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

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

Question 1.

(i) 2x – \(\frac {3}{4}\) y = 3, 5x – 2y = 7

Answer:

⇒ 8x – 3y = 3 × 4

⇒ 8x – 3y = 12 (1)

and 5x – 2y = 7 (2)

Multiplying equation (1) by 2 and equation (2) by 3, we get

Substituting the value of x in equation (2), we get

5 × 3 – 2y = 7

⇒ 15 – 2y = 7

⇒ 15 – 7 = 2y

⇒ 2y = 8

⇒ y = \(\frac {8}{2}\)

⇒ y = 4

Hence, x = 3, y = 4

(ii) 2(x – 4) = 9y + 2, x – 6y = 2

Answer:

2(x – 4) = 9y + 2 …(1)

x – 6y = 2 …(2)

Now, 2 (x – 4) = 9y + 2 ⇒ 2x – 8 = 9y + 2

⇒ 2x – 9y = 2 + 8 ⇒ 2x – 9y = 10

Multiplying equation (2) by 2, we get

Substituting the value of y in equation (2) we get

x – 6 × 2 = 2 ⇒ x – 12 = 2 ⇒ x = 2 + 12

⇒ x = 14

Hence, x = 14, y = 2.

Question 2.

(i) 97x + 53y = 177, 53x + 97y = 573

Answer:

97x + 53y = 177 ……..(1)

And 53x + 97y = 573 ………..(2)

Multiplying equation (1) by 53 and equation (2) by 97, we get

(ii) x + y = 5.5, x – y = 0.9

Answer:

Substituting the value of x in equation (1), we get

3.2 + y = 5.5 ⇒ y = 5.5 – 3.2 ⇒ y = 23

Hence, x = 3.2, y = 2.3

Question 3.

(i) x + y = 7xy, 2x – 3y + xy = 0

Answer:

x + y = 7xy

(ii)

Answer:

Question 4

(i) ax + by = a – b, bx – ay = a + b

Answer:

ax + by = a – b (1)

bx – ay = a + b (2)

Multiplying equa. (1), by a and equa. (2) by b, we get

(ii) 3x + 2y = 2xy

Answer:

Question 5.

Solve 2x = \(\frac{3}{y}\) =9, 3x + \(\frac{7}{y}\) = 2. Hence find the value of k if x = ky + 5.

Answer:

Given equations

Question 6.

Solve:

Hence f find the value of 2x^{2} – y^{2}.

Answer:

Let x + y = a, then

Question 7.

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

If so, then find

Answer:

which in true

Hence the given three equations are simulteneously