RBSE Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3

Rajasthan Board RBSE Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Textbook Exercise Questions and Answers.

RBSE Class 8 Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.3

Solve the following equations and check your results
Question 1.
3x = 2x + 18
Answer:
Given,
3x = 2x +18
or 3x - 2x = 18
(transposing 2x to LHS)
or x = 18

Verification:
Putting x = 18 in given equation
LHS = 3 × 18 = 54
and, RHS = 2 × 18 + 18
= 36 + 18 = 54
LHS = RHS
Hence x = 18 is right answer.

Question 2.
5t - 3 = 3t -5
Answer:
Given,
5t - 3 = 3t - 5
or 5t - 3t = - 5 + 3
(transposing 31 to LHS and - 3 to RHS)
or 2t = - 2
or \(\frac{2 t}{2}=\frac{-2}{2}\)
(Dividing both sides by 2)
or t = -1

Verification:
Putting t = -1 in given equation.
LHS = 5 × -1 - 3 = -5 - 3 = -8
and, RHS = 3 × -1 - 5 = -3-5 = -8
LHS = RHS
Hence t = - 1 is right answer.

Question 3.
5x + 9 = 5 + 3x
Answer:
Given,
5x + 9 = 5 + 3x
or 5x - 3x = 5 - 9
(transposing 3x to LHS and 9 to RHS)
or 2x = - 4
or \(\frac{2 x}{2}=\frac{-4}{2}\)
(Dividing both sides by 2)
or x = - 2

Verification:
Putting x = - 2 in the given equation,
LHS = 5(- 2) + 9 = - 10 + 9 = -1
RHS = 5 + 3(-2) = 5 - 6 = -1
LHS = RHS
Hence, x = - 2 is the right answer.

Question 4.
4z + 3 = 6 + 2z
Answer:
Given,
4z + 3 = 6 + 2z
or 4z - 2z - 6 - 3
(transposing 2z to LHS and 3 to RHS)
or 2x = 3
or \(\frac{2 z}{2}=\frac{3}{2}\)
(Dividing both sides by 2)
z = \(\frac{3}{2}\)

Verification:
Putting z = \(\frac{3}{2}\) in the given equation,
LHS = 4 × \(\frac{3}{2}\) + 3 = 6 + 3 = 9
and, RHS =6 + 2 × \(\frac{3}{2}\) = 6 + 3 = 9
LHS = RHS
Hence, z = 2 is the right answer.

Question 5.
2x - 1 = 14 - x
Answer:
Given,
2x - 1 = 14 - x
or 2x + x= 14 + 1
(transposing - x to LHS and - 1 to RHS)
or 3x = 15
or \(\frac{3 x}{3}=\frac{15}{3}\)
(Dividing both sides by 3)
or x = 5

Verification:
Putting x = 5 in the given equation,
LHS = 2 × 5 - 1 = 10 - 1 = 9
and, RHS = 14 - 5 = 9
LHS = RHS
Hence x = 5 is the right answer.

Question 6.
8x + 4 = 3 (x - 1) + 7
Answer:
Given,
8x + 4= 3(x - 1) + 7
or 8x + 4 = 3x - 3 + 7
or 8x - 3x= -3 + 7-4
(transposing 3x to LHS and 4 to RHS)
or 5x = 0
or \(\frac{5 x}{5}=\frac{0}{5}\)
(Dividing both sides by 5)
or x = 0

Verification:
Putting x = 0 in the given equation,
LHS = 8(0) + 4 = 0 + 4 = 4
and RHS = 3(0 - 1) + 7 = -3 + 7 = 4
LHS = RHS
Hence x = 0 is the right answer.

Question 7.
x = \(\frac{4}{5}\)(x + 10)
Answer:
Given, x = \(\frac{4}{5}\)(x + 10)
or 5x= 4 (x + 10)
(Multiplying both sides by 5)
or 5x= 4x + 40
or 5x - 4x= 40
(transposing 4x to LHS)
or x = 40

Verification:
Putting x = 40 is given equation,
LHS = 40
and, RHS = \(\frac{4}{5}\)(40 + 10)
= \(\frac{4}{5}\) × 50 = 4 × 10 = 40
LHS = RHS
Hence x = 40 is the right answer.

Question 8.
\(\frac{2 x}{3}\) + 1 = \(\frac{7 x}{15}\) + 3
Answer:
Given, \(\frac{2 x}{3}\) + 1 = \(\frac{7 x}{15}\) + 3
(\(\frac{2 x}{3}\) + 1) × 15 = (\(\frac{7 x}{15}\) + 3) × 15
(Multiplying both sides by 15)
or 10x + 15 = 7x + 45
or 10x - 7x = 45 - 15
(transposing 7x to LHS and 15 to RHS)
or 3x= 30
or \(\frac{3 x}{3}=\frac{30}{3}\)
(Dividing both sides by 3)
or x = 10

Verification:
Putting x = 10 in the given equation.
LHS = \(\frac{2 \times 10}{3}\) + 1 = \(\frac{20+3}{3}=\frac{23}{3}\)
RHS = \(\frac{7 \times 10}{15}\) + 3 = \(\frac{14}{3}\) + 3
= \(\frac{14+9}{3}=\frac{23}{3}\)
LHS = RHS
Hence, x = 10 is the right answer.

Question 9.
2y + \(\frac{5}{3} = \frac{26}{3}\) - y
Answer:
Given, 2y + \(\frac{5}{3} = \frac{26}{3}\) - y
or 2y + y = \(\frac{26}{3}-\frac{5}{3}\)
(transposing - y to LHS and \(\frac{5}{3}\) to RHS)
or 3y = \(\frac{21}{3}\)
or 3y = 7 
or \(\frac{3 y}{3}=\frac{7}{3}\)
(Dividing both sides by 3)
y = \(\frac{7}{3}\)

Verification:
Putting y = \(\frac{7}{3}\) in the given equation,
LHS = 2 × \(\frac{7}{3}+\frac{5}{3}=\frac{14}{3}+\frac{5}{3}\)
= \(\frac{14+5}{3}=\frac{19}{3}\)
and RHS = \(\frac{26}{3}-\frac{7}{3}\)
= \(\frac{26-7}{3}=\frac{19}{3}\)
LHS = RHS
Hence, y = \(\frac{7}{3}\) is the right answer.

Question 10.
3m = 5m - \(\frac{8}{5}\)
Answer:
Given, 3m = 5m - \(\frac{8}{5}\)
or 3m - 5m = -\(\frac{8}{5}\)
(transposing 5m to LHS)
or - 2m = -\(\frac{8}{5}\)
or -2m = \(\frac{\frac{-8}{5}}{-2}\)
(Dividing both sides by - 2)
or m = \(\frac{4}{5}\)

Verification:
Putting m = \(\frac{4}{5}\) in given equation,
LHS = 3 × \(\frac{4}{5}=\frac{12}{5}\)
and, RHS = 5 × \(\frac{4}{5}-\frac{8}{5}=\frac{20}{5}-\frac{8}{5}\)
= \(\frac{20-8}{5}=\frac{12}{5}\)
LHS = RHS
Hence, m = \(\frac{4}{5}\) is the right answer.