How do you solve the system?: $2(x-4) + y=6, 3x-2(y-3)=13$

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Answer 1 · 2015-11-02T03:05:28

$x = 5, y = 4$

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Answer 2 · 2015-11-02T03:06:31

First expand the system and then use simultaneous equations to get $x=5$ and $y=4$

Expanding and simplifying the system

$2(x-4) + y =2x-8+y=6$

Therefore,
$2x+y=14$ $(1)$

$3x -2(y-3) =3x-2y+6=13$

Therefore,
$3x-2y=7$ $(2)$

Simultaneous Equations

$($ 2 $times$ $(1)$ $)$ $+$ $(2)$

$4x+2y+28$ $(1)$
$3x -2y=7$ $(2)$

Cancel out the $y$ terms and equate:

$7x=35$
$x=5$

Sub $x$ back into $(1)$ :

$2(5)+y=14$
$y=4$

How do you solve the system?: 2(x-4) + y=6, 3x-2(y-3)=132(x-4) + y=6, 3x-2(y-3)=13