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Let's solve the system of equations using the substitution method. The system provided is:

\[
\begin{aligned}
1) & \quad x + 5y = 32 \\
2) & \quad -5x - 3y = -28
\end{aligned}
\]

### Step 1: Solve one equation for one variable

Let's solve the first equation for \( x \):

\[
x + 5y = 32 \quad \Rightarrow \quad x = 32 - 5y
\]

### Step 2: Substitute into the second equation

Now, substitute \( x = 32 - 5y \) into the second equation:

\[
-5(32 - 5y) - 3y = -28
\]

Distribute the \( -5 \):

\[
-160 + 25y - 3y = -28
\]

Simplify:

\[
-160 + 22y = -28
\]

### Step 3: Solve for \( y \)

Add 160 to both sides:

\[
22y = 132
\]

Now, divide by 22:

\[
y = \frac{132}{22} = 6
\]

### Step 4: Solve for \( x \)

Now that we have \( y = 6 \), substitute it back into the equation for \( x \):

\[
x = 32 - 5(6) = 32 - 30 = 2
\]

### Final Answer

The solution to the system is:

\[
(x, y) = (2, 6)
\]

Would you like further details on any part of this solution?

Here are five related questions you might explore:
1. How would the solution change if the first equation was \( x + 5y = 30 \)?
2. How does the substitution method compare to the elimination method?
3. What is the geometric interpretation of solving systems of linear equations?
4. Can you solve the system using matrix methods such as Gaussian elimination?
5. What happens if the system has no solutions or infinitely many solutions?

**Tip:** Always check your solution by substituting the values of \( x \) and \( y \) back into the original equations to verify.

Solve the following system of equations with the substitution method: {x + 5y = 32, -5x - 3y = -28}

Learn how to solve a system of linear equations using the substitution method. This example involves the equations x + 5y = 32 and -5x - 3y = -28, demonstrating step-by-step how to find the solution. Suitable for students in Grades 8-10.

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