Exercise 8.3
1. 6x + 5 = 2x + 17
Solution:
We have
6x + 5 = 2x+ 17
Transposing 2x to LHS and 5 to RHS, we get
6x – 2x = 17 – 5
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
x = 3
Verification:
Substituting x = 3 in the given equation, we get
6 × 3 + 5 = 2 × 3 + 17
18 + 5 = 6 + 17
23 = 23
LHS = RHS
Thus, verified.
2. 2(5x – 3) -3(2x – 1) = 9
Solution:
We have
2(5x – 3) – 3(2x – 1) = 9
Expanding the brackets, we get
2 × 5x – 2 × 3 – 3 × 2x + 3 × 1 = 9
10x – 6 – 6x + 3 = 9
10x – 6x – 6 + 3 = 9
4x – 3 = 9
Adding 3 to both sides, we get
4x – 3 + 3 = 9 + 3
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
Thus, x = 3.
Verification:
Substituting x = 3 in LHS, we get
= 2(5 × 3 – 3) – 3(2 × 3 – 1)
= 2 × 12 – 3 × 5
= 24 – 15
= 9
LHS = RHS
Thus, verified.
3. (x/2)= (x/2) + 1
Solution:
Given (x/2) = (x/2) + 1
Transposing (x/2) = (x/3) to LHS we get
(x/2) – (x/3) to LHS we get
(3x – 2x)/6 = 1 [CH of and 2 is 6]
x/6
Multiplying 6 to both sides we get,
X = 6
Verification:
Substituting x = 6 in given equation we get
(6/2) = ( 6/3) + 2
3 = 3+1
3 = 3
Thus LSH =RHS
Thus, verified.
.
6. 3(x – 3) = 5(2x + 1)
Solution:
3(x – 3) = 5(2x + 1)
On expanding the brackets on both sides, we get
= 3 × x – 3 × 3 = 5 × 2x + 5 × 1
= 3x – 9 = 10x + 5
Transposing 10x to LHS and 9 to RHS, we get
= 3x – 10x = 9 + 5
= -7x = 14
Dividing both sides by 7, we get
= –7x/7 = 14/7
=x = -2
Verification:
Substituting x = -2 on both sides, we get
3(-2 – 3) = 5{2(-2) +1}
3(-5) = 5(-3)
-15 = -15
LHS = RHS
Thus, verified.
7. 3x – 2(2x – 5) = 2(x + 3) – 8
Solution:
3x – 2(2x – 5) = 2(x + 3) – 8
On expanding the brackets on both sides, we get
= 3x – 2 × 2x + 2 × 5 = 2 × x + 2 × 3 – 8
= 3x – 4x + 10 = 2x + 6 – 8
= -x + 10 = 2x – 2
Transposing x to RHS and 2 to LHS, we have
= 10 + 2 = 2x + x
= 3x = 12
Dividing both sides by 3, we have
= 3x/3 = 12/3
= x = 4
Verification:
Substituting x = 4 on both sides, we have
3(4) – 2{2(4) – 5} = 2(4 + 3) – 8
12 – 2(8 – 5) = 14 – 8
12 – 6 = 6
6 = 6
LHS = RHS
Thus, verified.
8. x – (x/4) – (1/2) = 3 + (x/4)
Solution:
Given x – (x/4) – (1/2) = 3+ (x/4)
Transposing (x/4) – (1/2) = 3 + (x/4)
Transposing (x/4) to LHS and (1/2) to RHS
X – (X/4) – (X/4) = 3 + (1/2)
(4x – x – x) / 4 = (6+1)/ 2
2x/ 4 = 7/2
x/2 = 7/2
x = 7
Verification:
Substituting x = 7 in the given equation we get
7 – (7/4) – (1/2) = 3 + (7/4)
(28-7-2)/ 4 = (12+7)/ 4
Thus LHS = RHS
Thus, verified.
12. 0.6x + 4/5 = 0.28x + 1.16
Solution:
0.6x + 4/5 = 0.28x + 1.16
Transposing 0.28x to LHS and 45 to RHS, we get
= 0.6x – 0.28x = 1.16 – 45
= 0.32x = 1.16 – 0.8
= 0.32x = 0.36
Dividing both sides by 0.32, we get
= 0.32 x 0.32 = 0.360.32
= x = 98
Verification:
Substituting x = 98 on both sides, we get
0.6(9/8) + 45 = 0.28(9/8) + 1.16
5.4/8 + 4/5 = 2.52/8 + 1.16
0.675 + 0.8 = 0.315 + 1.16
1.475 = 1.475
LHS = RHS
Thus, verified.
13. 0.5x+ (x/3) = 0.25x+7
Solution:
Given 0.5x + (x/3) = 0.25x + 7
(5/10) x + (x/3) = (25x/100) + 7
(x/2) + (x/3) = (25x/100)+ 7
(x/2) + (x/3) = ( x/4) + 7
Transposing (x/4) to LHS we get
(x/2) + (x/3) – (x/4) = 7
(6x+ 4x – 3x) – (x/4) = 7
(7x/12) = 7
Multiplying both sides by 2 we get
(7x/12) × 12 = 7×12
7x = 84
Dividing both sides by 7 we ge t
(7x/7) = (84/7)
X= 12
Verification:
Substituting x = 12 in given equation we get
0.5 (12) + (12/3) = 0.25 (12) +7
6 + 4 = 3 + 7
10 = 10
Thus LHS = RHS
Thus, verified.