# How do you solve using elimination of 5m + 3n = 1.5 and -8m - 2n = 20?

Oct 31, 2015

First, scale the equations such that one of the variables for both equations will have the same coefficient (the sign need not be the same).

$\left[1\right] 5 m + 3 n = 1.5$
$\left[2\right] - 8 m - 2 n = 20$

For the above equations, let us multiply $\left[1\right]$ by $2$ and $\left[2\right]$ by $3$

$\left[1\right] \implies 2 \left(5 m + 3 n = 1.5\right)$
$\left[1\right] \implies 10 m + 6 n = 3$

$\left[2\right] \implies 3 \left(- 8 m - 2 n = 20\right)$
$\left[2\right] \implies - 24 m - 6 n = 60$

Now, let's add equations $\left[1\right]$ and $\left[2\right]$

$\left[1\right] \implies 10 m + 6 n = 3$
$\left[2\right] \implies - 24 m - 6 n = 60$

$\left[1\right] + \left[2\right] \implies - 14 m = 60$
$\implies m = - \frac{60}{14}$

$\implies m = - \frac{30}{7}$

To get $n$, simply substitute the value of $m$ to either $\left[1\right]$ or $\left[2\right]$ and solve for n

$\left[1\right] \implies 5 m + 3 n = 1.5$

$\left[1\right] \implies 10 \left(- \frac{30}{7}\right) + 6 n = 3$

$\left[1\right] \implies - \frac{300}{7} + 6 n = 3$

$\left[1\right] \implies 6 n = 3 + \frac{300}{7}$

$\left[1\right] \implies 6 n = \frac{21 + 300}{7}$

$\left[1\right] \implies 6 n = \frac{321}{7}$

$\left[1\right] \implies n = \frac{321}{7 \cdot 6}$

$\left[1\right] \implies n = \frac{107 \cdot 3}{7 \cdot 2 \cdot 3}$

$\left[1\right] \implies n = \frac{107}{14}$