# For the following system of equations, what is the x-value of the solution -x+2y=6, 6y=x+18?

Jan 10, 2017

$x = 0$

#### Explanation:

Step 1) Solve the first equation for $y$:

$- x + 2 y = 6$

$- x + \textcolor{red}{x} + 2 y = 6 + \textcolor{red}{x}$

$0 + 2 y = 6 - x$

$2 y = 6 - x$

$\frac{2 y}{\textcolor{red}{2}} = \frac{6 - x}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} y}{\cancel{\textcolor{red}{2}}} = \frac{6}{2} - \frac{x}{2}$

$y = 3 - \frac{x}{2}$

Step 2) Substitute $3 - \frac{x}{2}$ for $y$ in the second equation and solve for $x$:

$6 \left(3 - \frac{x}{2}\right) = x + 18$

$18 - \frac{6 x}{2} = x + 18$

$18 - 3 x = x + 18$

$18 - 3 x + \textcolor{red}{3 x} - \textcolor{b l u e}{18} = x + 18 + \textcolor{red}{3 x} - \textcolor{b l u e}{18}$

$18 - \textcolor{b l u e}{18} - 3 x + \textcolor{red}{3 x} = x + \textcolor{red}{3 x} + 18 - \textcolor{b l u e}{18}$

$0 - 0 = 4 x + 0$

$0 = 4 x$

$\frac{0}{\textcolor{red}{4}} = \frac{4 x}{\textcolor{red}{4}}$

$0 = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} x}{\cancel{\textcolor{red}{4}}}$

$0 = x$

$x = 0$