\begin{cases}

\frac{1}{2}x-3=\frac{5}{4}y\\

3x+5y+2=0

\end{cases}

We are given the system of linear equations:

\[
\begin{cases}
\frac{1}{2}x - 3 = \frac{5}{4}y \\
3x + 5y + 2 = 0
\end{cases}
\]

### Step 1: Solve the first equation for \( x \) or \( y \).
Start by solving the first equation for \( x \):

\[
\frac{1}{2}x - 3 = \frac{5}{4}y
\]

Add \( 3 \) to both sides:

\[
\frac{1}{2}x = \frac{5}{4}y + 3
\]

Multiply through by 2 to eliminate the fraction:

\[
x = 2 \left( \frac{5}{4}y + 3 \right)
\]

Simplify:

\[
x = \frac{5}{2}y + 6
\]

### Step 2: Substitute into the second equation.
Now, substitute \( x = \frac{5}{2}y + 6 \) into the second equation \( 3x + 5y + 2 = 0 \):

\[
3\left( \frac{5}{2}y + 6 \right) + 5y + 2 = 0
\]

Distribute the 3:

\[
\frac{15}{2}y + 18 + 5y + 2 = 0
\]

Simplify the equation:

\[
\frac{15}{2}y + 5y + 20 = 0
\]

Convert \( 5y \) to have a denominator of 2:

\[
\frac{15}{2}y + \frac{10}{2}y + 20 = 0
\]

Combine like terms:

\[
\frac{25}{2}y + 20 = 0
\]

### Step 3: Solve for \( y \).
Subtract 20 from both sides:

\[
\frac{25}{2}y = -20
\]

Multiply both sides by 2 to eliminate the denominator:

\[
25y = -40
\]

Solve for \( y \):

\[
y = -\frac{40}{25} = -\frac{8}{5}
\]

### Step 4: Solve for \( x \).
Substitute \( y = -\frac{8}{5} \) into \( x = \frac{5}{2}y + 6 \):

\[
x = \frac{5}{2} \left( -\frac{8}{5} \right) + 6
\]

Simplify:

\[
x = -4 + 6 = 2
\]

### Final Answer:
The solution to the system is:

\[
x = 2, \quad y = -\frac{8}{5}
\]

Would you like a more detailed explanation or have any questions?

Here are 5 related questions:
1. How do you solve a system of equations using substitution?
2. What is the elimination method in solving systems of equations?
3. How do you handle equations involving fractions?
4. How can you check if a solution is correct for a system of equations?
5. What are the applications of solving systems of linear equations in real life?

**Tip**: Always double-check your substitution in the second equation to avoid errors in solving systems of equations.

\begin{cases}
\frac{1}{2}x-3=\frac{5}{4}y\\
3x+5y+2=0
\end{cases}

Learn how to solve a system of linear equations using the substitution method with a detailed step-by-step solution. This problem covers algebraic concepts like substitution and handling fractions. Suitable for students in Grades 9-10.

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