Solve the system of equations.
 
\[\begin{aligned}
& -4x+3y = -2
\\\\
& y=x-1
\end{aligned}\]

To solve the given system of equations:

\[
\begin{aligned}
& -4x + 3y = -2 \tag{1} \\
& y = x - 1 \tag{2}
\end{aligned}
\]

### Step 1: Substitute \( y = x - 1 \) (from Equation (2)) into Equation (1)
Substitute \( y \) into the first equation:

\[
-4x + 3(x - 1) = -2
\]

Simplify:

\[
-4x + 3x - 3 = -2
\]

Combine like terms:

\[
- x - 3 = -2
\]

Add 3 to both sides:

\[
- x = 1
\]

Solve for \( x \):

\[
x = -1
\]

### Step 2: Substitute \( x = -1 \) into Equation (2)
Now substitute \( x = -1 \) into \( y = x - 1 \):

\[
y = -1 - 1
\]

\[
y = -2
\]

### Final Solution:
The solution to the system of equations is:

\[
\boxed{x = -1, y = -2}
\]

### Verification:
1. Substitute \( x = -1 \) and \( y = -2 \) into Equation (1):
   \[
   -4(-1) + 3(-2) = 4 - 6 = -2 \quad \text{(True)}.
   \]
2. Substitute \( x = -1 \) into Equation (2):
   \[
   y = -1 - 1 = -2 \quad \text{(True)}.
   \]

The solution satisfies both equations.

---

Do you want further explanation or have questions?  
Here are five related questions to explore:
1. How can substitution be used to solve other systems of equations?  
2. What is the elimination method for solving systems of equations?  
3. How can you check the solution graphically?  
4. What happens if the system of equations has no solution?  
5. How can matrices be used to solve systems of equations?

**Tip**: Always verify solutions by substituting them back into the original equations!

Solve the system of equations: -4x + 3y = -2 and y = x - 1.

Learn how to solve the system of linear equations -4x + 3y = -2 and y = x - 1 using the substitution method. This detailed explanation includes step-by-step calculations and verification, suitable for students in Grades 8-10.

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