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

Question

6514 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-02T03:05:28

$x = 5, y = 4$ $x = 5, y = 4$

enter image source here

Answer 2 · 2015-11-02T03:06:31

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

Expanding and simplifying the system

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

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

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

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

Simultaneous Equations

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

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

Cancel out the $y$ $y$ terms and equate:

$7 x = 35$ $7x=35$
$x = 5$ $x=5$

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

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

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