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

\[
-2x + y = -3 \tag{1}
\]
\[
5x - 3y = 5 \tag{2}
\]

### Step 1: Solve one equation for one variable
From equation (1), we can solve for \( y \) in terms of \( x \):

\[
y = 2x - 3 \tag{3}
\]

### Step 2: Substitute \( y \) into the second equation
Now, substitute equation (3) into equation (2):

\[
5x - 3(2x - 3) = 5
\]

### Step 3: Simplify and solve for \( x \)
Expand the equation:

\[
5x - 6x + 9 = 5
\]

Combine like terms:

\[
-x + 9 = 5
\]

Subtract 9 from both sides:

\[
-x = -4
\]

Solve for \( x \):

\[
x = 4
\]

### Step 4: Substitute \( x = 4 \) into the expression for \( y \)
Substitute \( x = 4 \) into equation (3):

\[
y = 2(4) - 3 = 8 - 3 = 5
\]

### Final Solution:
The solution to the system is:
\[
x = 4, \quad y = 5
\]

Would you like more details or have any further questions?

Here are 5 related questions to expand on this topic:
1. Can you solve the same system using the elimination method?
2. How does substitution differ from elimination in terms of strategy?
3. What happens if the system of equations has no solution or infinitely many solutions?
4. Can substitution be used for non-linear systems? If so, how?
5. How can you check if your solution is correct for any system of equations?

**Tip:** After solving a system, always substitute the values back into the original equations to verify your solution.

Use substitution to solve the system: -2x + y = -3 and 5x - 3y = 5.

Learn how to solve the system of linear equations -2x + y = -3 and 5x - 3y = 5 using the substitution method. This problem includes key algebraic concepts such as solving for one variable and substituting into another equation. Suitable for students in Grades 8-10.

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