GSEB Class 8 Maths Solutions Chapter 2 એકચલ સુરેખ સમીકરણ Exercise 2.5

Question 1. \( \frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4} \)
Answer:
\( \frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4} \)
We move the x-terms to one side and constant terms to the other side:
\( \frac{x}{2}-\frac{x}{3}=\frac{1}{4}+\frac{1}{5} \)
The least common multiple (LCM) of 2 and 3 is 6. The LCM of 4 and 5 is 20. We make the denominators common:
\( \frac{3x-2x}{6}=\frac{5+4}{20} \)
\( \frac{x}{6}=\frac{9}{20} \)
To find the value of x, we multiply both sides by 6:
\( x = \frac{9}{20} \times 6 \)
\( x = \frac{54}{20} \)
Simplifying the fraction gives:
\( x = \frac{27}{10} \)
In simple words: First, group all the terms with 'x' on one side and regular numbers on the other. Then, find a common bottom number (LCM) for each side and add or subtract. Finally, multiply to get 'x' by itself.

Exam Tip: Remember to group similar terms on separate sides of the equation. Always find the Least Common Multiple (LCM) for fractions to perform addition or subtraction accurately.

Question 2. \( \frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6} = 21 \)
Answer:
\( \frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6} = 21 \)
To combine the fractions on the left side, we find the LCM of the denominators 2, 4, and 6, which is 12. We multiply the numerator and denominator of each fraction to get 12 in the denominator:
\( \frac{n \times 6}{2 \times 6}-\frac{3 n \times 3}{4 \times 3}+\frac{5 n \times 2}{6 \times 2} = 21 \)
\( \frac{6 n-9 n+10 n}{12} = 21 \)
Combining the terms in the numerator:
\( \frac{7 n}{12} = 21 \)
To isolate n, we multiply both sides by 12:
\( 7n = 21 \times 12 \)
\( 7n = 252 \)
Now, we divide both sides by 7:
\( n = \frac{252}{7} \)
\( n = 36 \)
In simple words: To solve for 'n', first make all the fractions on the left side have the same bottom number (LCM). Add the top numbers, then multiply to move the bottom number to the other side. Finally, divide to find 'n'.

Exam Tip: When dealing with multiple fractions, always begin by finding the LCM of all denominators to simplify the equation effectively. This prevents errors in addition and subtraction.

Question 3. \( x + 7- \frac {8 x}{3} = \frac{17}{6}-\frac{5 x}{2} \)
Answer:
\( x + 7- \frac {8 x}{3} = \frac{17}{6}-\frac{5 x}{2} \)
The LCM of the denominators 3, 6, and 2 is 6. We multiply both sides of the equation by 6 to clear the fractions:
\( (6 \times x) + (6 \times 7) - 6\left(\frac{8 x}{3}\right) = 6\left(\frac{17}{6}\right) - 6\left(\frac{5 x}{2}\right) \)
\( 6x + 42 - 16x = 17 - 15x \)
Combine like terms on each side:
\( -10x + 42 = 17 - 15x \)
Move the x-terms to one side and constant terms to the other side. We bring -15x to the left side and 42 to the right side:
\( -10x + 15x = 17 - 42 \)
\( 5x = -25 \)
Divide both sides by 5:
\( x = \frac{-25}{5} \)
\( x = -5 \)
In simple words: Clear the fractions by multiplying everything by the smallest common multiple of the denominators. Then, gather all 'x' terms on one side and numbers on the other. Finally, divide to find the value of 'x'.

Exam Tip: Always distribute the multiplier (LCM) to every term on both sides of the equation, even to terms that are not fractions, to avoid calculation mistakes.

Question 4. \( \frac{x-5}{3}=\frac{x-3}{5} \)
Answer:
\( \frac{x-5}{3}=\frac{x-3}{5} \)
The LCM of the denominators 3 and 5 is 15. We multiply both sides of the equation by 15:
\( 15\left(\frac{x-5}{3}\right) = 15\left(\frac{x-3}{5}\right) \)
Simplifying both sides:
\( 5(x - 5) = 3(x - 3) \)
Distribute the numbers into the parentheses:
\( 5x - 25 = 3x - 9 \)
Move the x-terms to the left side and constant terms to the right side. We move 3x to the left side and -25 to the right side:
\( 5x - 3x = 25 - 9 \)
\( 2x = 16 \)
Divide both sides by 2:
\( \frac{2 x}{2}=\frac{16}{2} \)
\( x = 8 \)
In simple words: To solve this, multiply both sides by the smallest common multiple of the bottom numbers to remove fractions. Distribute and then move all 'x' terms to one side and numbers to the other. Finally, divide to find 'x'.

Exam Tip: When cross-multiplying or clearing denominators, remember to distribute the multiplied value to all terms within the parentheses, not just the first term.

Question 5. \( \frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t \)
Answer:
\( \frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t \)
The LCM of the denominators 4 and 3 is 12. We multiply both sides of the equation by 12:
\( 12\left(\frac{3 t-2}{4}\right) - 12\left(\frac{2t+3}{3}\right) = 12 \times \frac {2}{3} - 12t \)
Simplifying each term:
\( 3(3t - 2) - 4(2t + 3) = 8 - 12t \)
Distribute the numbers into the parentheses:
\( 9t - 6 - 8t - 12 = 8 - 12t \)
Combine like terms on the left side:
\( t - 18 = 8 - 12t \)
Move the t-terms to one side and constant terms to the other side. We bring -12t to the left side and -18 to the right side:
\( t + 12t = 8 + 18 \)
\( 13t = 26 \)
Divide both sides by 13:
\( \frac{13 t}{13}=\frac{26}{13} \)
\( t = 2 \)
In simple words: Get rid of all fractions by multiplying the entire equation by the LCM of the denominators. Expand any brackets. Group the 't' terms on one side and the numbers on the other. Then, divide to find 't'.

Exam Tip: Be very careful with negative signs when distributing, especially with terms like \( -4(2t+3) \), which correctly becomes \( -8t - 12 \), not \( -8t + 12 \).

Question 6. \( m-\frac{m-1}{2} = 1 - \frac{m-2}{3} \)
Answer:
\( m-\frac{m-1}{2} = 1 - \frac{m-2}{3} \)
The LCM of the denominators 2 and 3 is 6. We multiply both sides of the equation by 6:
\( 6m - 6\left(\frac{m-1}{2}\right) = 6 \times 1 - 6\left(\frac{m-2}{3}\right) \)
Simplifying each term:
\( 6m - 3(m - 1) = 6 - 2(m - 2) \)
Distribute the numbers into the parentheses:
\( 6m - 3m + 3 = 6 - 2m + 4 \)
Combine like terms on each side:
\( 3m + 3 = 10 - 2m \)
Move the m-terms to the left side and constant terms to the right side. We bring -2m to the left side and 3 to the right side:
\( 3m + 2m = 10 - 3 \)
\( 5m = 7 \)
Divide both sides by 5:
\( \frac{5m}{5}=\frac{7}{5} \)
\( m = \frac {7}{5} \)
In simple words: First, multiply the entire equation by the smallest common multiple of the bottom numbers to get rid of fractions. Then, open up any brackets carefully. Gather all 'm' terms on one side and all numbers on the other. Finally, divide to find 'm'.

Exam Tip: Always include the terms without fractions (like 'm' and '1' in this question) when multiplying by the LCM to clear denominators. Forgetting to do so is a common error.

Simplify and solve the following equations:

Question 7. \( 3(t-3) = 5(2t + 1) \)
Answer:
\( 3(t-3) = 5(2t + 1) \)
Distribute the numbers into the parentheses on both sides:
\( 3t - 9 = 10t + 5 \)
Move the t-terms to one side and constant terms to the other side. We bring 10t to the left side and -9 to the right side:
\( 3t - 10t = 5 + 9 \)
Combine like terms:
\( -7t = 14 \)
To make the coefficient of t positive, multiply both sides by (-1):
\( 7t = -14 \)
Divide both sides by 7:
\( \frac{7 t}{7}=\frac{-14}{7} \)
\( t = -2 \)
In simple words: First, multiply the numbers outside the brackets by everything inside them. Then, move all terms with 't' to one side and plain numbers to the other. Combine similar terms and then divide to find 't'.

Exam Tip: Pay close attention to signs when moving terms across the equals sign; remember to change their sign. Also, be careful with negative coefficients at the final step.

Question 8. \( 15(y – 4) -2 (y - 9) + 5(y + 6) = 0 \)
Answer:
\( 15(y – 4) -2 (y - 9) + 5(y + 6) = 0 \)
Distribute the numbers into the parentheses:
\( 15y - 60 - 2y + 18 + 5y + 30 = 0 \)
Group the y-terms and the constant terms separately:
\( (15y - 2y + 5y) + (-60 + 18 + 30) = 0 \)
Combine the y-terms and the constant terms:
\( 18y - 12 = 0 \)
Move the constant term to the right side. We bring -12 to the right side:
\( 18y = 12 \)
Divide both sides by 18:
\( \frac{18 y}{18}=\frac{12}{18} \)
\( y = \frac {2}{3} \)
In simple words: Open all the brackets by multiplying. Then, gather all the 'y' terms together and all the normal numbers together. Combine them, move the number to the other side, and divide to find 'y'.

Exam Tip: Take great care with distributing negative signs, like in \( -2(y-9) \), which becomes \( -2y + 18 \). A common error is writing \( -2y - 18 \).

Question 9. \( 3(5z - 7) -2 (9z - 11) = 4(8z - 13) - 17 \)
Answer:
\( 3(5z - 7) -2 (9z - 11) = 4(8z - 13) - 17 \)
Distribute the numbers into the parentheses on both sides:
\( 15z - 21 - 18z + 22 = 32z - 52 - 17 \)
Combine like terms on each side of the equation:
\( (15z - 18z) + (-21 + 22) = 32z + (-52 - 17) \)
\( -3z + 1 = 32z - 69 \)
Move the z-terms to the left side and constant terms to the right side. We bring 32z to the left side and 1 to the right side:
\( -3z - 32z = -69 - 1 \)
\( -35z = -70 \)
To make the coefficient of z positive, multiply both sides by (-1):
\( 35z = 70 \)
Divide both sides by 35:
\( \frac{35 z}{35}=\frac{70}{35} \)
\( z = 2 \)
In simple words: First, multiply numbers into brackets on both sides. Then, combine terms on each side. Move all 'z' terms to one side and numbers to the other. Combine them, and finally divide to get 'z'.

Exam Tip: When moving multiple terms, it helps to group them visually (e.g., all 'z' terms on the left, all constants on the right) to minimize errors in signs.

Question 10. \( 0.25 (4f-3) = 0.05(10f - 9) \)
Answer:
\( 0.25 (4f-3) = 0.05(10f - 9) \)
Distribute the numbers into the parentheses on both sides:
\( 0.25 \times 4f - 0.25 \times 3 = 0.05 \times 10f - 0.05 \times 9 \)
\( f - 0.75 = 0.5f - 0.45 \)
Move the f-terms to the left side and constant terms to the right side:
\( f - 0.5f = 0.75 - 0.45 \)
Combine like terms:
\( 0.5f = 0.3 \)
To find f, divide both sides by 0.5:
\( f = \frac{0.3}{0.5} \)
\( f = \frac{3}{5} \)
\( f = 0.6 \)
In simple words: First, multiply the decimal numbers by everything inside their brackets. Then, collect all the 'f' terms on one side and the normal numbers on the other side. Combine them and then divide to find 'f'.

Exam Tip: When working with decimals, ensure accurate multiplication. You can also convert decimals to fractions (e.g., 0.25 to \( \frac{1}{4} \)) at the beginning to simplify calculations if preferred.