# How do you solve 5y – x – 6 = 0 and -3x + 6y = -9?

May 19, 2016

$x = 9 , y = 3$

#### Explanation:

First Method, using substitution

By rearranging the first equation we have,

$x = 5 y - 6$

substitute value of $x$ in second equation

$- 3 \cdot \left(5 y - 6\right) + 6 y = - 9$

$- 15 y + 18 + 6 y = - 9$

$- 9 y + 18 = - 9$

$\Rightarrow y - 2 = 1$
$\mathmr{and} y = 3$

by substituting value of $y$ in either of the equations
we have

$x = 9$

Second Method,
Multiply first equation with 3,

$3 \cdot \left(5 y - x - 6\right) = 0$

$15 y - 3 x - 18 = 0$

Now subtract LHS of the second equation from the LHS above equation and do the same with the RHS,

$\left(15 y - 3 x - 18\right) - \left(- 3 x + 6 y\right) = 0 - \left(- 9\right)$

$15 y - 3 x - 18 + 3 x - 6 y = 9$

$9 y - 18 = 9$
$\Rightarrow y = 3$

and by substituting $y$ in either of the equations we have

$x = 9$