Question #a512b

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Answer 1 · 2017-06-26T19:06:01

Any integer multiple of the lowest solution: $(3,-2)$

Lets label the first equation

$2x+3y=0$

And the second equation

$x+4y=-5$

The first step is to multiply the second equation by $-2$

$-2(x+4y)=-2(-5)$

$-2x-8y=10$

Adding this second equation to the first equation

$2x+3y=0$
$-2x-8y=10$

$=> -5y=10$

$y=-2$

Plugging $y=-2$ back into one of the original two equations

$2x+3color(red)(y)=0$

$2x+3color(red)(-2)=0$

$2x-6=0$

$2x=6$

$x=3$

The solution is $(3,-2)$ . In fact, any integer multiple of this answer gives another solution:

$2xx(3,-2)=(6,-4)$

$3xx(3,-2)=(9,-6)$

$4xx(3,-2)=(12,-8)$