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

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

### Step 1: Solve one equation for one variable
From the first equation:
\[
-x + y = 1 \implies y = x + 1
\]
(we solved for \( y \)).

### Step 2: Substitute into the second equation
Now substitute \( y = x + 1 \) into the second equation:
\[
4x - 3(x + 1) = -2
\]

### Step 3: Solve for \( x \)
Distribute the \(-3\) on the left-hand side:
\[
4x - 3x - 3 = -2
\]
Simplify:
\[
x - 3 = -2
\]
Add 3 to both sides:
\[
x = 1
\]

### Step 4: Solve for \( y \)
Now that we know \( x = 1 \), substitute it back into \( y = x + 1 \):
\[
y = 1 + 1 = 2
\]

### Final Solution:
The solution to the system is \( x = 1 \) and \( y = 2 \).

### Conclusion:
Since we found a single unique solution, the answer is **"One solution"** with the solution pair \( (x, y) = (1, 2) \).

Do you have any questions or need further details on this method?

### Related Questions:
1. How can you check if a system has no solution or infinite solutions using the substitution method?
2. How does the substitution method differ from the elimination method?
3. What happens if you try solving a system where both equations represent the same line?
4. Can substitution be used in systems with three variables?
5. How do you recognize dependent and independent systems of equations?

### Tip:
Always substitute the solution back into both original equations to ensure it's correct.

Use the substitution method to solve the system of equations: -x + y = 1 and 4x - 3y = -2.

Learn how to solve a system of linear equations using the substitution method. In this problem, we solve the system -x + y = 1 and 4x - 3y = -2 step-by-step, demonstrating key algebra concepts. Suitable for students in Grades 9-10.

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