## ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 9 Linear Equations and Inequalities Ex 9.1

Solve the following (1 to 9) equations:

Question 1.

(i) 2 (3 – 2x) = 13

(ii) \(\frac { 3 }{ 5 }\) y – 2 = \(\frac { 7 }{ 10 }\)

Solution:

(i) 2 (3 – 2x) = 13 (Remove group symbol)

6 – 4x = 13 (Transposing 6 to R.H.S.)

-4x = 13 – 6

-4x = 7

x = \(\frac { 7 }{ -4 }\)

Question 2.

Solution:

Question 3.

(i) 7(x – 2) = 2 (2x – 4)

(ii) 21 – 3(x – 7) = x + 20

Solution:

(i) 7(x – 2) = 2 (2x – 4)

Removing group symbols,

7x – 14 = 4x – 8

7x – 4x = -8 + 14

(Transposing 4x to L.H.S. and -14 to R.H.S.)

3x = 6

x = 2.

(ii) 21 – 3 (x – 7) = x + 20

Removing group symbols,

21 – 3x + 21 = x + 20

⇒ 42 – 3x = x + 20

⇒ -3x – x = 20 – 42

(Transposing x to L.H.S. and 42 to R.H.S.)

⇒ -4x = -22

⇒ x = \(\frac { 22 }{ 4 }\) = \(\frac { 11 }{ 2 }\) = 5\(\frac { 1 }{ 2 }\)

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

x = 6

Question 7.

Solution:

Question 8.

(i) y + 1.2y = 4.4

(ii) 15% of x = 21

Solution:

Question 9.

(i) 2p + 20% of (2p – 1) = 7

(ii) 3 (2x – 1) + 25% of x = 97

Solution:

Question 10.

Find the value of p if the value of x^{4} – 3x^{3} – px – 5 is equal to 23 when x = -2.

Solution:

x = -2

x^{4} – 3x^{3} – px – 5 = 23

⇒ (-2)^{4} – 3(-2)^{3} – p(-2) – 5 = 23

⇒ 16 + 24 + 2p – 5 = 23

⇒ 35 + 2p = 23

⇒ 2p = 23 – 35 = -12

⇒ p = -6