GSEB Class 8 Maths Solutions Chapter 2 Linear Equations in One Variable Exercise 2.5

Question 1. Solve the following linear equations.
(i) \( \frac { x }{ 2 } - \frac { 1 }{ 5 } = \frac { x }{ 3 } + \frac { 1 }{ 4 } \)
(ii) \( \frac { n }{ 2 } - \frac { 3n }{ 4 } + \frac { 5n }{ 6 } = 21 \)
(iii) \( x + 7 - \frac { 8x }{ 3 } = \frac { 17 }{ 6 } - \frac { 5x }{ 2 } \)
(iv) \( \frac { x-5 }{ 3 } = \frac { x - 3 }{ 5 } \)
(v) \( \frac {3t - 2 }{ 4 } - \frac { 2t+3 }{ 3 } = \frac { 2 }{ 3 }-t \)
(vi) \( m - \frac { m - 1}{2} = 1 - \frac {m - 2 }{ 3 } \)
Answer:
(i) We need to solve the equation: \( \frac { x }{ 2 } - \frac { 1 }{ 5 } = \frac { x }{ 3 } + \frac { 1 }{ 4 } \)
By moving \( -\frac { 1 }{ 5 } \) to the right side and \( \frac { x }{ 3 } \) to the left side, we get:
\( \frac { x }{ 2 } - \frac { x }{ 3 } = \frac { 1 }{ 4 } + \frac { 1 }{ 5 } \)
Now, combine the terms on both sides:
\( \frac { 3x - 2x }{ 6 } = \frac { 5 + 4 }{ 20 } \)
This simplifies to:
\( \frac { x }{ 6 } = \frac { 9 }{ 20 } \)
To find \( x \), multiply both sides by 6:
\( x = \frac { 9 }{ 20 } \times 6 \)
\( x = \frac { 54 }{ 20 } \)
\( x = \frac { 27 }{ 10 } \)
Therefore, the value of \( x \) is \( \frac { 27 }{ 10 } \).
(ii) We need to solve the equation: \( \frac { n }{ 2 } - \frac { 3n }{ 4 } + \frac { 5n }{6} = 21 \)
To clear the denominators, we find the LCM of 2, 4, and 6, which is 12.
Multiply every term on both sides by 12:
\( 12 \times \frac {n}{2} - 12 \times \frac { 3n }{ 4 } + 12 \times \frac {5n }{6} = 12 \times 21 \)
This simplifies to:
\( 6n - 9n + 10n = 252 \)
Combine the terms with \( n \):
\( (6 - 9 + 10)n = 252 \)
\( 7n = 252 \)
Now, divide both sides by 7 to find \( n \):
\( n = \frac { 252 }{ 7 } \)
\( n = 36 \)
Thus, the value of \( n \) is 36.
(iii) We need to solve the equation: \( x + 7 - \frac { 8x }{ 3 } = \frac { 17 }{6} - \frac { 5x }{ 2 } \)
To remove the fractions, find the LCM of 3, 6, and 2, which is 6.
Multiply every term on both sides by 6:
\( 6 \times x + 6 \times 7 - 6 \times \frac { 8x }{ 3 } = 6 \times \frac {17}{6} - 6 \times \frac { 5x }{ 2 } \)
This simplifies to:
\( 6x + 42 - 16x = 17 - 15x \)
Combine the \( x \) terms on the left side:
\( (6 - 16)x + 42 = 17 - 15x \)
\( -10x + 42 = 17 - 15x \)
Now, move the constant 42 to the right side and \( -15x \) to the left side:
\( -10x + 15x = 17 - 42 \)
\( 5x = -25 \)
Divide both sides by 5 to find \( x \):
\( x = \frac { -25 }{ 5 } \)
\( x = -5 \)
So, the solution for \( x \) is -5.
(iv) We need to solve the equation: \( \frac { x-5 }{ 3 } = \frac { x - 3 }{ 5 } \)
To eliminate the denominators, find the LCM of 3 and 5, which is 15.
Multiply both sides of the equation by 15:
\( 15 \times \frac {x-5 }{ 3 } = 15 \times \frac {x-3 }{ 5 } \)
This leads to:
\( 5(x-5) = 3(x-3) \)
Distribute the numbers on both sides:
\( 5x - 25 = 3x - 9 \)
Now, move the constant \( -25 \) to the right side and \( 3x \) to the left side:
\( 5x - 3x = -9 + 25 \)
Combine the terms:
\( 2x = 16 \)
Divide both sides by 2 to find \( x \):
\( x = \frac { 16 }{ 2 } \)
\( x = 8 \)
Therefore, the value of \( x \) is 8.
(v) We need to solve the equation: \( \frac {3t - 2 }{ 4 } - \frac { 2t+3 }{ 3 } = \frac { 2 }{ 3 }-t \)
To remove the fractions, find the LCM of 4 and 3, which is 12.
Multiply every term on both sides by 12:
\( 12 \times \frac {3t-2}{4} - 12 \times \frac { 2t + 3 }{ 3 } = 12 \times \frac {2}{3} - 12 \times t \)
This simplifies to:
\( 3(3t - 2) - 4(2t + 3) = 4 \times 2 - 12t \)
Expand the brackets:
\( 9t - 6 - 8t - 12 = 8 - 12t \)
Combine the terms on the left side:
\( (9 - 8)t - (6 + 12) = 8 - 12t \)
\( t - 18 = 8 - 12t \)
Now, move the constant \( -18 \) to the right side and \( -12t \) to the left side:
\( t + 12t = 8 + 18 \)
Combine the terms:
\( 13t = 26 \)
Divide both sides by 13 to find \( t \):
\( t = \frac { 26 }{ 13 } \)
\( t = 2 \)
Hence, the value of \( t \) is 2.
(vi) We need to solve the equation: \( m - \frac { m - 1}{2} = 1 - \frac {m - 2 }{ 3 } \)
To eliminate the denominators, find the LCM of 2 and 3, which is 6.
Multiply every term on both sides by 6:
\( 6 \times m - 6 \times \frac { m-1 }{ 2 } = 6 \times 1 - 6 \times \frac { m-2 }{ 3 } \)
This simplifies to:
\( 6m - 3(m - 1) = 6 - 2(m - 2) \)
Expand the brackets:
\( 6m - 3m + 3 = 6 - 2m + 4 \)
Combine like terms on both sides:
\( (6 - 3)m + 3 = (6 + 4) - 2m \)
\( 3m + 3 = 10 - 2m \)
Now, move the constant 3 to the right side and \( -2m \) to the left side:
\( 3m + 2m = 10 - 3 \)
Combine the terms:
\( 5m = 7 \)
Divide both sides by 5 to find \( m \):
\( m = \frac { 7 }{ 5 } \)
Therefore, the value of \( m \) is \( \frac { 7 }{ 5 } \).
In simple words: For each equation, first simplify by clearing fractions (using LCM) and opening brackets. Then, gather all the variable terms on one side and constant numbers on the other side. Combine the terms and finally divide to isolate and find the value of the variable.

Exam Tip: Always double-check calculations involving negative numbers and fractions. A common mistake is forgetting to distribute a negative sign to all terms inside a bracket or incorrectly finding the LCM.

Question 2. Simplify and solve the following linear equations.
(vii) \( 3(t - 3) = 5(2t + 1) \)
(viii) \( 15(y - 4) - 2(y - 9) + 5(y + 6) = 0 \)
(ix) \( 3(5z - 7) - 2(9z - 11) = 4(8z - 13) - 17 \)
(x) \( 0.25(4f - 3) = 0.05(10f - 9) \)
Answer:
(vii) We need to solve the equation: \( 3(t - 3) = 5(2t + 1) \)
First, distribute the numbers into the brackets on both sides:
\( 3t - 9 = 10t + 5 \)
Now, move the constant \( -9 \) to the right side and \( 10t \) to the left side:
\( 3t - 10t = 5 + 9 \)
Combine the terms:
\( -7t = 14 \)
Divide both sides by \( -7 \) to find \( t \):
\( t = \frac { 14 }{ -7 } \)
\( t = -2 \)
Thus, the value of \( t \) is -2.
(viii) We need to solve the equation: \( 15(y - 4) - 2(y - 9) + 5(y + 6) = 0 \)
First, open all the brackets by distributing the numbers:
\( 15y - 60 - 2y + 18 + 5y + 30 = 0 \)
Next, gather all the like terms together:
\( (15 - 2 + 5)y + (-60 + 18 + 30) = 0 \)
Combine the coefficients of \( y \) and the constant terms:
\( 18y + (-12) = 0 \)
\( 18y - 12 = 0 \)
Now, move the constant \( -12 \) to the right side of the equation:
\( 18y = 12 \)
Divide both sides by 18 to determine \( y \):
\( y = \frac { 12 }{ 18 } \)
Simplify the fraction:
\( y = \frac { 2 }{ 3 } \)
Hence, the value of \( y \) is \( \frac { 2 }{ 3 } \).
(ix) We need to solve the equation: \( 3(5z - 7) - 2(9z - 11) = 4(8z - 13) - 17 \)
First, open all the brackets by distributing the multipliers:
\( 15z - 21 - 18z + 22 = 32z - 52 - 17 \)
Next, collect the like terms on both the left and right sides:
\( (15 - 18)z + (-21 + 22) = 32z + (-52 - 17) \)
This simplifies to:
\( -3z + 1 = 32z - 69 \)
Now, transpose the constant 1 to the right side and \( 32z \) to the left side:
\( -3z - 32z = -69 - 1 \)
Combine the terms:
\( -35z = -70 \)
Divide both sides by \( -35 \) to determine \( z \):
\( z = \frac { -70 }{ -35 } \)
\( z = 2 \)
Thus, the value of \( z \) is 2.
(x) We need to solve the equation: \( 0.25(4f - 3) = 0.05(10f - 9) \)
First, expand the brackets by multiplying the terms:
\( 0.25 \times 4f - 0.25 \times 3 = 0.05 \times 10f - 0.05 \times 9 \)
This simplifies to:
\( 1f - 0.75 = 0.5f - 0.45 \)
Now, move the constant \( -0.75 \) to the right side and \( 0.5f \) to the left side:
\( f - 0.5f = -0.45 + 0.75 \)
Combine the like terms:
\( 0.5f = 0.30 \)
Divide both sides by \( 0.5 \) to find \( f \):
\( f = \frac { 0.30 }{ 0.5 } \)
\( f = \frac { 30 }{ 50 } \)
\( f = \frac { 3 }{ 5 } \)
\( f = 0.6 \)
Thus, the value of \( f \) is 0.6.
In simple words: For each equation, expand all brackets. Then, group the variable terms together on one side and the constant numbers on the other. Simplify by combining similar terms, and finally, divide to find the solution for the variable.

Exam Tip: Be meticulous with calculations involving decimals and fractions. When expanding brackets, make sure to multiply every term inside the bracket by the outside factor, and be careful with negative signs.