# How do you solve the system of equations x+ y = 10 and y = x - 6?

Oct 18, 2016

(8,2)

#### Explanation:

To solve this equation, I would use substitution. The first equation is $x + y = 10$. The second equation is $y = x - 6$. Substitute the second equation into the first one and simplify.

$\textcolor{red}{x + y = 10}$ and $\textcolor{b l u e}{y = x - 6}$
$\textcolor{red}{x} + \left(\textcolor{b l u e}{x - 6}\right) = \textcolor{red}{10}$
$\textcolor{p u r p \le}{2 x - 6 = 10}$
$\textcolor{p u r p \le}{2 x - 6} + 6 = \textcolor{p u r p \le}{10} + 6$
$\textcolor{p u r p \le}{2 x = 16}$
$\frac{\textcolor{p u r p \le}{2 x}}{2} = \frac{\textcolor{p u r p \le}{16}}{2}$
$\textcolor{p u r p \le}{x = 8}$

Once you have solved for $x$, DO NOT FORGET to solve for $y$. Plug $\textcolor{p u r p \le}{x = 8}$ in for $x$ in either of the original equations and simplify for $y$. I am going to solve for $y$ using the second equation.
$\textcolor{b l u e}{y = x - 6}$
$\textcolor{b l u e}{y =} \textcolor{p u r p \le}{8} \textcolor{b l u e}{- 6}$
$\textcolor{b l u e}{y = 2}$