# How do you solve the simultaneous equations y = x^2- 2 and  y = 3x + 8?

Mar 2, 2018

x = 5 , y= 23 and x = -2 , y = 2.

#### Explanation:

Here, we have a quadratic and a linear equation which can be solved by the substitution method.

From the second equation, know $y = 3 x + 8$
put this in place of $y$ in the first equation. We get,
$3 x + 8 = {x}^{2} - 2$
Rearrange
${x}^{2} - 3 x - 10 = 0$
It can be factorized :
${x}^{2} - 5 x + 2 x - 10 = 0$
$x \left(x - 5\right) + 2 \left(x - 5\right) = 0$
$\left(x - 5\right) \left(x + 2\right) = 0$

Therefore, $x = 5$ and $x = - 2$ are the roots of equation.
Plugging it into the second equation we get $y$ as:
$y = 3 \left(5\right) + 8 = 23$
$y = 3 \left(- 2\right) + 8 = 2$

So we get the solution as $\left(5 , 23\right) \left(- 2 , 2\right)$

Mar 2, 2018

$\left(- 2 , 2\right) , \left(5 , 23\right)$

#### Explanation:

$\text{Since both equations give y in terms of x we can equate}$
$\text{them}$

$\Rightarrow {x}^{2} - 2 = 3 x + 8$

$\text{rearrange into standard form}$

y$\Rightarrow {x}^{2} - 3 x - 10 = 0 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\text{the factors of - 10 which sum to - 3 are - 5 and + 2}$

$\Rightarrow \left(x - 5\right) \left(x + 2\right) = 0$

$\text{equate each factor to zero and solve for x}$

$x - 5 = 0 \Rightarrow x = 5$

$x + 2 = 0 \Rightarrow x = - 2$

$\text{substitute these values into } y = 3 x + x$

$x = 5 \Rightarrow y = \left(3 \times 5\right) + 8 = 23$

$x = - 2 \Rightarrow y = \left(3 \times - 2\right) + 8 = 2$

$\text{the solutions are "(5,23)" and } \left(- 2 , 2\right)$
graph{(y-3x-8)(y-x^2+2)((x+2)^2+(y-2)^2-0.04)((x-5)^2+(y-23)^2-0.04)=0 [-10, 10, -5, 5]}