# How do you use elimination to solve the system of equations: 2x-4y=26 and 3x+2y=15?

May 31, 2016

color(green)(y= -3
$\textcolor{g r e e n}{x = + 7}$

#### Explanation:

Given:

$\text{ "2x-4y=26" }$..............................Equation (1)
$\text{ "3x+2y=15" }$..............................Equation (2)

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Multiply eqn(2) by 2 and then add

$\text{ "2x-4y=26" } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(1\right)$
$\text{ "underline(6x+4y=30)" } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left({2}_{a}\right)$
$\text{ "8x+0y=56" } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(3\right)$
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Divide eqn(3) by 8

$\textcolor{g r e e n}{\text{ } x = \frac{56}{8} = 7}$
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Multiply eqn(1) by 3 and eqn(2) by 2
Then $\text{eqn"(1_a)-"eqn} \left({2}_{b}\right)$

$6 x - 12 y = 78 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left({1}_{a}\right)$
$\underline{6 x + \textcolor{w h i t e}{.} 4 y = 30} \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left({2}_{b}\right)$
$0 x - 16 y = 48 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(4\right)$

Divide both sides by $- 16$

$\textcolor{g r e e n}{y = \frac{48}{- 16} = - 3}$