# How do you solve 9x -2y = 28 and y= -3x + 1?

Oct 13, 2015

$\left\{\begin{matrix}x = 2 \\ y = - 5\end{matrix}\right.$

#### Explanation:

Your starting system of equations looks like this

$\left\{\begin{matrix}9 x - 2 y = 28 \\ y = - 3 x + 1\end{matrix}\right.$

Notice that you can rearrange the second equation to get the $x$ and $y$ terms on the left-hand side of the equation

$\left\{\begin{matrix}9 x - 2 y = 28 \\ 3 x + y = 1\end{matrix}\right.$

Another thing to notice here is that if you multiply the second equation by $2$, and add the left-hand sides and right-hand sides of the two equations separately, you can cancel out the $y$ term.

This will allow you to find the value of $x$.

$\left\{\begin{matrix}9 x - 2 y = 28 \\ 3 x + \textcolor{w h i t e}{x} y = \textcolor{w h i t e}{x} 1 | \times 2\end{matrix}\right.$

$\left\{\begin{matrix}9 x - 2 y = 28 \\ 6 x + 2 y = \textcolor{w h i t e}{x} 2\end{matrix}\right.$

Now add the two equations as described above

$\left\{\begin{matrix}9 x - 2 y = 28 \\ 6 x + 2 y = \textcolor{w h i t e}{x} 2\end{matrix}\right.$
color(white)(x)stackrel("-------------------------------")

$9 x - \textcolor{red}{\cancel{\textcolor{b l a c k}{2 y}}} + 6 x + \textcolor{red}{\cancel{\textcolor{b l a c k}{2 y}}} = 28 + 2$

$15 x = 30 \implies x = \frac{30}{15} = \textcolor{g r e e n}{2}$

Now use this value of $x$ into one of the two equations to get the value of $y$

$3 \cdot \left(\textcolor{g r e e n}{2}\right) + y = 1$

$y = 1 - 6 = \textcolor{g r e e n}{- 5}$

The solution set for this system of equations will be

$\left\{\begin{matrix}x = 2 \\ y = - 5\end{matrix}\right.$