Solve the system of equations by the substitution method.
left brace Start 2 By 1 Matrix 1st Row 1st Column x plus y equals 15 2nd Row 1st Column y equals negative 4 x EndMatrix x+y=15	y=−4x

We are given the following system of equations:

\[
\begin{cases}
x + y = 15 \quad (1) \\
y = -4x \quad (2)
\end{cases}
\]

### Step 1: Substitute the expression for \( y \) from equation (2) into equation (1).

From equation (2), we know:
\[
y = -4x
\]
Substitute this into equation (1):
\[
x + (-4x) = 15
\]
Simplify the equation:
\[
x - 4x = 15 \implies -3x = 15
\]

### Step 2: Solve for \( x \).

To solve for \( x \), divide both sides by -3:
\[
x = \frac{15}{-3} = -5
\]

### Step 3: Substitute \( x = -5 \) back into equation (2) to find \( y \).

Using equation (2):
\[
y = -4x = -4(-5) = 20
\]

### Final Solution:
The solution to the system of equations is:
\[
x = -5, \quad y = 20
\]

Would you like further details or have any questions?

Here are 5 related questions:
1. How do you check if a solution satisfies both equations in a system?
2. Can a system of equations have no solution? If so, when?
3. What is the difference between substitution and elimination methods?
4. How would you graphically interpret the solution \( x = -5 \), \( y = 20 \)?
5. Can you solve a system of equations where both equations are nonlinear?

**Tip**: Always substitute the solution back into the original equations to verify its correctness.

Solve the system of equations by the substitution method: x + y = 15, y = -4x

This guide demonstrates how to solve the system of equations x + y = 15 and y = -4x using the substitution method. You'll learn step-by-step techniques for substituting and solving linear equations, ideal for Grades 8-10.

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