How do you solve $0.3x-0.2y=4$ and $0.4x+0.5y=63/19$ using substitution?

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Answer 1 · 2018-06-20T14:12:01

$x=220/19,y=-50/19$

writing your System in the form

$3x-2y=40$
$4x+5y=630/19$
so we get from the first equation

$y=3/2x-20$
plugging this in the second equation we get

$4x+5(3/2x-20)=630/19$
$x=220/19$

$y=-50/19$

How do you solve 0.3x-0.2y=40.3x-0.2y=4 and 0.4x+0.5y=63/190.4x+0.5y=63/19 using substitution?