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

May 3, 2016

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

#### Explanation:

Notice how easily we can isolate $x$ in the equation ${y}^{2} = x + 3$. We can subtract $3$ from both sides of the equation to reveal that $\textcolor{b l u e}{x = {y}^{2} - 3}$. With this, we can replace $\textcolor{b l u e}{x}$ in the second equation, $\textcolor{b l u e}{x} - 2 y = 12$, with the value that we know $\textcolor{b l u e}{x}$ is equal to: $\textcolor{b l u e}{{y}^{2} - 3}$.

This gives us: $\text{ "color(blue)x-2y=12" "=>" } \left(\textcolor{b l u e}{{y}^{2} - 3}\right) - 2 y = 12$

Continuing to solve the resulting equation, we obtain ${y}^{2} - 2 y - 15 = 0$, which can be factored into $\left(y - 5\right) \left(y + 3\right) = 0$ and solved giving two solutions for $y$, which are that $\textcolor{red}{y = 5}$ and $\textcolor{g r e e n}{y = - 3}$.

Each of the values of $y$ can be plugged into either one of the two equations to find their corresponding values of $x$. I will choose the first equation, although you will receive the same results should you choose to input your values of $y$ into the second.

Solving for $x$ when $\textcolor{red}{y = 5}$, we obtain:

${\textcolor{red}{y}}^{2} = x + 3 \text{ "=>" "color(red)5^2=x+3" "=>" } \textcolor{red}{x = 22}$

Since $\textcolor{red}{x = 22}$ and $\textcolor{red}{y = 5}$, we have a solution point at $\left(\textcolor{red}{22} , \textcolor{red}{5}\right)$.

And for when $\textcolor{g r e e n}{y = - 3}$, we obtain:

${\textcolor{g r e e n}{y}}^{2} = x + 3 \text{ "=>" "(color(green)(-3))^2=x+3" "=>" } \textcolor{g r e e n}{x = 6}$

Thus there is also a solution point at $\left(\textcolor{g r e e n}{6} , \textcolor{g r e e n}{- 3}\right)$.

Graphing the two equations should reveal a sideways parabola with a line intersecting it at the points $\left(22 , 5\right)$ and $\left(6 , - 3\right)$:

graph{(y^2-x-3)(x-2y-12)=0 [-5.14, 30.9, -7.66, 10.36]}