, s = 9 and t = 6.

**(iii)** 2x + 3y = 9 **…(1)**

3x + 4y = 5 **…(2)**

Equation (1) can be written as

2x = 9 – 3y

x = (9 – 3y)/2 **...(3)**

By substituting the value of x in equation (2)

3×(9 – 3y)/ 2 + 4y = 5

By further calculation

(27 – 9y)/2 + 4y = 5

By taking LCM

27 – 9y + 8y = 10

So we get

-y = – 17

y = 17

Substituting y in equation (3)

x = [9 – (3×17)]/2

By further calculation

x = (9 – 51)/ 2

x = – 21

Hence, x = – 21 and y = 17.

**(iv)** 3x – 5y = 4 **…(1)**

9x – 2y = 7 **…(2)**

Multiply equation (1) by 3

9x – 15y = 12

9x – 2y = 7

By subtracting both the equations

– 13y = 5

y = -5/13

Equation (1) can be written as

3x – 5y = 4

x = (4 + 5y)/3 **…(3)**

By substituting the value of x in equation (2)

9 [(4 + 5y)/3] – 2y = 7

By further calculation

12 + 15y – 2y = 7

13y = – 5

So we get

y = -5/13

Substituting y in equation (3)

Hence, x = 9/13 and y = – 5/13.

**2. (i) a + 3b = 5**

**7a – 8b = 6**

**(ii) 5x + 4y – 4 = 0**

**x – 20 = 12y**

**Solution**

**(i)** a + 3b = 5 **…(1)**

7a – 8b = 6 **…(2)**

Now multiply equation (1) by 7

7a + 21b = 35 **…(3)**

7a – 8b = 6 **…(4)**

By subtracting both the equations

29b = 29

So we get

b = 29/29 = 1

Now substituting b = 1 in equation (1)

a + 3 (1) = 5

By further calculation

a + 3 = 5

So we get

a = 5 – 3 = 2

Therefore, a = 2 and b = 1.

**(ii)** 5x + 4y – 4 = 0

x – 20 = 12y

We can write it as

5x + 4y = 4 **…(1)**

x – 12y = 20 **…(2)**

Now multiply equation (2) by 5

5x + 4y = 4 **…(3)**

5x – 60y = 100

By subtracting both the equations

64y = –96

So we get

y = -96/64 = –3/2

Now substitute the value of y in equation (1)

5x + 4 (-3/2) = 4

By further calculation

5x + 2 (-3) = 4

So we get

5x – 6 = 4

⇒ 5x = 4 + 6 = 10

By division

x = 10/5 = 2

Therefore, x = 2 and y = – 3/2.

**3. (i) 2x – 3y/4 = 3**

**5x – 2y – 7 = 0**

**(ii) 2x + 3y = 23**

**5x – 20 = 8y**

**Solution**

**(i)** 2x – 3y/4 = 3

5x – 2y – 7 = 0

We can write it as

2x/1 – 3y/4 = 3

By taking LCM

(8x – 3y)/ 4 = 3

By cross multiplication

8x – 3y = 12 **…(1)**

5x – 2y = 7 **…(2)**

Now multiply equation (1) by 2 and (2) by 3

16x – 6y = 24

15x – 6y = 21

By subtracting both the equations

x = 3

Now substituting the value of x in equation (1)

8×3 – 3y = 12

By further calculation

24 – 3y = 12

⇒ –3y = 12 – 24

So we get

– 3y = – 12

⇒ y = – 12/-3 = 4

Therefore, x = 3 and y = 4.

**(ii)** 2x + 3y = 23

5x – 20 = 8y

We can write it as

2x + 3y = 23 **…(1)**

5x – 8y = 20 **…(2)**

By multiplying equation (1) by 5 and equation (2) by 2

10x + 15y = 115

10x – 16y = 40

By subtracting both the equations

31y = 75

So we get

y = 75/31 = 2 13/31

By substituting the value of y in equation (1)

2x + 3 (75/31) = 23

By further calculation

2x + 225/31 = 23

We can write it as

2x = 23/1 – 225/31

Taking LCM

2x = (713 – 225)/31 = 488/31

So we get

x = 488/ (31×2) = 244/31 = 7 27/31

Therefore, x = 7 27/31 and y = 2 13/31.

**4. (i) mx – ny = m ^{2} + n^{2}**

**x + y = 2m**

**(ii) 2x/a + y/b = 2**

**x/a – y/b = 4**

**Solution**

**(i)** mx – ny = m^{2} + n^{2} **…(1)**

x + y = 2m **…(2)**

We can write it as

x = 2m – y **…(3)**

Now substitute the value of x in (1)

m(2m – y) – ny = m^{2} + n^{2}

By further calculation

2m^{2} – my – ny = = m^{2} + n^{2}

Taking out y as common

m^{2} – y(m + n) = n^{2}

It can be written as

m^{2} – n^{2} – y(m + n) = 0

Expanding using formula

(m – n) (m + n) – y(m + n) = 0

Taking (m + n) as common

(m + n)[(m – n) – y] = 0

So we get

m – n – y = 0

y = m – n

From equation (3)

x = 2m – (m – n)

By further calculation

x = 2m – m + n = m + n

Hence, x = m + n and y = m – n.

**(ii)** 2x/a + y/b = 2 **…(1)**

x/a – y/b = 4 **…(2)**

Adding both the equations

3x/a = 6

So we get

x = 6a/3 = 2a

Substituting x in equation (1)

2 (2a)/ a + y/b = 2

By further calculation

4a/a + y/b = 2

So we get

4 + y/b = 2

y/b = 2 – 4 = –2

Here

y = –2b

Therefore, x = 2a and y = –2b.

**5. Solve 2x + y = 35, 3x + 4y = 65. Hence, find the value of x/y.**

**Solution**

It is given that

2x + y = 35 **…(1)**

3x + 4y = 65 **…(2)**

Now multiply equation (1) by 4

8x + 4y = 140 **…(3)**

3x + 4y = 65 **…(4)**

By subtracting both the equations

5x = 75

⇒ x = 75/5 = 15

Now substituting the value of x in equation (1)

8×15 + 4y = 140

By further calculation

120 + 4y = 140

⇒ 4y = 140 – 120

So we get

4y = 20

⇒ y = 20/4 = 5

Here

x/y = 15/5 = 3

Therefore, x/y = 3.

**6. Solve the simultaneous equations 3x – y = 5, 4x – 3y = –1. Hence, find p, if y = px –3.**

**Solution**

It is given that

3x – y = 5 **…(1)**

4x – 3y = –1 **…(2)**

Now multiply equation (1) by 3

9x – 3y = 15 **…(3)**

4x – 3y = –1** …(4)**

Subtracting equation (3) and (4)

5x = 16

⇒ x = 16/5

Substitute the value of x in equation (3)

3× 16/5 – y = 5

By further calculation

48/5 – y = 5

⇒ 48/5 – 5 = y

Taking LCM

(48 – 25)/5 = y

So we get

y = 23/5

We know that

y = px – 3

⇒ 23/5 = p× 16/5 – 3

Substitute the value of x and y

23/5 + 3 = 16p/5

Taking LCM

(23 + 15)/5 = 16p/5

By further calculation

38/5 = 16p/5

So we get

16p = 38

⇒ p = 19/8

Therefore, x = 16/5, y = 23/5 and p = 19/8.

**Exercise 5.2**

**Solve the following systems of simultaneous linear equations by the elimination method (1 to 9):**

**1. (i) 3x + 4y = 10**

**2x – 2y = 2**

**(ii) 2x = 5y + 4**

**3x – 2y + 16 = 0**

**Solution**

**(i)** 3x + 4y = 10 **…(1)**

2x – 2y = 2 **…(2)**

Multiplying equation (1) by 1 and (2) by 2

3x + 4y = 10

4x – 4y = 4

By adding both the equations

7x = 14

By division

x = 14/7 = 2

Substituting the value of x in equation (2)

2×2 – 2y = 2

By further calculation

4 – 2y = 2

So we get

2y = 4 – 2 = 2

⇒ y = 2/2 = 1

Therefore, x = 2 and y = 1.

**(ii)** 2x = 5y + 4

3x – 2y + 16 = 0

We can write it as

2x – 5y = 4 **…(1)**

3x – 2y = – 16 **…(2)**

Now multiply equation (1) by 3 and (2) by 2

6x – 15y = 12 **…(3)**

6x – 4y = – 32 **…(4)**

By subtracting both the equations

–11y = 44

⇒ y = -44/11 = – 4

Substitute the value of y in equation (1)

2x – 5(-4) = 4

By further calculation

2x + 20 = 4

So we get

2x = 4 – 20 = –16

⇒ x = –16/2 = –8

Therefore, x = –8 and y = –4.

**2. (i) ¾ x – 2/3 y = 1**

**3/8 x – 1/6 y = 1**

**(ii) 2x – 3y – 3 = 0**

**2x/3 + 4y + ½ = 0.**

**Solution**

**(i)** ¾ x – 2/3 y = 1

3/8 x – 1/6 y = 1

We can write it as

¾ x – 2/3 y = 1

(9x – 8y)/12 = 1

By cross multiplication

9x – 8y = 12** …(1)**

3/8 x – 1/6 y = 1

⇒ (9x – 4y)/24 = 1

By cross multiplication

9x – 4y = 24 **…(2)**

Subtracting equations (1) and (2)

–4y = –12

By division

y = –12/–4 = 3

Substitute the value of y in (1)

9x – 8×3 = 12

By further calculation

9x – 24 = 12

⇒ 9x = 12 + 24 = 36

By division

x = 36/9 = 4

Therefore, x = 4 and y = 3.

**(ii)** 2x – 3y – 3 = 0

2x/3 + 4y + ½ = 0

We can write it as

2x – 3y – 3 = 0

⇒ 2x – 3y = 3 **…(1)**

⇒ 2x/3 + 4y + ½ = 0

⇒ 2x/3 + 4y = – ½

Taking LCM

(2x + 12y)/ 3 = – ½

By cross multiplication

2 (2x + 12y) = –1×3

So we get

4x + 24y = – 3 **…(2)**

Multiply equation (1) by 2

4x – 6y = 6

4x + 24y = –3

By subtracting both the equations

–30y = 9

So we get

y = -9/30 = –3/10

Substitute the value of y in equation (1)

2x – 3(-3/10) = 3

By further calculation

2x + 9/10 = 3

We can write it as

2x = 3 – 9/10

By taking LCM

2x = (30 – 9)/ 10

So we get

2x = 21/10

⇒ x = 21/20

Therefore, x = 21/20 and y = –3/10.

**3. (i) 15x – 14y = 117**

**14x – 15y = 115**

**(ii) 41x + 53y = 135**

**53x + 41y = 147.**

**Solution**

**(i)** 15x – 14y = 117 **…(1)**

14x – 15y = 115 **…(2)**

Now multiply equation (1) by 14 and (2) by 15

210x – 196y = 1638 **…(3)**

210x – 225y = 1725 **…(4)**

By subtracting both the equations

29y = –87

So we get

y = -87/29 = –3

Substitute the value of y in equation (1)

15x – 14(-3) = 117

By further calculation

15x + 42 = 117

So we get

15x = 117 – 42 = 75

By division

x = 75/15 = 5

Therefore, x = 5 and y = – 3.

**(ii)** 41x + 53y = 135 **…(1)**

53x + 41y = 147 **…(2)**

Now multiply equation (1) by 53 and (2) by 41

2173x + 2809y = 7155 **…(3)**

2173x + 1681y = 6027 **…(4)**

By subtracting both the equations

1128y = 1128

So we get

y = 1128/1128 = 1

Substitute the value of y in equation (1)

41x + 53×1 = 135

By further calculation

41x + 53 = 135

So we get

41x = 135 – 53 = 82

By division

x = 82/41 = 2

Therefore, x = 2 and y = 1.

**4. (i) x/6 = y – 6**

**3x/4 = 1 + y**

**(ii) x – 2/3 y = 8/3**

**2x/5 – y = 7/5.**

**Solution**

**(i)** x/6 = y – 6

3x/4 = 1 + y

We can write it as

x = 6 (y – 6)

⇒ x = 6y – 36

⇒ x – 6y = –36 **…(1)**

⇒ 3x/4 = 1 + y

By cross multiplication

3x = 4 (1 + y)

So we get

3x = 4 + 4y

⇒ 3x – 4y = 4 **...(2)**

Multiply equation (1) by 3

3x – 18y = – 108

⇒ 3x – 4y = 4

Subtracting both the equations

– 14y = – 112

So we get

y = – 112/-14 = 8

Substitute the value of y in equation (1)

x – 6×8 = – 36

By further calculation

x – 48 = – 36

⇒ x = – 36 + 48

⇒ x = 12

Therefore, x = 12 and y = 8.

**(ii)** x – 2/3 y = 8/3

⇒ 2x/5 – y = 7/5

We can write it as

x – 2/3 y = 8/3

Taking LCM

(3x – 2y)/3 = 8/3

By cross multiplication

3x – 2y = 8/3 ×3 = 8

⇒ 3x – 2y = 8 **…(1)**

⇒ 2x/5 – y = 7/5

Taking LCM

(2x – 5y)/ 5 = 7/5

By cross multiplication

2x – 5y = 7/5 ×5 = 7

⇒ 2x – 5y = 7 **...(2)**

Multiply equation (1) by 2 and (2) by 3

6x – 4y = 16 **…(3)**

6x – 15y = 21 **…(4)**

Subtracting both the equations

11y = –5

⇒ y = –5/11

Substitute the value of y in equation (1)

3x – 2(-5/11) = 8

By further calculation

3x + 10/11 = 8

We can write it as

3x = 8 – 10/11

Taking LCM

3x = (88 – 10)/11 = 78/11

By cross multiplication

x = 78/(11×3) = 26/11

Therefore, x = 26/11 and y = –5/11.

**5. (i) 9 – (x – 4) = y + 7**

**2 (x + y) = 4 – 3y**

**(ii) 2x + (x – y)/ 6 = 2**

**x – (2x + y)/3 = 1.**

**Solution**

**(i)** 9 – (x – 4) = y + 7

⇒ 2(x + y) = 4 – 3y

We can write it as

9 – (x – 4) = y + 7

⇒ 9 – x + 4 = y + 7

By further calculation

13 – x = y + 7

⇒ – x – y = 7 – 13 = –6

⇒ x + y = 6 **…(1)**

⇒ 2(x + y) = 4 – 3y

⇒ 2x + 2y = 4 – 3y

By further calculation

2x + 2y + 3y = 4

So we get

2x + 5y = 4 **…(2)**

Now multiply equation (1) by 5 and (2) by 1

5x + 5y = 30

⇒ 2x + 5y = 4

By subtracting both the equations

3x = 26

So we get

x = 26/3

Substitute the value of x in (1)

26/3 + y = 6

We can write it as

y = 6 – 26/3

Taking LCM

y = (18 – 26)/3

So we get

y = – 8/3

Therefore, x = 26/3 and y = – 8/3.

**(ii)** 2x + (x – y)/6 = 2

⇒ x – (2x + y)/3 = 1

⇒ 2x + (x – y)/6 = 2

Multiply by 6

12x + x – y = 12

By further calculation

13x – y = 12 **...(2)**

⇒ x – (2x + y)/3 = 1

Multiply by 3

3x – 2x – y = 3

By further calculation

x – y = 3 **…(2)**

So we get

x = 3 + y **…(3)**

Substitute the value of x in (1)

13 (3 + y) – y = 12

By further calculation

39 + 13y – y = 12

So we get

12y = 12 – 39 = – 27

By division

y = –27/12 = – 9/4

Substitute the value of y in (3)

x = 3 + y

⇒ x = 3 + (-9)/4

By further calculation

x = 3 – 9/4

Taking LCM

x = (12 – 9)/ 4

⇒ x = ¾

Therefore, x = ¾ and y = –9/4.

**6. x – 3y = 3x – 1 = 2x – y.**

**Solution**

It is given that

x – 3y = 3x – 1 = 2x – y

Here,

x- 3y = 3x – 1

⇒ x – 3x – 3y = –1

By further calculation

⇒ –2x – 3y = –1

⇒ 2x + 3y = 1 **…(1)**

3x – 1 = 2x – y

⇒ 3x – 2x + y = 1

By further simplification

⇒ x + y = 1 **…(2)**

Multiply equation (2) by 2 and subtract from equation (1)

2x + 3y = 1

2x + 2y = 2

So we get

y = –1

Substitute the value of y in equation (1)

2x + 3(-1) = 1

So we get

2x – 3 = 1

⇒ 2x = 1 + 3 = 4

By division

⇒ x = 4/2 = 2

Therefore, x = 2 and y = –1.

**7. (i) 4x + (x – y)/8 = 17**

**2y + x – (5y + 2)/3 = 2**

**(ii) (x + 1)/2 + (y – 1)/3 = 8**

**(x – 1)/3 + (y + 1)/ 2 = 9.**

**Solution**

**(i)** 4x + (x–y)/8 = 17

2y + x – (5y+2)/3 = 2

We can write it as

4x + (x–y)/8 = 17

⇒ (32 + x – y)/8 = 17

By further calculation

⇒ (33x – y)/8 = 17

By cross multiplication

⇒ 33x – y = 136 **…(1)**

2y + x – (5y + 2)/3 = 2

Taking LCM

⇒ [3(2y + x) – 5(5y + 2)]/3 = 2

By further calculation

6y + 3x – 5y – 2 = 2×3

So we get

y + 3x – 2 = 6

⇒ 3x + y = 6 + 2

⇒ 3x + y = 8 **…(2)**

By adding both the equations

36x = 144

By division

x = 144/36 = 4

Substitute the value of x in equation (1)

33×4 – y = 136

By further calculation

132 – y = 136

⇒ –y = 136 – 132

So we get

–y = 4

⇒ y = –4

Therefore, x = 4 and y = –4.

**(ii)** (x + 1)/2 + (y – 1)/3 = 8

(x – 1)/3 + (y + 1)/2 = 9

We can write it as

(x + 1)/2 + (y – 1)/3 = 8

Taking LCM

⇒ (3x + 3 + 2y – 2)/6 = 8

By further calculation

3x + 2y + 1 = 48

So we get

3x + 2y = 47** …(1)**

(x – 1)/3 + (y + 1)/2 = 9

Taking LCM

⇒ (2x – 2 + 3y + 3)/6 = 9

By further calculation

⇒ 2x + 3y + 1 = 54

So we get

⇒ 2x + 3y = 53 **…(2)**

By adding equation (1) and (2)

5x + 5y = 100

Dividing by 5

⇒ x + y = 20 **...(3)**

By subtracting equation (1) and (2)

x – y = –6 …(4)

Now add equation (3) and (4)

2x = 14

⇒ x = 14/2 = 7

Subtracting equation (4) and (3)

2y = 26

⇒ y = 26/2 = 13

Therefore, x = 7 and y = 13.

**8. (i) 3/x + 4y = 7**

**5/x + 6y = 13**

**(ii) 5x – 9 = 1/y**

**x + 1/y = 3.**

**Solution**

**(i)** 3/x + 4y = 7 **…(1)**

5/x + 6y = 13 **…(2)**

Substitute 1/x = a in equation (1) and (2)

3a + 4y = 7 **…(3)**

5a + 6y = 13 **…(4)**

Multiply equation (3) by 5 and (4) by 3

15a + 20y = 35

15a + 18y = 39

Subtracting both the equations

2y = -4

So we get

⇒ y = –4/2 = –2

Substitute the value of y in equation (3)

3a + 4 (-2) = 7

By further calculation

3a – 8 = 7

⇒ 3a = 7 + 8 = 15

So we get

3a = 15

⇒ a = 15/3 = 5

Here x = 1/a = 1/5

Therefore, x = 1/5 and y = –2.

**(ii)** 5x – 9 = 1/y **…(1)**

x + 1/y = 3 **…(2)**

Substitute 1/y = b in (1) and (2)

5x – 9 = b

⇒ 5x – b = 9 **…(3)**

⇒ x + b = 3 **…(4)**

By adding equation (3) and (4)

5x – b = 9 **…(3)**

x + b = 3 **…(4)**

So we get

6x = 12

By division

⇒ x = 12/6 = 2

Substitute the value of x in equation (4)

2 + b = 3

⇒ b = 3 – 2

⇒ b = 1

Here 1/y = b

b = 1/y

⇒ y = 1

Therefore, x = 2 and y = 1.

**9. (i) px + qy = p – q**

**qx – py = p + q**

**(ii) x/a – y/b = 0**

**ax + by = a ^{2} + b^{2}.**

**Solution**

(i) px + qy = p – q **…(1)**

qx – py = p + q **…(2)**

Now multiply equation (1) by p and (2) by q

p^{2}x + pqy = p^{2} – pq

q^{2}x – pqy = pq + q^{2}

By adding both the equations

(p^{2} + q^{2}) x = p^{2} + q^{2}

By further calculation

x = (p^{2}+q^{2})/(p^{2}+q^{2}) = 1

From equation (1)

p×1 + qy = p – q

By further calculation

⇒ p – qy = p – q

So we get

⇒ qy = p – q – p = –q

Here

⇒ y = -q/q = – 1

Therefore, x = 1 and y = –1.

**(ii)** x/a – y/b = 0

ax + by = a^{2} + b^{2}

We can write it as

x/a – y/b = 0

Taking LCM

(bx – ay)/ab = 0

By cross multiplication

⇒ bx – ay = 0 **...(1)**

ax + by = a^{2} + b^{2} **…(2)**

Multiply equation (1) by b and equation (2) by a

b^{2}x – aby = 0

a^{2}x + aby = a^{2} + ab^{2}

By adding both the equations

(a^{2} + b^{2})x = a^{2}+ ab^{2} = a (a^{2} + b^{2})

So we get

⇒ x = a(a^{2} + b^{2})/(a^{2}+b^{2}) = a

From equation (2)

b×a – ay = 0

By further calculation

ab – ay = 0

⇒ ay = ab

So we get

y = ab/a = b

Therefore, x = a and y = b.

**10. Solve 2x + y = 23, 4x – y = 19. Hence, find the values of x – 3y and 5y – 2x.**

**Solution**

It is given that

2x + y = 23 **…(1)**

4x – y = 19 **…(2)**

Adding both the equations

6x = 42

⇒ x = 42/6 = 7

Substitute the value of x in equation (1)

2×7 + y = 23

By further calculation

14 + y = 23

So we get

y = 23 – 14 = 9

Therefore, x = 7 and y = 9.

**x – 3y** = 7 – 3×9 = 7 – 27 = –20

**5y – 2x **= 5 × 9 – 2×7 = 45 – 14 = 31

**11. The expression ax + by has value 7 when x = 2, y = 1. When x = –1, y = 1, it has value 1, find a and b.**

**Solution**

It is given that

ax + by = 7 when x = 2 and y = 1

Substituting the values

a(2) + b(1) = 7

⇒ 2a + b = 7 **…(1)**

Here

ax + by = 1 when x = –1 and y = 1

Substituting the values

a(-1) + b(1) = 1

⇒ –a + b = 1 **…(2)**

By subtracting both the equations

–3a = –6

So we get

a = – 6/–3 = 2

Substituting the value of a in equation (1)

2×2 + b = 7

By further calculation

⇒ 4 + b = 7

⇒ b = 7 – 4 = 3

Therefore, a = 2 and b = 3.

**12. Can the following equations hold simultaneously?**

**3x – 7y = 7**

**11x + 5y = 87**

**5x + 4y = 43.**

**If so, find x and y.**

**Solution**

3x – 7y = 7 **…(1)**

11x + 5y = 87 **…(2)**

5x + 4y = 43 **…(3)**

Now multiply equation (1) by 5 and (2) by 7

15x – 35y = 35

77x + 35y = 609

By adding both the equations

92x = 644

By division

x = 644/92 = 7

Substitute the value of x in equation (1)

3×7 – 7y = 7

By further calculation

21 – 7y = 7

So we get

– 7y = 7 – 21 = – 14

⇒ y = – 14/–7 = 2

Therefore, x = 7 and y = 2.

If x = 7 and y – 2 satisfy the equation (3) then we can say that the equations hold simultaneously

Substitute the value of x and y in equation (3)

5x + 4y = 43

By further calculation

⇒ 5×7 + 4×2 = 43

So we get

35 + 8 = 43

43 = 43 which is true.

Therefore, the equations hold simultaneously.

**Exercise 5.3**

**1. Solve the following systems of simultaneous linear equations by cross-multiplication method:**

**(i) 3x + 2y = 4**

**8x + 5y = 9**

**(ii) 3x – 7y + 10 = 0**

**y – 2x = 3**

**Solution**

**(i)** 3x + 2y = 4

8x + 5y = 9

We can write it as

3x + 2y – 4 = 0

8x + 4y – 9 = 0

By cross multiplication method

x/(-18 + 20) = y/(-32 + 27) = 1/(15 – 16)

By further calculation

x/2 = y/-5 = 1/-1

So we get

x/2 = –1

⇒ x = –2

⇒ y = –5(-1) = 5

Therefore, x = – 2 and y = 5.

**(ii)** 3x – 7y + 10 = 0

y – 2x = 3

We can write it as

3x – 7y + 10 = 0

y – 2x – 3 = 0

By cross multiplication method

x/(21 – 10) = y/(-20 + 9) = 1/(3 – 14)

By further calculation

x/11 = y/-11 = 1/-11

So we get

x/11 = 1/-11

⇒ x = – 1

Similarly

y/-11 = 1/-11

⇒ y = 1

Therefore, x = –1 and y = 1.

**2. Solve the following pairs of linear equations by cross-multiplication method:**

**(i) x – y = a + b**

**ax + by = a ^{2} – b^{2}**

**(ii) 2bx + ay = 2ab**

**bx – ay = 4ab.**

**Solution**

**(i)** x – y = a + b

ax + by = a^{2} – b^{2}

We can write it as

x – y – (a + b) = 0

ax + by – (a^{2} – b^{2}) = 0

By cross multiplication method

x/[a^{2} – b^{2} + b (a + b)] = y/[-a(a+b) + a^{2} – b^{2}] = 1/(b + a)

By further calculation

⇒ x/(a^{2} – b^{2} + ab + b^{2}) = y/(-a^{2} – ab + a^{2} – b^{2}) = 1/(a + b)

So we get

⇒ x/[a(a + b)] = y/[-b(a + b)] = 1/(a + b)

⇒ x = a (a + b)/(a + b) = a

⇒ y = [-b(a + b)]/(a + b) = – b

Therefore, x = a and y = – b.

**(ii)** 2bx + ay = 2ab

bx – ay = 4ab

We can write it as

2bx + ay – 2an = 0

bx – ay – 4ab = 0

By cross multiplication method

x/(-4a^{2}b – 2a^{2}b) = y/(-2ab^{2} + 8ab^{2}) = 1/(-2ab – ab)

By further calculation

x/-6a^{2}b = y/6ab^{2}= 1/-3ab

So we get

⇒ x = -6a^{2}b/-3ab = 2a

⇒ y = 6ab^{2}/-3ab = –2b

Therefore, x = 2a and b = –2b.

**Exercise 5.4**

**Solve the following pairs of linear equations (1 to 5):**

**1. (i) 2/x + 2/3y = 1/6**

**2/x – 1/y = 1**

**(ii) 3/2x + 2/3y = 5**

**5/x – 3/y = 1.**

**Solution**

(i) 2/x + 2/3y = 1/6 **…(1)**

2/x – 1/y = 1 **…(2)**

By subtracting both the equations

5/3y = -5/6

By cross multiplication

–15y = 30

By division

y = 30/ -15 = – 2

Substitute the value of y in equation (1)

2/x + 2/(3×(-2)) = 1/6

By further calculation

2/x – 1/3 = 1/6

So we get

2/x = 1/6 + 1/3

Taking LCM

2/x = (1+2)/6 = 3/6

By cross multiplication

x = (2×6)/3 = 12/3 = 4

Therefore, x = 4 and y = –2.

**(ii)** 3/2x + 2/3y = 5 **…(1)**

5/x – 3/y = 1 **…(2)**

Multiply equation (1) by 1 and (2) by 2/9

3/2x + 2/3y = 5

10/9x – 2/3y = 2/9

By adding both the equations

(3/2 + 10/9)1/x = 5 + 2/9

Taking LCM

⇒ (27 + 20)/18 × 1/x = (45+2)/9

By further calculation

⇒ 47/18x = 47/9

By cross multiplication

⇒ x = (47×9)/(47×18) = ½

Substitute the value of x in equation (2)

5/½ – 3/y = 1

By further calculation

10 – 3/y = 1

⇒ 3/y = 10 – 1 = 9

So we get

y = 3/9 = 1/3

Therefore, x = ½ and y = 1/3

**2. (i) (7x – 2y)/ xy = 5**

**(8x + 7y)/ xy = 15**

**(ii) 99x + 101y = 499xy**

**101x + 99y = 501xy.**

**Solution**

**(i)** (7x – 2y)/xy = 5

(8x + 7y)/xy = 15

We can write it as

7x/xy – 2y/xy = 5

8x/xy + 7y/xy = 15

By further simplification

7/y – 2/x = 5 **…(1)**

8/y + 7/x = 15 **…(2)**

Now multiply equation (1) by 7 and (2) by 2

49/y – 14/x = 35

16/y + 14/x = 30

By adding both the equations

65/y = 65

So we get

⇒ y = 65/65 = 1

Substitute the value of y in equation (1)

7/1 – 2/x = 5

By further calculation

⇒ 2/x = 7–5 = 2

So we get

⇒ x = 2/2 = 1

Therefore, x = 1 and y = 1.

**(ii)** 99x + 101y = 499xy

101x + 99y = 501xy

Now divide each term by xy

99 x/xy + 101 y/xy = 499 xy/xy

101 y/xy + 99 x/xy = 501 xy/xy

By further calculation

99/y + 101/x = 499 **…(1)**

101/y + 99/x = 501 **…(2)**

By adding both the equations

200/y + 200/x = 1000

Divide by 200

1/y + 1/x = 5 **...(3)**

Subtracting both the equations

-2/y + 2/x = – 2

Divide by 2

⇒ -1/y + 1/x = – 1 **…(4)**

By adding equation (3) and (4)

2/x = 4

So we get

x = 2/4 = ½

By subtracting equation (3) and (4)

2/y = 6

So we get

⇒ y = 2/6 = 1/3

Therefore, x = ½ and y = 1/3 if x ≠ 0, y ≠ 0.

**3. (i) 3x + 14y = 5xy**

**21y – x = 2xy**

**(ii) 3x + 5y = 4xy**

**2y – x = xy.**

**Solution**

**(i)** 3x + 14y = 5xy

21y – x = 2xy

Now dividing each equation by xy of x ≠ 0, y ≠ 0

3x/xy + 14y/xy = 5xy/xy

By further calculation

3/y = 14/x = 5 **…(1)**

**(ii)** 3x + 5y = 4xy

2y – x = xy

We can write it as

3x + 5y = 4xy

– x + 2y = xy

Divide each equation by xy if x≠ 0 and y ≠ 0

3x/xy + 5y/xy = 4xy/xy

So we get

3/y + 5/x = 4 **…(1)**

-x/xy + 2y/xy = xy/xy

So we get

-1/y + 2/x = 1 **…(2)**

Now multiply equation (1) by 1 and (2) by 3

3/y + 5/x = 4

-3/y + 6/x = 3

By adding both the equations

11/x = 7

So we get

⇒ x = 11/7

Substitute the value of x in equation (2)

-1/y + 2/11/7 = 1

By further calculation

-1/y + (2×7)/11 = 1

⇒ -1/y + 14/11 = 1

We can write it as

⇒ -1/y = 1 – 14/11

Taking LCM

⇒ -1/y = (11 – 14)/11

So we get

⇒ -1/y = -3/11

By cross multiplication

⇒ -3y = –11

⇒ y = –11/-3 = 11/3

Therefore, x = 11/7 and y = 11/3.

**4. (i) 20/ (x + 1) + 4/ (y – 1) = 5**

**10/ (x + 1) – 4/ (y – 1) = 1**

**(ii) 3/ (x + y) + 2/ (x – y) = 3**

**2/ (x + y) + 3/ (x – y) = 11/3.**

**Solution**

(i) 20/(x+1) + 4/(y–1) = 5 **…(1)**

10/(x+1) – 4/(y–1) = 1 **…(2)**

Add equation (1) and (2)

30/(x + 1) = 6

By cross multiplication

⇒ 30 = 6(x+1)

By further calculation

⇒ 30/6 = x + 1

⇒ 5 = x + 1

So we get

⇒ x = 5 – 1 = 4

Substitute the value of x in equation (1)

20/(x+1) + 4/(y–1) = 5

⇒ 20/(4+1) + 4/(y–1) = 5

By further calculation

20/5 + 4/(y–1) = 5

⇒ 4 + 4/(y–1) = 5

We can write it as

⇒ 4/(y–1) = 5 – 4 = 1

⇒ 4/(y–1) = 1

By cross multiplication

⇒ 4 = 1(y–1)

So we get

⇒ 4 = y – 1

⇒ y = 4 + 1 = 5

Therefore, x = 4 and y = 5.

**(ii)** 3/(x+y) + 2/(x–y) = 3 **…(1)**

2/(x+y) + 3/(x–y) = 11/3 **…(2)**

Multiply equation (1) by 3 and (2) by 2

9/(x+y) + 6/(x–y) = 9 **…(3)**

4/(x+y) + 6/(x–y) = 22/3 **…(4)**

Subtracting both the equations

5/(x+y) = 9 – 22/3

Taking LCM

⇒ 5/(x+y) = 5/3

By cross multiplication

⇒ 5×3 = 5(x + y)

By further calculation

⇒ (5×3)/5 = x + y

⇒ x + y = (3×1)/3

⇒ x + y = 3 **...(5)**

Substitute equation (5) in (1)

3/3 + 2/(x–y) = 3

By further calculation

1 + 2/(x–y) = 3

⇒ 2/(x–y) = 3 – 1 = 2

So we get

⇒ 2/2 = x – y

Here

1 = x – y **…(6)**

We can write it as

x – y = 1

x + y = 3

By adding both the equations

2x = 4

⇒ x = 4/2 = 2

Substitute x = 2 in equation (5)

2 + y = 3

⇒ y = 3 – 2 = 1

Therefore, x = 2 and y = 1.

**5. (i) 1/2(2x+3y) + 12/7(3x–2y) = ½**

**7/(2x+3y) + 4/(3x–2y) = 2**

**(ii) 1/2(x+2y) + 5/3(3x–2y) = –3/2**

**5/ 4(x 2y) – 3/ 5(3x–2y) = 61/60.**

**Solution**

**(i)** 1/ 2(2x+3y) + 12/ 7(3x–2y) = ½

7/ (2x+3y) + 4/ (3x–2y) = 2

Consider 2x + 3y = a and 3x – 2y = b

We can write it as

1/2a + 12/7b = ½

7/a + 4/b = 2

Now multiply equation (1) by 7 and (2) by ½

7/2a + 12/b = 7/2

7/2a + 2/b = 1

Subtracting both the equations

10/b = 5/2

So we get

b = (10×2)/5 = 4

Substitute the value of b in equation (2)

7/a + 4/4 = 2

7/a + 1 = 2

So we get

⇒ 7/a = 2 – 1 = 1

⇒ a = 7

Here

2x + 3y = 7 **…(3)**

3x – 2y = 4 **…(4)**

Multiply equation (3) by 2 and (4) by 3

4x + 6y = 14

9x – 6y = 12

So we get

13x = 26

⇒ x = 26/13 = 2

Substitute the value of x in (3)

2×2 + 3y = 7

By further calculation

⇒ 4 + 3y = 7

So we get

3y = 7 – 4 = 3

⇒ y = 3/3 = 1

Therefore, x = 2 and y = 1.

**(ii)** 1/ 2(x + 2y) + 5/ 3(3x – 2y) = – 3/2

5/ 4(x + 2y) – 3/ 5 (3x – 2y) = 61/60

Consider x + 2y = a and 3x – 2y = b

1/2a + 5/3b = –3/2 **…(1)**

5/4a – 3/5b = 61/60 **…(2)**

Now, multiply equation (1) by 5/2 and (2) by (1)

5/4a + 25/6b = –15/4

5/4a – 3/5b = 61/60

Subtracting both the equations

25/6b + 3/5b = –15/4 – 61/60

Taking LCM

(125 + 18)/ 30b = (-225 – 61)/ 60

By further calculation

143/30b = –286/60

By cross multiplication

30b×(-286) = 60×143

So we get

⇒ b = (60×143)/(30×-286) = –1

Substitute the value of b in equation (1)

1/2a + 5/(3×–1) = -3/2

By further calculation

1/2a – 5/4 = –3/2

We can write it as

1/2a = –3/2 + 5/3

Taking LCM

1/2a = (-9+ 10)/ 6 = 1/6

So we get

a = 6/2 = 3

Here

x + 2y = 3 **…(3)**

3x – 2y = – 1 **…(4)**

Adding both the equations

4x = 2

⇒ x = 2/4 = ½

Substitute the value of x in equation (3)

½ + 2y = 3

By further calculation

2y = 3 – ½

Taking LCM

2y = 5/2

⇒ y = 5/(2×2) = 5/4

Therefore, x = ½ and y = 5/4.

**Chapter Test**

**Solve the following simultaneous linear equations (1 to 4):**

**1.(i) 2x – (¾)y = 3,**

**5x – 2y = 7**

**Solution**

8x-3y = 12 **…(i)**

5x-2y = 7 **…(ii)**

Multiply (i) by 5 and (ii) by 8, we get

40x-15y = 60 **...(iii)**

40x -16y = 56 **...(iv)**

Subtract (iv) from (iii), we get

y = 4

Substitute y in (i)

8x- 3×4 = 12

⇒ 8x = 12+12

⇒ 8x = 24

⇒ x = 24/8

⇒ x = 3

Hence, x = 3 and y = 4.

**(ii) 2(x-4) = 9y+2**

**x – 6y = 2**

**Solution**

2(x-4) = 9y+2

⇒ 2x-8 = 9y+2

⇒ 2x-9y = 2+8

⇒ 2x-9y = 10 **…(i)**

⇒ x-6y = 2 **…(ii)**

Multiply (ii) by 2, we get

2x -12y = 4 **…(iii)**

Subtract (iii) from (i), we get

2x-9y - (2x -12y) = 10 -4

⇒ 2x - 9y -2x +12y = 6

⇒ 0+3y = 6

⇒ 3y = 6

⇒ y = 6/3

⇒ y = 2

Substitute the value of y in (i)

2x- 9×2 = 10

⇒ 2x-18 = 10

⇒ 2x = 10+18

⇒ 2x = 28

⇒ x = 28/2

⇒ x = 14

Hence, x = 14 and y = 2.

**2. (i) 97x+53y = 177**

**53x+97y = 573**

**Solution**

Given equations are as follows.

97x+53y = 177 **…(i)**

53x+97y = 573 **…(ii)**

Multiply (i) by 53 and (ii) by 97

53(97x+53y) = 53×177

⇒ 5141x + 2809y = 9381 **…(iii)**

97(53x+97y) = 97×573

⇒ 5141x+9409y = 55581 **…(iv)**

Subtract (iv) from (iii)

(5141x+2809y) - (5141x+9409y) = 9381-55581

⇒ 5141x + 2809y - 5141x - 9409y = -46200

⇒ 0x -6600y = -46200

⇒ -6600y = -46200

⇒ y = -46200/-6600

⇒ y = 7

Substitute the value of y in (i)

97x + 53×7 = 177

⇒ 97x+371 = 177

⇒ 97x = 177-371

⇒ 97x = –194

⇒ x = -194/97

⇒ x = -2

Hence, x = -2 and y = 7.

**(ii) x+y = 5.5**

**x-y = 0.9**

**Solution**

x+y = 5.5 **…(i)**

x-y = 0.9 **…(ii)**

Adding (i) and (ii), we get

2x = 5.5+0.9

⇒ 2x = 6.4

⇒ x = 6.4/2

⇒ x = 3.2

Substitute value of x in (i)

3.2+y = 5.5

⇒ y = 5.5-3.2

⇒ y = 2.3

Hence, x = 3.2 and y = 2.3.

**3. (i) x+y = 7xy**

**2x-3y+xy = 0**

**Solution**

x + y = 7xy **…(i)**

2x - 3y + xy = 0 **...(ii)**

Divide (i) by xy, we get

Divide (ii) by xy, we get

Multiplying (iii) by 3, we get

Adding (v) and (iv), we get

Substitute value of y in (iv)

Hence x = 1/3 and y = 1/4.

**(ii) 30/(x-y) + 44/(x+y) = 10**

**40/(x-y) + 55/(x+y) = 13**

**Solution**

Multiply (i) by 4 and (ii) by 3, we get

Subtracting (iv) from (iii), we get

Substitute (v) in (i), we get

Now solve for (v) and (vi)

x+y = 11

x-y = 5

Add (v) and (vi)

2x = 16

⇒ x = 16/2 = 8

Substitute x in (v)

8+y = 11

⇒ y = 11-8

⇒ y = 3

Hence, x = 8 and y = 3.

**4. (i) ax+by = a-b**

**bx-ay = a+b**

**Solution**

ax + by = a-b **…(i)**

bx - ay = a+b **…(ii)**

Multiplying (i) by a and (ii) by b, we get

a(ax+by) = a(a-b)

⇒ a^{2}x +aby = a^{2}-ab **…(iii)**

b(bx-ay) = b(a+b)

⇒ b^{2}x -aby = ab+b^{2} **…(iv)**

Adding (iii) and (iv)

a^{2}x +aby + b^{2}x -aby = a^{2}-ab + ab+b^{2}

⇒ (a^{2}+b^{2})x = (a^{2}+b^{2})

⇒ x = (a^{2}+b^{2})/(a^{2}+b^{2})

⇒ x = 1

Substitute the value of x in (i), we get

a×1 + by = a-b

⇒ a + by = a-b

⇒ by = -b

⇒ y = -b/b

⇒ y = -1

Hence, x = 1 and y = -1.

**(ii)** **3x + 2y = 2xy**

**Solution**

3x + 2y = 2xy **…(i)**

Divide (i) by xy

Multiply (ii) by 2, we get

Subtract (iii) from (iv)

—————

Substitute y in (iii)

(3/3) + (2/x) = 2

⇒ 1+(2/x) = 2

⇒ (2/x) = 1

⇒ x = 2

Hence, x = 2 and y = 3.

**5. Solve**

**2x -(3/y) =9**

**3x + (7/y) = 2.**

**Hence find the value of k if x = ky + 5.**

**Solution**

2x -(3/y) = 9 **…(i)**

3x + (7/y) = 2 **…(ii)**

Multiply (i) by 3 and (ii) by 2, we get

6x- (9/y) = 27 **...(iii)**

6x+ (14/y) = 4 **…(iv)**

Subtracting (iv) from (iii), we get

-23/y = 23

⇒ y = 23/-23

⇒ y = -1

Substitute y in (i)

2x-(3/-1) = 9

⇒ 2x+3 = 9

⇒ 2x = 9-3

⇒ 2x = 6

⇒ x = 6/2

x = 3

Hence, x = 3 and y = -1.

Given x = ky+5

Substitute x and y in above equation.

3 = k×-1+5

⇒ 3 = -k+5

⇒ k = 5-3

⇒ k = 2

Hence the value of k is 2.

**6.** **Solve,**

**Hence find the value of 2x ^{2}-y^{2}.**

**Solution**

Let (x+y) = a

Multiply (iii) by 5

Subtracting (ii) from (iv)

Substitute x in (iii)

(1/a) -1/(2×3) = 1/30

⇒ (1/a) – (1/6) = 1/30

⇒ 1/a = (1/30)+(1/6)

⇒ 1/a = (1+5)/30

⇒ 1/a = 6/30

⇒ a = 30/6

⇒ a = 5

Substitute a in x+y = a

3+y = 5

⇒ y = 5-3

⇒ y = 2

Hence, x = 3, y = 2.

2x^{2}-y^{2} = 2×3^{2 }- 2^{2}

= 2×9 - 4

= 18-4

= 14

Hence, the value of 2x^{2}-y^{2} is 14.

**7. Can x, y be found to satisfy the following equations simultaneously ?**

**If so, find them.**

**Solution**

Multiply (i) by 5 and (ii) by 2, we get

Subtract (v) from (iv)

31/x = 95-2

⇒ 31/x = 93

⇒ x = 31/93

⇒ x = 1/3

Substitute x in (i)

(2/y)+ 5÷(1/3) = 19

⇒ (2/y)+ 5×3 = 19

⇒ (2/y) = 19-15

⇒ (2/y) = 4

⇒ y = 2/4

⇒ y = 1/2

Substitute x and y in (iii)

3×(1/3) + 8×(1/2) = 5

⇒ 1+4 = 5

The value of x and y satisfies (iii).

Hence the given equations are simultaneous.