# How do you solve the system of equations 2x - 5y = 3 and 3x - 6y = 9?

Apr 7, 2018

$x = 9$
$y = 3$

#### Explanation:

To solve this system of equations, we will first manipulate the first equation so that $y$ is alone on the left-hand side of the equation.

$2 x - 5 y = 3$,
$- 5 y = 3 - 2 x$,
$5 y = - 3 + 2 x$,
$5 y = 2 x - 3$,
$y = \frac{2}{5} x - \frac{3}{5}$.

We now plug this value of $y$ into the second equation.

$3 x - 6 y = 9$,
$3 x - 6 \left(\frac{2}{5} x - \frac{3}{5}\right) = 9$,
$3 x - \frac{12}{5} x + \frac{18}{5} = 9$.

Move all the $x$'s to one side of the equation and the constants to the other.

$3 x - \frac{12}{5} x + \frac{18}{5} = 9$,
$3 x - \frac{12}{5} x = 9 - \frac{18}{5}$.

Multiplying both sides of the equation by $5$ makes it easier to manipulate.

$5 \left(3 x - \frac{12}{5} x\right) = 5 \left(9 - \frac{18}{5}\right)$,
$15 x - 12 x = 45 - 18$,
$3 x = 27$,
$x = 9$.

Now plug this $x$ back into the first equation.

$2 x - 5 y = 3$,
$2 \left(9\right) - 5 y = 3$,
$18 - 5 y = 3$,
$18 = 3 + 5 y$,
$15 = 5 y$,
$3 = y$.

Thus, we have $x = 9$ and $y = 3$.