How do you solve the following system?: $5 x + 2 y = 1, 3 x - 7 y = - 12$ $5x + 2y =1 ,3x-7y=-12$

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Answer 1 · 2016-03-06T16:21:01

$x = - \frac{17}{41}$ $x=-17/41$

$y = \frac{63}{41}$ $y= 63/41$

$5 \cdot x + 2 \cdot y = 1$ $5 * x + 2 * y = 1$

$3 \cdot x - 7 \cdot y = - 12$ $3 * x - 7 * y=-12$

Multiply the first equation by $3.5$ $3.5$

$17.5 \cdot x + 7 \cdot y = 3.5$ $17.5 * x + 7 * y = 3.5$

$3 \cdot x - 7 \cdot y = - 12$ $3 * x - 7 * y=-12$

Add the two equations to get

$20.5 * x + cancel(7y) - cancel(7y) = -8.5$

$x = -8.5/20.5 = -17/41$

This means that you have

$3 * (-17/41) - 7y = -12$

$y = (12 - 3 * 17/41)/7 = 63/41$

How do you solve the following system?: 5x + 2y =1 ,3x-7y=-125x+2y=1,3x−7y=−125x + 2y =1 ,3x-7y=-12