# How do you solve the following system: 4x-y=6, 5x+2y=20 ?

Jul 15, 2018

$x = \frac{32}{13} \mathmr{and} y = \frac{50}{13}$

#### Explanation:

$\text{ "4xcolor(blue)(-y)=6" } \ldots \ldots \ldots A$
$\text{ "5xcolor(blue)(+2y)=20" } \ldots . . B$

One of the easiest methods to eliminate one of the variables is to make them into additive inverses. (Their sum is $0$)

Multiplying equation $A$ by $2$ will achieve this,:

$A \times 2 : \textcolor{w h i t e}{\times} 8 x \textcolor{b l u e}{- 2 y} = 12 \text{ } \ldots \ldots \ldots C$
$\textcolor{w h i t e}{\times \times \times x} 5 x \textcolor{b l u e}{+ 2 y} = 20 \text{ } \ldots . . B$

$C + B : \textcolor{w h i t e}{\times} 13 x \text{ "=32" } \ldots \ldots \ldots C$
$C + B : \textcolor{w h i t e}{x . . x} x \text{ "=32/13" }$

If $x = \frac{32}{13}$, substitute into one of the original equations to find $y$

$5 \left(\frac{32}{13}\right) + 2 y = 20 \text{ } \ldots . . B$

$\frac{160}{13} + 2 y = 20$

$\text{ } 2 y = 20 - \frac{160}{13}$

$\text{ } 2 y = \frac{100}{13}$

$\text{ } y = \frac{50}{13}$

Check in $A$

$4 \left(\frac{32}{13}\right) - \frac{50}{13}$
$= \frac{78}{13}$
$= 6$