How do you solve $2x - 4y = 32$ and $8x - 8y = 96$ ?

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How do you solve $2x - 4y = 32$ and $8x - 8y = 96$ ?

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Answer 1 · 2018-02-06T16:36:52

Notice that we're gonna solve for $x$ first because if we solve for $x$ in the first equation... we can factorize and get $x$ without doing any kind of multiplication or something

Explanation:

We're gonna solve for $x$ first

First equation
$2x-4y=32$
$2x=32+4y$
$x=(32+4y)/2$
$x=32/2+(4y)/2$
$x=16+2y$

Second equation
$8x-8y =96$
Factorize
$8(x-y)=96$
Transfer the 8 and put value of $x$
$16+2y-y=96/8$
$16+y=12$
$y=12-16$
$color(red)(y=-4)$
Get the value of $x$
$x=16+2y$
$x=16+2xx-4$
$x=16+(-8)$
$color(red)(x=8)$

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How do you solve $2x - 4y = 32$ and $8x - 8y = 96$ ?

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How do you solve $2x - 4y = 32$ and $8x - 8y = 96$ ?