# How do you solve the following system?: -8x -7y =7, 5x -3y = -4

Mar 26, 2016

#### Answer:

$\left(x , y\right) = \left(- \frac{49}{59} , - \frac{3}{59}\right)$

#### Explanation:

Using matrices and Cramer's Rule
(with a bit of practice this can be done mentally, but I will lay out the steps in detail)

Converting
$\textcolor{w h i t e}{\text{XXX}} - 8 x - 7 y = 7$
and
$\textcolor{w h i t e}{\text{XXX}} 5 x - 3 y = - 4$
into matrix form:
$\textcolor{w h i t e}{\text{XXXXX")xcolor(white)("XX")ycolor(white)(} X X} c$
$\textcolor{w h i t e}{\text{XXX}} \left(\begin{matrix}- 8 & - 7 & 7 \\ 5 & - 3 & - 4\end{matrix}\right)$
Using the determinants
$\textcolor{w h i t e}{\text{XXX}} {D}_{c}$ (the original matrix without the $c$ column)
$\textcolor{w h i t e}{\text{XXXXX}} = | \left(- 8 , - 7\right) , \left(5 , - 3\right) | = \left(- 8\right) \cdot \left(- 3\right) - \left(- 7\right) \cdot \left(5\right) = 59$
Similarly
$\textcolor{w h i t e}{\text{XXX}} {D}_{x} = | \left(7 , - 7\right) , \left(- 4 , - 3\right) | = 7 \cdot \left(- 3\right) - \left(- 7\right) \cdot \left(- 4\right) = - 49$
and
$\textcolor{w h i t e}{\text{XXX}} {D}_{y} = | \left(- 8 , 7\right) , \left(5 , - 4\right) | = \left(- 8\right) \cdot \left(- 4\right) - 5 \cdot 7 = - 3$

By Cramer's Rule
$\textcolor{w h i t e}{\text{XXX}} x = \frac{{D}_{x}}{{D}_{c}} = - \frac{49}{59}$
and
$\textcolor{w h i t e}{\text{XXX}} y = \frac{{D}_{y}}{{D}_{c}} = - \frac{3}{59}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Alternately: By Elimination
[1]$\textcolor{w h i t e}{\text{XXX}} - 8 x - 7 y = 7$
[2]$\textcolor{w h i t e}{\text{XXX}} 5 x - 3 y = - 4$

[3]=[1]$\times 5$$\textcolor{w h i t e}{\text{XXX}} - 40 x - 35 y = 35$
[4]=[2]$\times 8$$\textcolor{w h i t e}{\text{XXX}} 40 x - 24 y = - 32$

[5]=[3]+[4]$\textcolor{w h i t e}{\text{XXX}} - 59 y = 3$
[6]$\textcolor{w h i t e}{\text{XXXXXXXXX}} y = - \frac{3}{59}$

...and similarly for $x$