# How do you solve y=sqrt x and y=x-6 using substitution?

Sep 29, 2016

When $x = 9 , y = 2 \text{ or " 3}$
When $x = 4 , y = 3 \text{ or } - 2$

#### Explanation:

You can solve the equations by equating the y's and then substituting for each y.

$\textcolor{red}{y = \sqrt{x}} \textcolor{w h i t e}{\times \times} \mathmr{and} \text{ } \textcolor{b l u e}{y = x - 6}$

$\textcolor{w h i t e}{\times \times \times \times x} \textcolor{red}{y} = \textcolor{b l u e}{y}$

Therefore..$\textcolor{red}{\sqrt{x}} = \textcolor{b l u e}{x - 6} \text{ } \leftarrow$ only x terms

${\left(\sqrt{x}\right)}^{2} = {\left(x - 6\right)}^{2} \text{ } \leftarrow$ square both sides.

$x = {x}^{2} - 12 x + 36 \text{ } \leftarrow$ make the quadratic = 0

${x}^{2} - 13 x + 36 = 0 \text{ } \leftarrow$ factorize

Find factors of 36 which ADD to 13. Signs are both negative.

$\left(x - 9\right) \left(x - 4\right) = 0$

If $x - 9 = 0 \rightarrow x = 9 \text{ }$ OR If $\text{ } x - 4 = 0 \rightarrow x = 4$
Now find y.
$\textcolor{red}{y = \sqrt{x}} \textcolor{w h i t e}{\times \times} \mathmr{and} \text{ } \textcolor{b l u e}{y = x - 6}$

$y = \sqrt{9} = 3 \text{ } \leftarrow$ only the principal square root was indicated
$y = \sqrt{4} = 2$
$y = x - 6 \rightarrow y = 9 - 6 = 3$
$y = x - 6 \rightarrow y = 4 - 6 = - 2$