# How do you solve y² = x + 3 and x - 2y = 12?

Jun 10, 2018

If $y = 5$, $x = 22$
If $y = - 3$, $x = 6$

#### Explanation:

${y}^{2} = x + 3$ --- (1)
$x - 2 y = 12$ --- (2)

From (2)
$x - 2 y = 12$
$x = 12 + 2 y$ --- (3)

Sub (3) into (1)
${y}^{2} = 12 + 2 y + 3$
${y}^{2} = 15 + 2 y$
${y}^{2} - 2 y - 15 = 0$
$\left(y - 5\right) \left(y + 3\right) = 0$
$y = 5$ or $y = - 3$

If $y = 5$,
$x = 12 + 2 y$
$x = 12 + 10$
$x = 22$

If $y = - 3$,
$x = 12 + 2 y$
$x = 12 - 6$
$x = 6$

Jun 10, 2018

$\left(6 , - 3\right) , \left(22 , 5\right)$

#### Explanation:

${y}^{2} = x + 3 \to \left(1\right)$

$x - 2 y = 12 \to \left(2\right)$

$\text{from equation } \left(2\right) \to x = 12 + 2 y \to \left(3\right)$

$\text{substitute "x=12+2y" in equation } \left(1\right)$

${y}^{2} = 12 + 2 y + 3$

${y}^{2} - 2 y - 15 = 0 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\left(y - 5\right) \left(y + 3\right) = 0$

$y + 3 = 0 \Rightarrow y = - 3$

$y - 5 = 0 \Rightarrow y = 5$

$\text{substitute these values into equation } \left(3\right)$

$y = - 3 \to x = 12 - 6 = 6 \to \left(6 , - 3\right)$

$y = 5 \to x = 12 + 10 = 22 \to \left(22 , 5\right)$

A graphical solution below:

#### Explanation:

Solving this graphically:

Here's one point of intersection:

graph{(y^2-x-3)(x-2y-12)=0}

And here's the other:

graph{(y^2-x-3)(x-2y-12)=0[10,30,0,10]}