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

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

1 answer

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:

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