Question #df22a

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Question #df22a

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Answer 1 · 2017-02-25T20:07:56

I got:
$x_{1} = 3 and y_{1} = 6$ $x_1=3 and y_1=6$
$x_{2} = - 1 and y_{2} = - 2$ $x_2=-1 and y_2=-2$

Explanation:

We can try substituting the first into the second equation for $y$ $y$ to get:
$2 x = x^{2} - 3$ $2x=x^2-3$
$x^{3} - 2 x - 3 = 0$ $x^3-2x-3=0$
solve using the Quadratic Formula:
$x_{1, 2} = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm 4}{2}$ $x_(1,2)=(2+-sqrt(4+12))/2=(2+-4)/2$
we get two solutions:
$x_{1} = \frac{6}{2} = 3$ $x_1=6/2=3$
$x_{2} = - \frac{2}{3} = - 1$ $x_2=-2/3=-1$

These two values, substituted into the first equation will give us:
$x_{1} = 3$ $x_1=3$ then $y_{1} = 2 \cdot 3 = 6$ $y_1=2*3=6$
$x_{2} = - 1$ $x_2=-1$ then $y_{2} = 2 \cdot (- 1) = - 2$ $y_2=2*(-1)=-2$

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