# How do you solve the system of equations 2x + 3y = 13 and 3x - 5y = - 9?

Apr 4, 2018

#(x,y)=(2,3)

#### Explanation:

$2 x = 13 - 3 y$
$x = \frac{13}{2} - \frac{3}{2} y$ solve for $x$ in terms of $y$ in the first equation
$3 \left(\frac{13}{2} - \frac{3}{2} y\right) - 5 y = - 9$ plug into the other equation
$\frac{39}{2} - \frac{9}{2} y - \frac{10}{2} y = - \frac{18}{2}$ distribute and set a common denominator
$\frac{39}{2} - \frac{19}{2} y = - \frac{18}{2}$
$39 - 19 y = - 18$ this is a matter of personal preference, but I like to multiply the whole equation by the denominator to get rid of fractions (for now)
$- 19 y = - 57$
$y = \frac{57}{19}$
$y = 3$

Now, going back to the other equation
$2 x + 3 \left(\frac{57}{19}\right) = 13$ plug in value of y
$2 x + \frac{171}{19} = 13$
$2 x + 9 = 13$
$2 x = 4$
$x = 2$