Let's solve the system of equations:

\[
\begin{aligned}
1) & \quad -3x + y = -5 \\
2) & \quad 5x + 2y = 23
\end{aligned}
\]

### Step 1: Solve the first equation for \( y \)
From the first equation:
\[
y = 3x - 5
\]

### Step 2: Substitute this expression for \( y \) into the second equation
Substitute \( y = 3x - 5 \) into the second equation:
\[
5x + 2(3x - 5) = 23
\]

### Step 3: Solve for \( x \)
Expand the equation:
\[
5x + 6x - 10 = 23
\]
\[
11x - 10 = 23
\]
Add 10 to both sides:
\[
11x = 33
\]
Divide by 11:
\[
x = 3
\]

### Step 4: Substitute \( x = 3 \) back into the equation for \( y \)
Now, substitute \( x = 3 \) into \( y = 3x - 5 \):
\[
y = 3(3) - 5 = 9 - 5 = 4
\]

### Final Answer:
The solution to the system of equations is:
\[
x = 3, \quad y = 4
\]

Would you like any more details on this process or have any questions?

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Here are 5 related questions you could explore:
1. How can you verify the solution to a system of equations?
2. What is the graphical interpretation of the solution to this system?
3. How do you solve a system of equations using substitution versus elimination?
4. Can a system of linear equations have no solution or infinitely many solutions?
5. What happens when the lines represented by the equations are parallel?

**Tip:** Always check your solution by substituting the values back into the original equations.

Learn how to solve the system of linear equations −3x + y = −5 and 5x + 2y = 23 using the substitution method. This step-by-step guide covers key algebraic concepts such as solving for one variable and substitution. Suitable for students in Grades 8-10.

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