GSEB Class 8 Maths Solutions Chapter 2 Linear Equations in One Variable Exercise 2.3

Question 1. Solve the following equations and check your results.
(i) \( 3x = 2x + 18 \)
(ii) \( 5t - 3 = 3t - 5 \)
(iii) \( 5x + 9 = 5 + 3x \)
(iv) \( 4z + 3 = 6 + 2z \)
(v) \( 2x - 1 = 14 - x \)
(vi) \( 8x + 4 = 3(x - 1) + 7 \)
(vii) \( x = \frac{4}{5}(x + 10) \)
(viii) \( \frac{2x}{3} + 1 = \frac{7x}{15} + 3 \)
(ix) \( 2y + \frac{5}{3} = \frac{26}{3} - y \)
(x) \( 3m = 5m - \frac{8}{5} \)
Answer:
(i) We need to solve the equation \( 3x = 2x + 18 \).
First, we move the term \( 2x \) from the Right Hand Side (RHS) to the Left Hand Side (LHS) of the equation. This gives us:
\( 3x - 2x = 18 \)
So, \( x = 18 \).
To verify our answer, we substitute \( x = 18 \) into both sides of the original equation.
LHS \( = 3x = 3 \times 18 = 54 \)
RHS \( = 2x + 18 = 2 \times 18 + 18 = 36 + 18 = 54 \)
Since the LHS equals the RHS, our solution is correct.
(ii) We need to solve the equation \( 5t - 3 = 3t - 5 \).
First, we move the constant term \( -3 \) from the LHS to the RHS:
\( 5t = 3t - 5 + 3 \)
\( 5t = 3t - 2 \)
Next, we move the term \( 3t \) from the RHS to the LHS:
\( 5t - 3t = -2 \)
\( 2t = -2 \)
To find \( t \), we divide both sides by 2:
\( \frac{2t}{2} = \frac{-2}{2} \)
So, \( t = -1 \).
To verify our answer, we substitute \( t = -1 \) into both sides of the original equation.
LHS \( = 5(-1) - 3 = -5 - 3 = -8 \)
RHS \( = 3(-1) - 5 = -3 - 5 = -8 \)
Since the LHS equals the RHS, our solution is correct.
(iii) We need to solve the equation \( 5x + 9 = 5 + 3x \).
First, we move the constant term \( 9 \) from the LHS to the RHS:
\( 5x = 5 + 3x - 9 \)
\( 5x = 3x - 4 \)
Next, we move the term \( 3x \) from the RHS to the LHS:
\( 5x - 3x = -4 \)
\( 2x = -4 \)
To find \( x \), we divide both sides by 2:
\( \frac{2x}{2} = \frac{-4}{2} \)
So, \( x = -2 \).
To verify our answer, we substitute \( x = -2 \) into both sides of the original equation.
LHS \( = 5(-2) + 9 = -10 + 9 = -1 \)
RHS \( = 5 + 3(-2) = 5 - 6 = -1 \)
Since the LHS equals the RHS, our solution is correct.
(iv) We need to solve the equation \( 4z + 3 = 6 + 2z \).
First, we move the constant term \( 3 \) from the LHS to the RHS:
\( 4z = 6 + 2z - 3 \)
\( 4z = 3 + 2z \)
Next, we move the term \( 2z \) from the RHS to the LHS:
\( 4z - 2z = 3 \)
\( 2z = 3 \)
To find \( z \), we divide both sides by 2:
\( z = \frac{3}{2} \).
To verify our answer, we substitute \( z = \frac{3}{2} \) into both sides of the original equation.
LHS \( = 4(\frac{3}{2}) + 3 = 2 \times 3 + 3 = 6 + 3 = 9 \)
RHS \( = 6 + 2(\frac{3}{2}) = 6 + 3 = 9 \)
Since the LHS equals the RHS, our solution is correct.
(v) We need to solve the equation \( 2x - 1 = 14 - x \).
First, we move the constant term \( -1 \) from the LHS to the RHS:
\( 2x = 14 - x + 1 \)
\( 2x = 15 - x \)
Next, we move the term \( -x \) from the RHS to the LHS:
\( 2x + x = 15 \)
\( 3x = 15 \)
To find \( x \), we divide both sides by 3:
\( x = \frac{15}{3} \)
So, \( x = 5 \).
To verify our answer, we substitute \( x = 5 \) into both sides of the original equation.
LHS \( = 2 \times 5 - 1 = 10 - 1 = 9 \)
RHS \( = 14 - 5 = 9 \)
Since the LHS equals the RHS, our solution is correct.
(vi) We need to solve the equation \( 8x + 4 = 3(x - 1) + 7 \).
First, we simplify the RHS by distributing the 3:
\( 8x + 4 = 3x - 3 + 7 \)
\( 8x + 4 = 3x + 4 \)
Next, we move the constant term \( 4 \) from the LHS to the RHS:
\( 8x = 3x + 4 - 4 \)
\( 8x = 3x \)
Now, we move the term \( 3x \) from the RHS to the LHS:
\( 8x - 3x = 0 \)
\( 5x = 0 \)
To find \( x \), we divide both sides by 5:
\( x = 0 \).
To verify our answer, we substitute \( x = 0 \) into both sides of the original equation.
LHS \( = 8 \times 0 + 4 = 0 + 4 = 4 \)
RHS \( = 3(0 - 1) + 7 = 3(-1) + 7 = -3 + 7 = 4 \)
Since the LHS equals the RHS, our solution is correct.
(vii) We need to solve the equation \( x = \frac{4}{5}(x + 10) \).
First, we simplify the RHS by distributing \( \frac{4}{5} \):
\( x = \frac{4}{5}x + \frac{4}{5} \times 10 \)
\( x = \frac{4}{5}x + 8 \)
Next, we move the term \( \frac{4}{5}x \) from the RHS to the LHS:
\( x - \frac{4}{5}x = 8 \)
To combine the terms on the LHS, we find a common denominator:
\( \frac{5x - 4x}{5} = 8 \)
\( \frac{x}{5} = 8 \)
To find \( x \), we multiply both sides by 5:
\( x = 8 \times 5 \)
So, \( x = 40 \).
To verify our answer, we substitute \( x = 40 \) into both sides of the original equation.
LHS \( = 40 \)
RHS \( = \frac{4}{5}(40 + 10) = \frac{4}{5} \times 50 = 4 \times 10 = 40 \)
Since the LHS equals the RHS, our solution is correct.
(viii) We need to solve the equation \( \frac{2x}{3} + 1 = \frac{7x}{15} + 3 \).
First, we move the constant term \( 1 \) from the LHS to the RHS:
\( \frac{2x}{3} = \frac{7x}{15} + 3 - 1 \)
\( \frac{2x}{3} = \frac{7x}{15} + 2 \)
Next, we move the term \( \frac{7x}{15} \) from the RHS to the LHS:
\( \frac{2x}{3} - \frac{7x}{15} = 2 \)
To combine the terms on the LHS, we find a common denominator, which is 15:
\( \frac{5 \times 2x}{15} - \frac{7x}{15} = 2 \)
\( \frac{10x - 7x}{15} = 2 \)
\( \frac{3x}{15} = 2 \)
\( \frac{x}{5} = 2 \)
To find \( x \), we multiply both sides by 5:
\( x = 2 \times 5 \)
So, \( x = 10 \).
To verify our answer, we substitute \( x = 10 \) into both sides of the original equation.
LHS \( = \frac{2 \times 10}{3} + 1 = \frac{20}{3} + 1 = \frac{20 + 3}{3} = \frac{23}{3} \)
RHS \( = \frac{7 \times 10}{15} + 3 = \frac{70}{15} + 3 = \frac{14}{3} + 3 = \frac{14 + 9}{3} = \frac{23}{3} \)
Since the LHS equals the RHS, our solution is correct.
(ix) We need to solve the equation \( 2y + \frac{5}{3} = \frac{26}{3} - y \).
First, we move the constant term \( \frac{5}{3} \) from the LHS to the RHS:
\( 2y = \frac{26}{3} - y - \frac{5}{3} \)
\( 2y = \frac{26 - 5}{3} - y \)
\( 2y = \frac{21}{3} - y \)
\( 2y = 7 - y \)
Next, we move the term \( -y \) from the RHS to the LHS:
\( 2y + y = 7 \)
\( 3y = 7 \)
To find \( y \), we divide both sides by 3:
\( y = \frac{7}{3} \).
To verify our answer, we substitute \( y = \frac{7}{3} \) into both sides of the original equation.
LHS \( = 2(\frac{7}{3}) + \frac{5}{3} = \frac{14}{3} + \frac{5}{3} = \frac{19}{3} \)
RHS \( = \frac{26}{3} - \frac{7}{3} = \frac{26 - 7}{3} = \frac{19}{3} \)
Since the LHS equals the RHS, our solution is correct.
(x) We need to solve the equation \( 3m = 5m - \frac{8}{5} \).
First, we move the term \( 5m \) from the RHS to the LHS:
\( 3m - 5m = - \frac{8}{5} \)
\( -2m = - \frac{8}{5} \)
To find \( m \), we divide both sides by \( -2 \):
\( m = \frac{-8}{5} \div (-2) \)
\( m = \frac{-8}{5} \times \frac{1}{-2} \)
\( m = \frac{8}{10} \)
So, \( m = \frac{4}{5} \).
To verify our answer, we substitute \( m = \frac{4}{5} \) into both sides of the original equation.
LHS \( = 3(\frac{4}{5}) = \frac{12}{5} \)
RHS \( = 5(\frac{4}{5}) - \frac{8}{5} = 4 - \frac{8}{5} = \frac{20 - 8}{5} = \frac{12}{5} \)
Since the LHS equals the RHS, our solution is correct.
In simple words: To solve these equations, move all terms with the variable to one side and all numbers to the other. Then, simplify to find the variable's value. Always check your answer by putting it back into the original equation to make sure both sides are equal.

Exam Tip: Remember to perform operations on both sides of the equation to maintain balance. When checking your answer, clearly show the calculation for both the LHS and RHS.