How do you solve $3 x - y = 4$ $3x-y= 4$ and $2 x - 3 y = - 9$ $2x-3y= -9$ ?

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Answer 1 · 2017-08-08T15:16:41

See below.

Multiplying the first equation by $3$ $3$ ,

$9 x - 3 y = 12$ $9x-3y=12$

$2 x - 3 y = - 9$ $2x-3y=-9$

Subtracting the two equations,

$7 x = 21$ $7x=21$

So $x = 3$ $x=3$

Plugging this in,

$2 (3) - 3 y = - 9$ $2(3)-3y=-9$

$6 + 9 = 3 y$ $6+9=3y$

$y = 5$ $y=5$

So, our final solution is $(3, 5)$ $(3,5)$ .

How do you solve 3x−y=43x-y= 4 and 2x−3y=−92x-3y= -9?