How do you find the exact solutions to the system $y = x + 2$ $y=x+2$ and $y = x^{2}$ $y=x^2$ ?

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How do you find the exact solutions to the system $y = x + 2$ $y=x+2$ and $y = x^{2}$ $y=x^2$ ?

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Answer 1 · 2017-02-03T13:30:56

I got:
$(2, 4)$ $(2,4)$
$(- 1, 1)$ $(-1,1)$

Explanation:

We can substitute the first equation into the second for $y$ $y$ :
$x + 2 = x^{2}$ $x+2=x^2$
rearrange and solve for $x$ $x$ using the Quadratic Formula:
$x^{2} - x - 2 = 0$ $x^2-x-2=0$
$x_{1, 2} = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2}$ $x_(1,2)=(1+-sqrt(1+8))/2=(1+-3)/2$
giving:
$x_{1} = \frac{4}{2} = 2$ $x_1=4/2=2$
$x_{2} = - \frac{2}{2} = - 1$ $x_2=-2/2=-1$
that substituted into the original first equation give:
$y_{1} = 2 + 2 = 4$ $y_1=2+2=4$
$y_{2} = - 1 + 2 = 1$ $y_2=-1+2=1$

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How do you find the exact solutions to the system $y = x + 2$ $y=x+2$ and $y = x^{2}$ $y=x^2$ ?

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Explanation:

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How do you find the exact solutions to the system $y = x + 2$ $y=x+2$ and $y = x^{2}$ $y=x^2$ ?