Question

solve the system of equations using Elimination, substitution, then graph the equation.

{3x - 2y = 6
{x + y = -8

Answer 1

To solve the given system of equations using elimination, substitution, and then graph them, we follow these steps:

### Given Equations:
1. \( 3x - 2y = 6 \) (Equation 1)
2. \( x + y = -8 \) (Equation 2)

### Step 1: Solving by Elimination
We will eliminate one variable by manipulating the equations.

First, we can express \( y \) from Equation 2:

\[
y = -8 - x
\]

Now we can substitute this expression for \( y \) into Equation 1:

\[
3x - 2(-8 - x) = 6
\]

Now simplify:

\[
3x + 16 + 2x = 6
\]

Combine like terms:

\[
5x + 16 = 6
\]

Now, isolate \( x \):

\[
5x = 6 - 16
\]
\[
5x = -10
\]
\[
x = -2
\]

Now that we have \( x \), we can substitute \( x = -2 \) back into Equation 2 to find \( y \):

\[
-2 + y = -8
\]
\[
y = -8 + 2
\]
\[
y = -6
\]

So the solution to the system of equations is:

\[
(x, y) = (-2, -6)
\]

### Step 2: Solving by Substitution
We already solved for \( y \) in terms of \( x \) using Equation 2. The solution has been confirmed as \( x = -2 \) and \( y = -6 \).

### Step 3: Graphing the Equations
Let's graph the equations using the intercept method.

1. **For Equation 1: \( 3x - 2y = 6 \)**:
- To find the intercepts:
- Set \( x = 0 \):
\[
3(0) - 2y = 6 \implies -2y = 6 \implies y = -3 \quad \text{(y-intercept is (0, -3))}
\]
- Set \( y = 0 \):
\[
3x - 2(0) = 6 \implies 3x = 6 \implies x = 2 \quad \text{(x-intercept is (2, 0))}
\]

- Plot the points (2, 0) and (0, -3) and draw the line.

2. **For Equation 2: \( x + y = -8 \)**:
- To find the intercepts:
- Set \( x = 0 \):
\[
0 + y = -8 \implies y = -8 \quad \text{(y-intercept is (0, -8))}
\]
- Set \( y = 0 \):
\[
x + 0 = -8 \implies x = -8 \quad \text{(x-intercept is (-8, 0))}
\]

- Plot the points (0, -8) and (-8, 0) and draw the line.

### Conclusion
The graph of the two equations will intersect at the point \((-2, -6)\), which is the solution to the system of equations.

You can visualize this by plotting both lines on a coordinate plane. The intersection point will confirm that both methods (elimination and substitution) yield the same solution:

\[
\boxed{(-2, -6)}
\]