# Question #df22a

##### 1 Answer
Feb 25, 2017

I got:
${x}_{1} = 3 \mathmr{and} {y}_{1} = 6$
${x}_{2} = - 1 \mathmr{and} {y}_{2} = - 2$

#### Explanation:

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

These two values, substituted into the first equation will give us:
${x}_{1} = 3$ then ${y}_{1} = 2 \cdot 3 = 6$
${x}_{2} = - 1$ then ${y}_{2} = 2 \cdot \left(- 1\right) = - 2$

Graphically: