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

Apr 3, 2018

$x = - \frac{4}{17}$

$y = \textcolor{w h i t e}{\text{dd}} \frac{12}{17}$

#### Explanation:

This is one of two approaches:

Given:
$2 x - 5 y = - 4 \text{ } \ldots \ldots \ldots . E q u a t i o n \left(1\right)$
 color(white)("d") x+6y=+4" "..........Equation(2)

Multiply both sides of $E q n \left(2\right)$ by 2

2x-color(white)("d")5y=-4" ".........Equation(1)
$\underline{2 x + 12 y = + 8 \text{ ".........Equation(2_a)larr" Subtract}}$
$\textcolor{w h i t e}{\text{d}} 0 - 17 y = - 12$

By choosing negative it will change the $y$ term into positive.
Divide both sides by $\left(- 17\right)$

$\textcolor{b l u e}{\textcolor{w h i t e}{\text{ddddddddddd}} \underline{\overline{| \textcolor{w h i t e}{. .} y = \frac{12}{17} \textcolor{w h i t e}{. .} |}}}$

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Substitute for $\textcolor{red}{y = \frac{12}{17}}$ into $E q n \left(2\right)$

$\textcolor{g r e e n}{x + 6 \textcolor{red}{y} = 4 \textcolor{w h i t e}{\text{dddd")->color(white)("dddd}} x + \left(6 \textcolor{red}{\times \frac{12}{17}}\right) = 4}$

$\textcolor{w h i t e}{\text{ddddddddddddd")->color(white)("dddd")x+color(white)("ddd")72/17color(white)("d.d}} = 4$

$\textcolor{b l u e}{\textcolor{w h i t e}{\text{ddddddddddd}} \underline{\overline{| \textcolor{w h i t e}{. .} x = - \frac{4}{17} \textcolor{w h i t e}{. .} |}}}$
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$\textcolor{b l u e}{\text{Check}}$

$2 x - 5 y = - 4 \text{ } \ldots \ldots \ldots . E q u a t i o n \left(1\right)$
color(white)("d")x+6y=+4" "..........Equation(2)

$2 \left(- \frac{4}{17}\right) - 5 \left(\frac{12}{17}\right) = - 4 \text{ } \ldots \ldots \ldots . E q u a t i o n \left(1\right)$
color(white)("d")(-4/17)+6(12/17)=+4" "..........Equation(2)

$- \frac{8}{17} - \frac{60}{17} = - 4 \text{ } \ldots \ldots \ldots . E q u a t i o n \left(1\right)$
$- \frac{4}{17} + \frac{72}{17} = + 4 \text{ } \ldots \ldots \ldots . E q u a t i o n \left(2\right)$

$- 4 = - 4 \text{ } \ldots \ldots \ldots . E q u a t i o n \left(1\right)$
$+ 4 = + 4 \text{ } \ldots \ldots \ldots . E q u a t i o n \left(2\right)$