GSEB Class 8 Maths Solutions Chapter 2 Linear Equations in One Variable Exercise 2.1

Question 1. Solve the following equations.
(i) \( x - 2 = 7 \)
(ii) \( y + 3 = 10 \)
(iii) \( 6 = z + 2 \)
(iv) \( \frac { 3 }{ 7 } + x = \frac { 17 }{ 7 } \)
(v) \( 6x = 12 \)
(vi) \( \frac { t }{ 5 } = 10 \)
(vii) \( \frac { 2x }{ 3 } = 18 \)
(viii) \( 1.6 = \frac { y }{ 1.5 } \)
(ix) \( 7x - 9 = 16 \)
(x) \( 14y - 8 = 13 \)
(xi) \( 17 + 6p = 9 \)
(xii) \( \frac { x }{ 3 } + 1 = \frac { 7 }{ 15 } \)
Answer:
(i) Given equation: \( x - 2 = 7 \)
Moving -2 to the right side (RHS), we get:
\( x = 7 + 2 \)
\( x = 9 \)
(ii) Given equation: \( y + 3 = 10 \)
Moving 3 to the right side (RHS), we find:
\( y = 10 - 3 \)
\( y = 7 \)
(iii) Given equation: \( 6 = z + 2 \)
Moving 2 to the left side (LHS), we obtain:
\( 6 - 2 = z \)
\( 4 = z \)
Therefore, \( z = 4 \)
(iv) Given equation: \( \frac { 3 }{ 7 } + x = \frac { 17 }{ 7 } \)
Moving \( \frac{3}{7} \) to the right side (RHS), we have:
\( x = \frac { 17 }{ 7 } - \frac { 3 }{ 7 } \)
\( x = \frac { 17 - 3 }{ 7 } \)
\( x = \frac { 14 }{ 7 } \)
\( x = 2 \)
(v) Given equation: \( 6x = 12 \)
Dividing both sides of the equation by 6, we get:
\( \frac { 6x }{ 6 } = \frac { 12 }{ 6 } \)
\( x = 2 \)
(vi) Given equation: \( \frac { t }{ 5 } = 10 \)
Multiplying both sides of the equation by 5, we have:
\( \frac { t }{ 5 } \times 5 = 10 \times 5 \)
\( t = 50 \)
(vii) Given equation: \( \frac { 2x }{ 3 } = 18 \)
Multiplying both sides by 3, we get:
\( \frac { 2x }{ 3 } \times 3 = 18 \times 3 \)
\( 2x = 54 \)
Dividing both sides by 2, we have:
\( \frac { 2x }{ 2 } = \frac { 54 }{ 2 } \)
\( x = 27 \)
(viii) Given equation: \( 1.6 = \frac { y }{ 1.5 } \)
To solve for \( y \), we multiply both sides by 1.5:
\( 1.6 \times 1.5 = \frac { y }{ 1.5 } \times 1.5 \)
\( 2.4 = y \)
Therefore, \( y = 2.4 \)
(ix) Given equation: \( 7x - 9 = 16 \)
Moving -9 to the right side (RHS), we obtain:
\( 7x = 16 + 9 \)
\( 7x = 25 \)
Dividing both sides by 7, we get:
\( \frac { 7x }{ 7 } = \frac { 25 }{ 7 } \)
\( x = \frac { 25 }{ 7 } \)
(x) Given equation: \( 14y - 8 = 13 \)
Moving -8 to the right side (RHS), we find:
\( 14y = 13 + 8 \)
\( 14y = 21 \)
Dividing both sides by 14, we have:
\( \frac { 14y }{ 14 } = \frac { 21 }{ 14 } \)
\( y = \frac { 3 }{ 2 } \)
(xi) Given equation: \( 17 + 6p = 9 \)
Moving 17 to the right side (RHS), we get:
\( 6p = 9 - 17 \)
\( 6p = -8 \)
Dividing both sides by 6, we obtain:
\( \frac { 6p }{ 6 } = \frac { -8 }{ 6 } \)
\( p = -\frac { 4 }{ 3 } \)
(xii) Given equation: \( \frac { x }{ 3 } + 1 = \frac { 7 }{ 15 } \)
First, move 1 to the right side (RHS):
\( \frac { x }{ 3 } = \frac { 7 }{ 15 } - 1 \)
\( \frac { x }{ 3 } = \frac { 7 - 15 }{ 15 } \)
\( \frac { x }{ 3 } = \frac { -8 }{ 15 } \)
Then, multiply both sides by 3:
\( \frac { x }{ 3 } \times 3 = \frac { -8 }{ 15 } \times 3 \)
\( x = \frac { -8 }{ 5 } \)
In simple words: For each equation, our goal is to get the variable (like x, y, t, or p) by itself on one side. We do this by moving numbers to the other side using opposite operations. If a number is subtracted, we add it to the other side. If it's added, we subtract it. If it's multiplying, we divide. If it's dividing, we multiply. Make sure to perform the same operation on both sides to keep the equation balanced. Simplify any fractions at the end.

Exam Tip: Always double-check your answer by substituting the obtained value of the variable back into the original equation to ensure both sides are equal.