date: ______ per: ______ homework 12: solving nonlinear systems graphically
** this is a 2 - page document! **
solve each system by graphing. be sure to identify the solution(s).
1.
$\begin{cases}y = 4x + 5\\y = x^2 + 8x + 9\end{cases}$
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2.
$\begin{cases}y = -x^2 - 4x + 2\\y = x^2 + 8x + 12\end{cases}$
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3.
$\begin{cases}y = x^2 + 12x + 26\\y = -x^2 - 4x - 6\end{cases}$
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4.
$\begin{cases}y = -x^2 + 10x - 28\\x - y = 7\end{cases}$
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5.
$\begin{cases}y = x^2 - 2x - 7\\y = -3x - 5\end{cases}$
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6.
$\begin{cases}y = x^2 - 6x + 10\\y = -2x^2 + 4x\end{cases}$
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© gina wilson (all things algebra®, llc), 2022

Question

Accepted Answer

### Problem 1:
# Explanation:
## Step 1: Set the equations equal
Set $ 4x + 5 = x^2 + 8x + 9 $
Rearrange to $ x^2 + 4x + 4 = 0 $
## Step 2: Solve the quadratic
Factor: $ (x + 2)^2 = 0 $
So $ x = -2 $
## Step 3: Find y
Substitute $ x = -2 $ into $ y = 4x + 5 $: $ y = 4(-2) + 5 = -3 $
# Answer:
Solution is $ (-2, -3) $

### Problem 2:
# Explanation:
## Step 1: Set equations equal
$ -x^2 - 4x + 2 = x^2 + 8x + 12 $
Rearrange: $ 2x^2 + 12x + 10 = 0 $
Simplify: $ x^2 + 6x + 5 = 0 $
## Step 2: Solve quadratic
Factor: $ (x + 1)(x + 5) = 0 $
So $ x = -1 $ or $ x = -5 $
## Step 3: Find y for each x
For $ x = -1 $: $ y = -(-1)^2 - 4(-1) + 2 = 5 $
For $ x = -5 $: $ y = -(-5)^2 - 4(-5) + 2 = -3 $
# Answer:
Solutions are $ (-1, 5) $ and $ (-5, -3) $

### Problem 3:
# Explanation:
## Step 1: Set equations equal
$ x^2 + 12x + 26 = -x^2 - 4x - 6 $
Rearrange: $ 2x^2 + 16x + 32 = 0 $
Simplify: $ x^2 + 8x + 16 = 0 $
## Step 2: Solve quadratic
Factor: $ (x + 4)^2 = 0 $
So $ x = -4 $
## Step 3: Find y
Substitute $ x = -4 $ into $ y = x^2 + 12x + 26 $: $ y = (-4)^2 + 12(-4) + 26 = 2 $
# Answer:
Solution is $ (-4, 2) $

### Problem 4:
# Explanation:
## Step 1: Rewrite linear equation
From $ x - y = 7 $, we get $ y = x - 7 $
## Step 2: Set equations equal
$ -x^2 + 10x - 28 = x - 7 $
Rearrange: $ -x^2 + 9x - 21 = 0 $ or $ x^2 - 9x + 21 = 0 $
## Step 3: Check discriminant
Discriminant $ D = (-9)^2 - 4(1)(21) = 81 - 84 = -3 $ (negative, no real solutions)
Wait, maybe graphing shows? Wait, let's re - check. Wait, maybe I made a mistake. Wait, $ y=-x^{2}+10x - 28$ and $ y=x - 7$
Set $ -x^{2}+10x - 28=x - 7$
$ -x^{2}+9x - 21 = 0$
$ x^{2}-9x + 21 = 0$
Discriminant $ D = 81-84=-3<0$, so no real solutions. But maybe graphing? Wait, maybe I miscalculated. Wait, let's check the original equations again. $ y=-x^{2}+10x - 28$, vertex at $ x =-\frac{10}{2(-1)} = 5$, $ y=-(25)+50 - 28=-3$. The line $ y=x - 7$, when $ x = 5$, $ y=-2$. The parabola at $ x = 5$ is $ y=-3$, line at $ x = 5$ is $ y=-2$. The parabola opens downward, the line has slope 1. Let's check if there are solutions. Wait, maybe the problem expects us to solve by graphing, but algebraically, discriminant is negative, so no real solutions. But maybe I made a mistake in rearrangement. Wait, $ x - y=7$ implies $ y=x - 7$. Substitute into the first equation: $ x - 7=-x^{2}+10x - 28$
$ x^{2}-9x + 21 = 0$, discriminant $ D = 81 - 84=-3$, so no real solutions.
# Answer:
No real solutions (if we consider real - valued solutions)

### Problem 5:
# Explanation:
## Step 1: Set equations equal
$ x^2 - 2x - 7 = -3x - 5 $
Rearrange: $ x^2 + x - 2 = 0 $
## Step 2: Solve quadratic
Factor: $ (x + 2)(x - 1) = 0 $
So $ x=-2$ or $ x = 1$
## Step 3: Find y for each x
For $ x=-2$: $ y=-3(-2)-5 = 1$
For $ x = 1$: $ y=-3(1)-5=-8$
# Answer:
Solutions are $ (-2, 1) $ and $ (1, -8) $

### Problem 6:
# Explanation:
## Step 1: Set equations equal
$ x^2 - 6x + 10 = -2x^2 + 4x $
Rearrange: $ 3x^2 - 10x + 10 = 0 $
## Step 2: Solve quadratic
Discriminant $ D=(-10)^2 - 4(3)(10)=100 - 120=-20$ (negative, no real solutions? Wait, no, maybe graphing? Wait, let's re - check. Wait, $ y=x^{2}-6x + 10=(x - 3)^{2}+1$, vertex at $ (3,1) $, opens upward. $ y=-2x^{2}+4x=-2(x - 1)^{2}+2$, vertex at $ (1,2) $, opens downward. Let's set them equal: $ x^{2}-6x + 10=-2x^{2}+4x$
$ 3x^{2}-10x + 10 = 0$
Discriminant $ D = 100-120=-20<0$, so no real solutions. But maybe the problem has a typo? Wait, maybe I made a mistake. Wait, let's check the equations again. $ y=x^{2}-6x + 10$ and $ y=-2x^{2}+4x$
Set $ x^{2}-6x + 10=-2x^{2}+4x$
$ 3x^{2}-10x + 10 = 0$
Discriminant $ D = 100 - 120=-20$, so no real solutions. But if we consider graphing, the parabola $ y = x^{2}-6x + 10$ is always above $ y = 0$ (since $ (x - 3)^{2}+1\geq1$) and $ y=-2x^{2}+4x=-2(x - 1)^{2}+2$ has maximum at $ y = 2$ when $ x = 1$. At $ x = 1$, $ y=x^{2}-6x + 10=1 - 6 + 10 = 5$, and $ y=-2x^{2}+4x=2$. So the two graphs do not intersect, so no real solutions.
# Answer:
No real solutions (if we consider real - valued solutions)

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Explanation:

Problem 1:

Step 1: Set the equations equal

Step 2: Solve the quadratic

Step 3: Find y

Step 1: Set equations equal

Step 2: Solve quadratic

Step 3: Find y for each x

Step 1: Set equations equal

Step 2: Solve quadratic

Step 3: Find y

Answer:

Problem 2: