1/2x - y = 3/2x - 3;  x=7
3/4x + 4/7y = 5/4x - 1; x=8 
3/5y - 4x = 3 - 2y; x=9

To solve these equations with the given values of \( x \), we will substitute the values for \( x \) and solve for \( y \) in each equation.

### 1. Solving for \( y \) in the first equation:
Given:
\[
\frac{1}{2}x - y = \frac{3}{2}x - 3, \quad x = 7
\]

Substitute \( x = 7 \):

\[
\frac{1}{2} \cdot 7 - y = \frac{3}{2} \cdot 7 - 3
\]

Simplify and solve for \( y \):

\[
\frac{7}{2} - y = \frac{21}{2} - 3
\]

Calculate the right side:

\[
\frac{7}{2} - y = \frac{21}{2} - \frac{6}{2}
\]

\[
\frac{7}{2} - y = \frac{15}{2}
\]

Now, solve for \( y \):

\[
-y = \frac{15}{2} - \frac{7}{2}
\]

\[
-y = \frac{8}{2}
\]

\[
-y = 4 \implies y = -4
\]

So, for the first equation, \( y = -4 \).

### 2. Solving for \( y \) in the second equation:
Given:
\[
\frac{3}{4}x + \frac{4}{7}y = \frac{5}{4}x - 1, \quad x = 8
\]

Substitute \( x = 8 \):

\[
\frac{3}{4} \cdot 8 + \frac{4}{7}y = \frac{5}{4} \cdot 8 - 1
\]

Simplify:

\[
6 + \frac{4}{7}y = 10 - 1
\]

\[
6 + \frac{4}{7}y = 9
\]

Now, solve for \( y \):

\[
\frac{4}{7}y = 9 - 6
\]

\[
\frac{4}{7}y = 3
\]

Multiply both sides by \( \frac{7}{4} \):

\[
y = 3 \cdot \frac{7}{4}
\]

\[
y = \frac{21}{4}
\]

So, for the second equation, \( y = \frac{21}{4} \).

### 3. Solving for \( y \) in the third equation:
Given:
\[
\frac{3}{5}y - 4x = 3 - 2y, \quad x = 9
\]

Substitute \( x = 9 \):

\[
\frac{3}{5}y - 4 \cdot 9 = 3 - 2y
\]

Simplify:

\[
\frac{3}{5}y - 36 = 3 - 2y
\]

Rearrange to solve for \( y \):

\[
\frac{3}{5}y + 2y = 3 + 36
\]

\[
\frac{3}{5}y + \frac{10}{5}y = 39
\]

\[
\frac{13}{5}y = 39
\]

Multiply both sides by \( \frac{5}{13} \):

\[
y = 39 \cdot \frac{5}{13}
\]

\[
y = 15
\]

So, for the third equation, \( y = 15 \).

### Summary of Solutions:
1. \( y = -4 \) for the first equation.
2. \( y = \frac{21}{4} \) for the second equation.
3. \( y = 15 \) for the third equation.

Would you like more details on any step, or do you have any other questions?

### Follow-up Questions:
1. How do we handle equations when there are multiple variables?
2. What is the method to check if the solutions are correct?
3. How do we solve these equations if we didn't know the value of \( x \)?
4. Can you explain how to solve equations involving fractions in more depth?
5. How can we solve equations graphically?

#### Tip:
When solving equations with fractions, clearing the fractions by multiplying through by a common denominator can simplify the process significantly.

1/2x - y = 3/2x - 3;  x=7
3/4x + 4/7y = 5/4x - 1; x=8
3/5y - 4x = 3 - 2y; x=9

Explore how to solve three linear equations by substituting given values for x and solving for y. This example covers core algebraic methods and demonstrates step-by-step solutions for each equation. Suitable for students in Grades 8-10.

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