## ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 9 Algebra Ex 9.5

Question 1.
State which of the following are equations with a variable. In case of an equation with a variable, identify the variable.
(i) 17 + x = 5
(ii) 2b – 3 = 7
(iii) (y – 7) > 5
(iv) $$\frac{9}{3}=3$$
(v) 7 × 3 – 19 = 2
(vi) 5 × 4 – 8 = 31
(vii) 2p < 15
(viii) 7 = 11 × 5 – 12 × 4
(ix) $$\frac{3}{2} q=5$$
Solution:
(i) 17 + x= 5 Is an equation → L.H.S. = R.H.S. → Related variable x.
(ii) 2b – 3 = 7 Is an equation → L.H.S. = R.H.S. → Related variable b.
(iii) (y- 7) >5
Is not an equation → L.H.S. ≠ R.H.S.
It has no sign of equality (=).
(iv) $$\frac{9}{3}=3$$
Is an equation = L.H.S. = R.H.S.
It has no variable.
(v) 7 × 3 – 19 = 2
Is an numerical equation = L.H.S. = R.H.S. It has no variable.
(vi) 5 × 4 – 8 = 31
Is an equation = L.H.S. = R.H.S. → Related variable t.
(vii) 2p < 15
Is not an equation = L.H.S. ≠ R.H.S.
It has no sign of equality.
(viii) 7 = 11 × 5 – 12 × 4
Is an numerical equation = L.H.S. = R.H.S. It has no variable.
(ix) $$\frac{3}{2} q=5$$
Is an equation → L.H.S. = R.H.S. → Related variable q.

Question 2.
Solve each of the following equations :
(i) x + 6 = 8
(ii) 2 – x = 5
(iii) 4x = -6
(iv) $$\frac{x}{2}=5$$
(v) 2y – 3 = 2
(vi) 4 – 5y = 2
Solution:
(i) x + 6 = 8
= x = 8 – 6 ⇒ x = 2
(ii) 2 – x = 5
= -x = 5 – 2 ⇒ -x = 3 ⇒ x = – 3
(iii) 4x = -6
$$=x=\frac{-6}{4}=\frac{-3}{2}$$
(iv) $$\frac{x}{2}=5$$
= x = 5 × 2 ⇒ x = 10
(v) 2y – 3 = 2
= 2y = 2 + 3 ⇒ 2y = 5 ⇒ y = $$\frac{5}{2}$$
(vi) 4 – 5y = 2
= 4 – 2 = 5y ⇒ 5y = 2
⇒ y = $$\frac{2}{5}$$

Question 3.
Solve the following linear equations:
(i) 5(x + 1) = 25
(ii)2(3x – 1) = 10
(iii) $$\frac{3 x-1}{4}=11$$
Solution:
(i) Givrn 5(x + 1) = 25
⇒ $$\frac{5(x+1)}{5}=\frac{25}{5}$$ (dividing both sides by 5)
⇒ x + 1 = 5
⇒ x + 1 – 1 = 5 – 1 (Subtracting 1 from both sides)
⇒ x = 4

(ii) 2(3x – 1) = 10
⇒ $$\left(\frac{2(3 x-1)}{2}\right)=\frac{10}{2}$$ (dividing both sides by 2)
⇒ 3x – 1 = 5
⇒ 3x – 1 + 1 = 5 + 1 (adding 1 to both sides)
⇒ 3x = 6
⇒ $$\frac{3 x}{3}=\frac{6}{3}$$ (dividing both sides by 3)
⇒ x = 2

(iii) Given $$\frac{3 x-1}{4}=11$$
⇒ $$4 \times \frac{3 x-1}{4}=4 \times 11$$ (multiplying both sides by 4)
⇒ 3x – 1 = 44
⇒ 3x – 1 + 1 = 44 + 1 (adding 1 to both sides)
⇒ 3x = 45
⇒ $$\frac{3 x}{3}=\frac{45}{3}$$ (dividing both sides by 3)
⇒ x = 15

Question 4.
Solve the following linear equations:

Solution:
(i) 5x – 6 = 12 – x
⇒ 5x + x = 12 + 6
$$6 x=18 \Rightarrow x=\frac{18}{6}=3$$
Verification
5x – 6 = 12 – x ⇒ 5(3) – 6 = 12 – 3
⇒ 15 – 6 = 9 ⇒ 9 = 9

⇒ 3(4 – n) = 1 (n + 3)
⇒ 12 – 3n = n + 3
⇒ -3 n – n = 3 – 12 ⇒ -4n = – 9
⇒ n = $$\frac{-9}{-4}=\frac{9}{4}$$
Verification

(iii) 5p + 7 = 19 – 2p
⇒ 5p + 2p = 19 – 7 ⇒ 7p = 12
⇒ $$p=\frac{12}{7}$$
Verification

(iv) $$2 x+\frac{5}{2}=\frac{2}{3}-x$$
$$\frac{4 x+5}{2}=\frac{2-3 x}{3}$$
⇒ 3(4x + 5) = 2(2 – 3x)
⇒ 12x + 15 = 4 – 6x
⇒ 12x + 6x = 4 – 15
⇒ 18x = -11 ⇒ x = $$\frac{-11}{18}$$
Verification

(v) $$\frac{x}{2}-5=\frac{x}{3}-4$$
⇒ $$\frac{x-10}{2}=\frac{x-12}{3}$$
⇒ 3(x – 10) = 2(x – 12)
⇒ 3x – 30 = 2x – 24
⇒ 3x – 2x = -24 + 30 ⇒ x = 6
Verification

Question 5.
Solve the following equations and verify your answers:
(i) 3(x + 7) = 18
(ii) 2(x- 1) = x + 2
(iii) $$3 x-\frac{1}{3}=2\left(x-\frac{1}{2}\right)+5$$
(iv) 4(2x – 1) -2(x – 5) = 5(x + 1) + 3
Solution:
(i) 3(x + 7) = 18
⇒ 3x + 21 = 18
⇒ 3x = 18 – 21 ⇒ 3x = -3 ⇒ x = $$\frac{-3}{3}$$
⇒ x = -1
Verification
⇒ 3(x + 7) = 18 ⇒ 3(-1 + 7) = 18 ⇒ 3(6)= 18
⇒ 18 = 18

(ii) 2(x – 1) = x + 2
⇒ 2x – 2 = x + 2
⇒ 2x – x = 2 + 2
⇒ x = 4
Verification
⇒ 2(x – 1) = x + 2
⇒ 2(4- 1) = 4 + 2
⇒ 2(3) = 6 ⇒ 6 = 6

⇒ 2(9x – 1) = 3(4x – 2 + 10)
⇒ 18x – 2 = 3(4x + 8)
⇒ 18x – 2 = 12x + 24
⇒ 18x – 12x = 24 + 2
⇒ 6x = 26 ⇒ x = $$\frac{26}{6}=\frac{13}{3}=4 \frac{1}{3}$$
Verification

(iv) 4(2x – 1) -2(x – 5) = 5(x + 1) + 3
8x – 4 – 2x + 10 = 5x + 5 + 3
⇒ 8x – 2x – 4 + 10 = 5x + 5 + 3
⇒ 6x – 5x = 8 – 6 ⇒ x = 2 ⇒ x = 2
Verification
4(2x – 1) -2(x – 5) = 5(x + 1) + 3
⇒ 4(2 × 2 – 1) -2(2 – 5) = 5(2 + 1) + 3
⇒ 4(4 – 1) -2 (-3) = 5(3) + 3
⇒ 4 × 3 + 6 = 15 + 3
⇒12 + 6 = 18
⇒ 18 = 18