# How do you solve this system of equations: -\frac { 3} { 5} x + \frac { 1} { 2} y = - 2 and - x + y = - 3?

Nov 26, 2017

$x = 5 \setminus \quad , \setminus \quad y = 2$

#### Explanation:

We’ll use elimination for this one to cancel out the $y$ terms, in order to first solve for $x$.

We’ll multiply the first equation by $- 2$:

$- \setminus \frac{3}{5} x + \setminus \frac{1}{2} y = - 2$

$\setminus \rightarrow - 2 \left(- \setminus \frac{3}{5} x + \setminus \frac{1}{2} y\right) = \left(- 2\right) - 2$

$\setminus \rightarrow \setminus \frac{6}{5} x - y = 4$

Now we can add this equation to the other equation to solve for $x$:

$\left(\setminus \frac{6}{5} x - y = 4\right) \setminus \quad + \setminus \quad \left(- x + y = - 3\right)$

$\setminus \rightarrow \setminus \frac{1}{5} x = 1$

$x = 5$

Now we can plug the value of $x$ into an equation to find the value of $y$:

$- x + y = - 3$

$\setminus \rightarrow - 5 + y = - 3$

$\setminus \rightarrow y = 2$

Having solved for both variables, we can check our work by plugging those values into one of the equations:

$- \setminus \frac{3}{5} x + \setminus \frac{1}{2} y = - 2$

$- \setminus \frac{3}{5} \left(5\right) + \setminus \frac{1}{2} \left(2\right) = - 2$

$- 3 + 1 = - 2$

$- 2 = - 2$