GSEB Class 8 Maths Solutions Chapter 2 એકચલ સુરેખ સમીકરણ Exercise 2.3

Solve the following equations and check the answer:

Question 1. \( 3x = 2x + 18 \)
Answer:
\( 3x = 2x + 18 \)
\( 3x - 2x = 18 \) (by moving \( 2x \) to the left side)
\( x = 18 \)

To check the answer:

Left Hand Side (LHS) of the equation \( = 3x \)
\( = 3 \times 18 \)
\( = 54 \)

Right Hand Side (RHS) of the equation \( = 2x + 18 \)
\( = 2 \times 18 + 18 \)
\( = 36 + 18 \)
\( = 54 \)

Since LHS \( = \) RHS, our answer is correct.
In simple words: First, move all the 'x' terms to one side to find the value of x. Then, put this value back into both sides of the original equation to see if they are equal, which confirms your answer is right.

Exam Tip: Always verify your solution by substituting the calculated variable value back into the original equation to ensure both sides balance, thereby confirming its accuracy.

Question 2. \( 5t - 3 = 3t - 5 \)
Answer:
\( 5t - 3 = 3t - 5 \)
\( 5t - 3t - 3 = -5 \) (by moving \( 3t \) to the left side)
\( 2t - 3 = -5 \)
\( 2t = -5 + 3 \) (by moving -3 to the right side)
\( 2t = -2 \)
\( \frac{2t}{2} = \frac{-2}{2} \) (by dividing both sides by 2)
\( t = -1 \)

To check the answer:

LHS of the equation \( = 5t - 3 \)
\( = 5 \times (-1) - 3 \)
\( = -5 - 3 \)
\( = -8 \)

RHS of the equation \( = 3t - 5 \)
\( = 3 \times (-1) - 5 \)
\( = -3 - 5 \)
\( = -8 \)

Since LHS \( = \) RHS, our answer is correct.
In simple words: Gather all the 't' terms on one side and numbers on the other side. Then, simplify to find 't'. Always check your answer by putting 't' back into the first equation.

Exam Tip: Be careful with signs when moving terms across the equals sign; a positive term becomes negative and vice-versa. Double-check calculations involving negative numbers.

Question 3. \( 5x + 9 = 5 + 3x \)
Answer:
\( 5x + 9 = 5 + 3x \)
\( 5x - 3x + 9 = 5 \) (by moving \( 3x \) to the left side)
\( 2x + 9 = 5 \)
\( 2x = 5 - 9 \) (by moving 9 to the right side)
\( 2x = -4 \)
\( \frac{2x}{2} = \frac{-4}{2} \) (by dividing both sides by 2)
\( x = -2 \)

To check the answer:

LHS of the equation \( = 5x + 9 \)
\( = 5 \times (-2) + 9 \)
\( = -10 + 9 \)
\( = -1 \)

RHS of the equation \( = 5 + 3x \)
\( = 5 + 3 \times (-2) \)
\( = 5 - 6 \)
\( = -1 \)

Since LHS \( = \) RHS, our answer is correct.
In simple words: Collect all terms with 'x' on one side and constant numbers on the other. Simplify the equation to solve for 'x'. Confirm your answer by placing the value of 'x' back into the first equation to ensure both sides are equal.

Exam Tip: Consistently group variable terms on one side and constant terms on the other before simplifying. This reduces errors in solving linear equations.

Question 4. \( 4z + 3 = 6 + 2z \)
Answer:
\( 4z + 3 = 6 + 2z \)
\( 4z - 2z + 3 = 6 \) (by moving \( 2z \) to the left side)
\( 2z + 3 = 6 \)
\( 2z = 6 - 3 \) (by moving 3 to the right side)
\( 2z = 3 \)
\( \frac{2z}{2} = \frac{3}{2} \) (by dividing both sides by 2)
\( z = \frac{3}{2} \)

To check the answer:

LHS of the equation \( = 4z + 3 \)
\( = 4 \times (\frac{3}{2}) + 3 \)
\( = 6 + 3 \)
\( = 9 \)

RHS of the equation \( = 6 + 2z \)
\( = 6 + 2 \times (\frac{3}{2}) \)
\( = 6 + 3 \)
\( = 9 \)

Since LHS \( = \) RHS, our answer is correct.
In simple words: Move the terms with 'z' to one side and plain numbers to the other. Work out 'z'. After that, check your work by plugging 'z' back into the original equation to ensure both sides match up.

Exam Tip: When dealing with fractions in the solution, simplify them where possible before substitution to avoid complex calculations during verification.

Question 5. \( 2x - 1 = 14 - x \)
Answer:
\( 2x - 1 = 14 - x \)
\( 2x + x - 1 = 14 \) (by moving \( -x \) to the left side)
\( 3x - 1 = 14 \)
\( 3x = 14 + 1 \) (by moving -1 to the right side)
\( 3x = 15 \)
\( \frac{3x}{3} = \frac{15}{3} \) (by dividing both sides by 3)
\( x = 5 \)

To check the answer:

LHS of the equation \( = 2x - 1 \)
\( = 2 \times 5 - 1 \)
\( = 10 - 1 \)
\( = 9 \)

RHS of the equation \( = 14 - x \)
\( = 14 - 5 \)
\( = 9 \)

Since LHS \( = \) RHS, our answer is correct.
In simple words: First, move all the 'x' terms to one side of the equation and all the numbers to the other side. Then, simplify to find the value of 'x'. Finally, substitute this 'x' value back into both sides of the original equation to make sure they are equal.

Exam Tip: Remember that when a term changes sides of the equation, its sign must be reversed. This is a common point of error.

Question 6. \( 8x + 4 = 3(x - 1) + 7 \)
Answer:
\( 8x + 4 = 3(x - 1) + 7 \)
\( 8x + 4 = 3x - 3 + 7 \)
\( 8x + 4 = 3x + 4 \)
\( 8x - 3x + 4 = 4 \) (by moving \( 3x \) to the left side)
\( 5x + 4 = 4 \)
\( 5x = 4 - 4 \) (by moving 4 to the right side)
\( 5x = 0 \)
\( x = 0 \)

To check the answer:

LHS of the equation \( = 8x + 4 \)
\( = 8 \times 0 + 4 \)
\( = 0 + 4 \)
\( = 4 \)

RHS of the equation \( = 3(x - 1) + 7 \)
\( = 3(0 - 1) + 7 \)
\( = 3(-1) + 7 \)
\( = -3 + 7 \)
\( = 4 \)

Since LHS \( = \) RHS, our answer is correct.
In simple words: Distribute the numbers outside the brackets, then move 'x' terms to one side and constants to the other. Simplify to find 'x'. Always check your solution by plugging 'x' back into the initial equation.

Exam Tip: Always expand brackets carefully before attempting to move terms. Pay attention to the signs during multiplication and rearrangement.

Question 7. \( x = \frac{4}{5}(x + 10) \)
Answer:
\( x = \frac{4}{5}(x + 10) \)
\( x = \frac{4x}{5} + \frac{40}{5} \)
\( x = \frac{4x}{5} + 8 \)
\( x - \frac{4x}{5} = 8 \) (by moving \( \frac{4x}{5} \) to the left side)
\( \frac{5x - 4x}{5} = 8 \) (taking LCM as 5)
\( \frac{x}{5} = 8 \)
\( x = 8 \times 5 \) (multiplying both sides by 5)
\( x = 40 \)

To check the answer:

LHS of the equation \( = x \)
\( = 40 \)

RHS of the equation \( = \frac{4}{5}(x + 10) \)
\( = \frac{4}{5}(40 + 10) \)
\( = \frac{4}{5}(50) \)
\( = 4 \times 10 \)
\( = 40 \)

Since LHS \( = \) RHS, our answer is correct.
In simple words: First, multiply the fraction into the bracket. Then, gather all the 'x' terms on one side and the numbers on the other. Find a common denominator to combine the 'x' terms, and then solve for 'x'. Always check your answer by placing the 'x' value back into the first equation.

Exam Tip: When solving equations with fractions, it's often easiest to multiply both sides of the equation by the least common multiple (LCM) of the denominators to eliminate the fractions, simplifying the process.

Question 8. \( \frac{2x}{3} + 1 = \frac{7x}{15} + 3 \)
Answer:
\( \frac{2x}{3} + 1 = \frac{7x}{15} + 3 \)
\( \frac{2x}{3} = \frac{7x}{15} + 3 - 1 \) (moving 1 to RHS)
\( \frac{2x}{3} = \frac{7x}{15} + 2 \)
\( \frac{2x}{3} - \frac{7x}{15} = 2 \) (moving \( \frac{7x}{15} \) to LHS)
To combine the fractions on the left, find the LCM of 3 and 15, which is 15.
\( \frac{2x \times 5 - 7x}{15} = 2 \)
\( \frac{10x - 7x}{15} = 2 \)
\( \frac{3x}{15} = 2 \)
\( \frac{x}{5} = 2 \)
\( x = 2 \times 5 \) (multiplying both sides by 5)
\( x = 10 \)

To check the answer:

LHS of the equation \( = \frac{2x}{3} + 1 \)
\( = \frac{2(10)}{3} + 1 \)
\( = \frac{20}{3} + 1 \)
\( = \frac{20 + 3}{3} \)
\( = \frac{23}{3} \)

RHS of the equation \( = \frac{7x}{15} + 3 \)
\( = \frac{7(10)}{15} + 3 \)
\( = \frac{70}{15} + 3 \)
\( = \frac{14}{3} + 3 \)
\( = \frac{14 + 9}{3} \)
\( = \frac{23}{3} \)

Since LHS \( = \) RHS, our answer is correct.
In simple words: Move 'x' terms to one side and constant numbers to the other. Find a common denominator for fractions and then simplify. Solve for 'x' by isolating it. Check your final answer by replacing 'x' in the original equation to see if both sides are equal.

Exam Tip: When fractions are involved, always find the least common multiple (LCM) of the denominators to simplify the equation before solving for the variable. This will help avoid errors.

Question 9. \( 2y + \frac{5}{3} = \frac{26}{3} - y \)
Answer:
\( 2y + \frac{5}{3} = \frac{26}{3} - y \)
\( 2y + y + \frac{5}{3} = \frac{26}{3} \) (moving \( -y \) to LHS)
\( 3y + \frac{5}{3} = \frac{26}{3} \)
\( 3y = \frac{26}{3} - \frac{5}{3} \) (moving \( \frac{5}{3} \) to RHS)
\( 3y = \frac{26 - 5}{3} \)
\( 3y = \frac{21}{3} \)
\( 3y = 7 \)
\( \frac{3y}{3} = \frac{7}{3} \) (dividing both sides by 3)
\( y = \frac{7}{3} \)

To check the answer:

LHS of the equation \( = 2y + \frac{5}{3} \)
\( = 2(\frac{7}{3}) + \frac{5}{3} \)
\( = \frac{14}{3} + \frac{5}{3} \)
\( = \frac{14 + 5}{3} \)
\( = \frac{19}{3} \)

RHS of the equation \( = \frac{26}{3} - y \)
\( = \frac{26}{3} - \frac{7}{3} \)
\( = \frac{26 - 7}{3} \)
\( = \frac{19}{3} \)

Since LHS \( = \) RHS, our answer is correct.
In simple words: First, bring all 'y' terms to one side and all the numbers to the other. Combine the fractions and simplify to get the value of 'y'. Check your answer by putting 'y' back into the original equation to make sure both sides match.

Exam Tip: When fractions share the same denominator, combining them is straightforward; simply add or subtract their numerators while keeping the denominator unchanged.

Question 10. \( 3m = 5m - \frac{8}{5} \)
Answer:
\( 3m = 5m - \frac{8}{5} \)
\( 3m - 5m = -\frac{8}{5} \) (moving \( 5m \) to LHS)
\( -2m = -\frac{8}{5} \)
\( 2m = \frac{8}{5} \) (multiplying by -1)
\( \frac{2m}{2} = \frac{8}{5} \times \frac{1}{2} \) (dividing both sides by 2)
\( m = \frac{4}{5} \)

To check the answer:

LHS of the equation \( = 3m \)
\( = 3 \times (\frac{4}{5}) \)
\( = \frac{12}{5} \)

RHS of the equation \( = 5m - \frac{8}{5} \)
\( = 5 \times (\frac{4}{5}) - \frac{8}{5} \)
\( = 4 - \frac{8}{5} \)
\( = \frac{20 - 8}{5} \)
\( = \frac{12}{5} \)

Since LHS \( = \) RHS, our answer is correct.
In simple words: Move the 'm' terms to one side of the equation. Combine them and then solve for 'm'. After finding 'm', place it back into the original equation to confirm that both sides are equal.

Exam Tip: Pay close attention to negative signs, especially when multiplying or dividing both sides of the equation. A common mistake is to lose track of the sign.