Solve the system by substitution.

minus, 3, x, plus, 4, y, equals, minus, 1

−3*x*+4*y*=

−1

x, equals, y, plus, 1

*x*=

*y*+1

To solve the system of equations using substitution, we have:

1. \(-3x + 4y = -1\) (Equation 1)
2. \(x = y + 1\) (Equation 2)

### Step 1: Substitute Equation 2 into Equation 1

Substitute \(x\) from Equation 2 into Equation 1:

\[
-3(y + 1) + 4y = -1
\]

### Step 2: Simplify the equation

Distribute \(-3\):

\[
-3y - 3 + 4y = -1
\]

Combine like terms:

\[
y - 3 = -1
\]

### Step 3: Solve for \(y\)

Add 3 to both sides:

\[
y = 2
\]

### Step 4: Substitute \(y\) back to find \(x\)

Now, substitute \(y\) back into Equation 2 to find \(x\):

\[
x = y + 1 = 2 + 1 = 3
\]

### Solution

The solution to the system is:

\[
x = 3, \quad y = 2
\]

### Summary

The values of \(x\) and \(y\) are \(3\) and \(2\) respectively.

If you need further details or have any questions, let me know! Here are some related questions you might consider:

1. How can you check if the solution is correct?
2. What is the graphical representation of these equations?
3. How would you solve this system using elimination instead?
4. What happens if you change the coefficients in the equations?
5. Can you solve a similar system with three variables? 

**Tip:** Always substitute carefully to avoid errors, and check your solutions by plugging them back into the original equations.

Solve the system by substitution: -3x + 4y = -1 and x = y + 1.

Explore the step-by-step solution to the system of equations -3x + 4y = -1 and x = y + 1 using substitution. This tutorial covers key algebraic concepts and methods, suitable for students in Grades 8-10.

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