Question

ear and quadratic systems of equations ws
th linear and quadratic system by graphing. state the solution(s) on the line. must be accurate!
2.)
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$$\begin{cases} y = -x^2 - 6x - 6 \\\\ y = 3 \\end{cases}$$

3.)
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$$\begin{cases} y = -(x - 2)^2 + 5 \\\\ y = -x + 1 \\end{cases}$$

4.)
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$$\begin{cases} y = x^2 - 4x + 2 \\\\ y = -\\frac{3}{4}x - 1 \\end{cases}$$

solution(s):
solution(s):

Explanation:

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System 1:

$$\begin{cases} y=x^2+2x-3 \\ y=2x+1 \end{cases}$$

Step1: Set equations equal

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

Step2: Simplify to solve for $x$

$x^2-4=0 \implies x^2=4 \implies x=2 \text{ or } x=-2$

Step3: Find $y$ for each $x$

For $x=2$: $y=2(2)+1=5$
For $x=-2$: $y=2(-2)+1=-3$
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System 2:

$$\begin{cases} y=-x^2-6x-6 \\ y=3 \end{cases}$$

Step1: Set equations equal

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

Step2: Simplify to solve for $x$

$-x^2-6x-9=0 \implies x^2+6x+9=0 \implies (x+3)^2=0 \implies x=-3$

Step3: Find corresponding $y$

$y=3$
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System 3:

$$\begin{cases} y=-(x-2)^2+5 \\ y=-x+1 \end{cases}$$

Step1: Set equations equal

$-(x-2)^2+5=-x+1$

Step2: Expand and simplify

$-(x^2-4x+4)+5=-x+1 \implies -x^2+4x-4+5=-x+1$
$-x^2+5x=0 \implies x(-x+5)=0 \implies x=0 \text{ or } x=5$

Step3: Find $y$ for each $x$

For $x=0$: $y=-0+1=1$
For $x=5$: $y=-5+1=-4$
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System 4:

$$\begin{cases} y=x^2-4x+2 \\ y=-\frac{3}{4}x-1 \end{cases}$$

Step1: Set equations equal

$x^2-4x+2=-\frac{3}{4}x-1$

Step2: Multiply by 4, simplify

$4x^2-16x+8=-3x-4 \implies 4x^2-13x+12=0$

Step3: Solve quadratic equation

Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, where $a=4, b=-13, c=12$
$\Delta=(-13)^2-4(4)(12)=169-192=-23<0$
No real solutions.

Answer:

1. Solutions: $(-2, -3)$ and $(2, 5)$
2. Solution: $(-3, 3)$
3. Solutions: $(0, 1)$ and $(5, -4)$
4. No real solutions