Step-by-step solutions for linear equations from RD Sharma Class 8 Maths, Chapter 9, Exercise 9.1.

Problem and Solution Explanation

The image contains two linear equations in one variable, and the task is to solve each equation and verify the solution. Below is a detailed explanation of the solutions provided.

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#### Problem 1: Solve \( 9 \frac{1}{4} = y - 1 \frac{1}{3} \)

##### Step-by-Step Solution:

1. Convert mixed fractions to improper fractions:
- \( 9 \frac{1}{4} = \frac{9 \times 4 + 1}{4} = \frac{36 + 1}{4} = \frac{37}{4} \)
- \( 1 \frac{1}{3} = \frac{1 \times 3 + 1}{3} = \frac{3 + 1}{3} = \frac{4}{3} \)

So, the equation becomes:
\[
\frac{37}{4} = y - \frac{4}{3}
\]

2. Isolate \( y \):
Add \( \frac{4}{3} \) to both sides of the equation:
\[
y = \frac{37}{4} + \frac{4}{3}
\]

3. Find a common denominator:
The denominators are 4 and 3. The least common multiple (LCM) of 4 and 3 is 12.
- Convert \( \frac{37}{4} \) to a fraction with denominator 12:
\[
\frac{37}{4} = \frac{37 \times 3}{4 \times 3} = \frac{111}{12}
\]
- Convert \( \frac{4}{3} \) to a fraction with denominator 12:
\[
\frac{4}{3} = \frac{4 \times 4}{3 \times 4} = \frac{16}{12}
\]

Now, add the fractions:
\[
y = \frac{111}{12} + \frac{16}{12} = \frac{111 + 16}{12} = \frac{127}{12}
\]

Therefore, the solution is:
\[
y = \frac{127}{12}
\]

##### Verification:
Substitute \( y = \frac{127}{12} \) back into the original equation:
\[
9 \frac{1}{4} = y - 1 \frac{1}{3}
\]

1. Convert \( 9 \frac{1}{4} \) and \( 1 \frac{1}{3} \) to improper fractions:
- \( 9 \frac{1}{4} = \frac{37}{4} \)
- \( 1 \frac{1}{3} = \frac{4}{3} \)

2. Calculate the right-hand side (RHS):
\[
\text{RHS} = y - 1 \frac{1}{3} = \frac{127}{12} - \frac{4}{3}
\]

3. Convert \( \frac{4}{3} \) to a fraction with denominator 12:
\[
\frac{4}{3} = \frac{4 \times 4}{3 \times 4} = \frac{16}{12}
\]

4. Subtract the fractions:
\[
\text{RHS} = \frac{127}{12} - \frac{16}{12} = \frac{127 - 16}{12} = \frac{111}{12}
\]

5. Simplify \( \frac{111}{12} \):
\[
\frac{111}{12} = \frac{37}{4}
\]

Since \( \frac{37}{4} \) is equal to the left-hand side (LHS), the solution is verified.

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#### Problem 2: Solve \( \frac{5x}{3} + \frac{2}{5} = 1 \)

##### Step-by-Step Solution:

1. Isolate the term involving \( x \):
Subtract \( \frac{2}{5} \) from both sides:
\[
\frac{5x}{3} = 1 - \frac{2}{5}
\]

2. Simplify the right-hand side:
Convert 1 to a fraction with denominator 5:
\[
1 = \frac{5}{5}
\]
So,
\[
1 - \frac{2}{5} = \frac{5}{5} - \frac{2}{5} = \frac{5 - 2}{5} = \frac{3}{5}
\]

Now the equation is:
\[
\frac{5x}{3} = \frac{3}{5}
\]

3. Cross-multiply to solve for \( x \):
Multiply both sides by 3 and 5 to eliminate the denominators:
\[
5x \cdot 5 = 3 \cdot 3
\]
\[
25x = 9
\]

4. Solve for \( x \):
Divide both sides by 25:
\[
x = \frac{9}{25}
\]

Therefore, the solution is:
\[
x = \frac{9}{25}
\]

##### Verification:
Substitute \( x = \frac{9}{25} \) back into the original equation:
\[
\frac{5x}{3} + \frac{2}{5} = 1
\]

1. Calculate \( \frac{5x}{3} \):
\[
\frac{5x}{3} = \frac{5 \cdot \frac{9}{25}}{3} = \frac{45}{75} = \frac{3}{5}
\]

2. Add \( \frac{2}{5} \):
\[
\frac{5x}{3} + \frac{2}{5} = \frac{3}{5} + \frac{2}{5} = \frac{3 + 2}{5} = \frac{5}{5} = 1
\]

Since the left-hand side equals the right-hand side, the solution is verified.

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Final Answers:

1. \( y = \frac{127}{12} \)
2. \( x = \frac{9}{25} \)

\[
\boxed{\frac{127}{12}, \frac{9}{25}}
\]