We are given three systems of equations and asked to find the solutions using graphing. System 31: $y = x^2 - 3x - 6$ $y + x = 2$ System 32: $y = -x^2 - 4x + 8$ $y = -x - 2$ System 33: $y = -x^2 - 4x + 1$ $y = -x - 3$

1. Problem Description

We are given three systems of equations and asked to find the solutions using graphing.

System 31:

$y = x^2 - 3x - 6$

$y + x = 2$

System 32:

$y = -x^2 - 4x + 8$

$y = -x - 2$

System 33:

$y = -x^2 - 4x + 1$

$y = -x - 3$

2. Solution Steps

System 31:

First, rewrite the second equation as $y = 2 - x$.

Now, set the two equations equal to each other:

$x^2 - 3x - 6 = 2 - x$

$x^2 - 2x - 8 = 0$

$(x - 4)(x + 2) = 0$

$x = 4$ or $x = -2$

If $x = 4$, then $y = 2 - 4 = -2$.

If $x = -2$, then $y = 2 - (-2) = 4$.

So the solutions are $(4, -2)$ and $(-2, 4)$.

System 32:

Set the two equations equal to each other:

$-x^2 - 4x + 8 = -x - 2$

$0 = x^2 + 3x - 10$

$0 = (x + 5)(x - 2)$

$x = -5$ or $x = 2$

If $x = -5$, then $y = -(-5) - 2 = 5 - 2 = 3$.

If $x = 2$, then $y = -2 - 2 = -4$.

So the solutions are $(-5, 3)$ and $(2, -4)$.

System 33:

Set the two equations equal to each other:

$-x^2 - 4x + 1 = -x - 3$

$0 = x^2 + 3x - 4$

$0 = (x + 4)(x - 1)$

$x = -4$ or $x = 1$

If $x = -4$, then $y = -(-4) - 3 = 4 - 3 = 1$.

If $x = 1$, then $y = -1 - 3 = -4$.

So the solutions are $(-4, 1)$ and $(1, -4)$.

3. Final Answer

System 31: $(4, -2), (-2, 4)$

System 32: $(-5, 3), (2, -4)$

System 33: $(-4, 1), (1, -4)$