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We are given a system of equations and asked to solve it using the substitution method. The system is:

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

### Step 1: Solve one equation for one variable
We can solve the first equation for \(y\):

\[
-x + y = -1 \implies y = x - 1
\]

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

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

Simplify:

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

\[
x + 3 = 3
\]

\[
x = 0
\]

### Step 3: Solve for \(y\)
Now that we know \(x = 0\), substitute this back into the equation for \(y\):

\[
y = x - 1 = 0 - 1 = -1
\]

### Final Solution:
The solution to the system is \(x = 0\) and \(y = -1\).

Thus, the solution is:

\[
(0, -1)
\]

Would you like further details or have any questions?

Here are 5 related questions to consider:
1. How do you apply the substitution method in different systems of equations?
2. What happens if both equations represent the same line?
3. How can you check if a system has no solution or infinite solutions?
4. What are the differences between substitution and elimination methods?
5. Can all systems of equations be solved using the substitution method?

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

Use the substitution method to solve the system: -x + y = -1, 4x - 3y = 3

Learn how to solve the system of equations -x + y = -1 and 4x - 3y = 3 using the substitution method. This step-by-step guide covers key algebra concepts and is suitable for students in Grades 8-10.

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