To solve the given system of equations using elimination, substitution, and then graph them, we follow these steps:
Given Equations:
- \( 3x - 2y = 6 \) (Equation 1)
- \( x + y = -8 \) (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.
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For Equation 1: \( 3x - 2y = 6 \):
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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 \( y = 0 \): \[ 3x - 2(0) = 6 \implies 3x = 6 \implies x = 2 \quad \text{(x-intercept is (2, 0))} \]
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Plot the points (2, 0) and (0, -3) and draw the line.
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For Equation 2: \( x + y = -8 \):
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To find the intercepts:
- Set \( x = 0 \): \[ 0 + y = -8 \implies y = -8 \quad \text{(y-intercept is (0, -8))} \]
- Set \( y = 0 \): \[ x + 0 = -8 \implies x = -8 \quad \text{(x-intercept is (-8, 0))} \]
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Plot the points (0, -8) and (-8, 0) and draw the line.
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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:
\[ \boxed{(-2, -6)} \]