How do you solve the following system: $2x-y/2=4, 5x+2y=12$ ?

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Answer 1 · 2016-02-03T21:25:04

$x = 28/13 , y = 8/13$

This is done by the process of elimination but it is not the only method.

Begin by multiplying the first equation by 4 to get the coefficients in front of the $y$ equal. Multiplying the first equation by 4 gives:

$8x -2y =16$

Now we can add both equations together

$(5x +2y) + (8x-2y) = (16)+(12)$

So we get:

$13x = 28 -> x = 28/13$

Now replace $x$ with this value in either of the equations and solve for $y$ . So using the first equation: $2x-y/2 =4$ .

$-> 2(28/13) - y/2 =4$
$56/13 - y/2 = 4$

Rearrange and you should arrive at:

$y = 2(56/13 - 4) = 2(56-52)/13=8/13$
$therefore x = 28/13, y = 8/13$

How do you solve the following system: 2x-y/2=4, 5x+2y=12 2x-y/2=4, 5x+2y=12 ?