Question

examples - round to three decimal places if needed
solve the systems by graphing.
y=-2x² -x + 3
y=-x + 6
solve the systems by graphing.
y=½x² - 2x - 7
y=3x - 15

Explanation:

System 1: $y=-2x^2-x+3$ and $y=-x+6$

Step1: Set equations equal

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

Step2: Simplify to solve for $x$

Cancel $-x$ on both sides:
$-2x^2 + 3 = 6$
$-2x^2 = 3$
$x^2 = -\frac{3}{2}$
Since a square of a real number cannot be negative, there are no real solutions for this system.

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System 2: $y=\frac{1}{2}x^2-2x-7$ and $y=3x-15$

Step1: Set equations equal

$\frac{1}{2}x^2 - 2x - 7 = 3x - 15$

Step2: Multiply by 2 to eliminate fraction

$x^2 - 4x - 14 = 6x - 30$

Step3: Rearrange to standard quadratic form

$x^2 - 10x + 16 = 0$

Step4: Factor the quadratic

$(x-2)(x-8)=0$
Solve for $x$:
$x=2$ or $x=8$

Step5: Find corresponding $y$-values

For $x=2$:
$y=3(2)-15=6-15=-9$
For $x=8$:
$y=3(8)-15=24-15=9$

Answer:

1. System 1 ($y=-2x^2-x+3$ and $y=-x+6$): No real solutions
2. System 2 ($y=\frac{1}{2}x^2-2x-7$ and $y=3x-15$): $(2, -9)$ and $(8, 9)$