How do you solve the following system: 3x+2y =0, 7y − 6x − 5 = 0 ?

1 Answer
Apr 3, 2016

$y = \frac{5}{11} , x = - \frac{10}{33}$

Explanation:

Rearrange the first equation to give a value of $y$ with respect to $x$.

$3 x + 2 y = 0$
$3 x = - 2 y$
$- \frac{3}{2} x = y$

Now substitute your new value of $y$ into the second equation

$7 \left(- \frac{3}{2} x\right) - 6 x - 5 = 0$
$- \frac{21}{2} x - 6 x - 5 = 0$
#-33x - 5 = 0

Rearrange and solve for $x$

$- \frac{33}{2} x = 5$
$x = - \frac{10}{33}$

And because we know $y$ with respect to $x$, we can solve

$y = - \frac{3}{2} x$
$y = - \frac{3}{2} \cdot - \frac{10}{33} = \frac{30}{66} = \frac{5}{11}$

Substitute both values into either original equation and check if it's right.

$3 x + 2 y = 0$
$- \frac{30}{33} + \frac{10}{11} = 0$

It checks out mathematically, so it must be correct.