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How do you solve this

Find two numbers such that twice the first plus three times the second is 7, and three times the first minus four times the second is 36.

This is what you have:
$2x + 3y = 7$
$3x - 4y = 36$

Solve for X in terms of Y in one problem, then take that number and replace the X in the other problem. Then solve for Y, replace to solve for X

$3x - 4y = 36$
$3x = 36 - 4y$
$x = 12 - (4/3)y$

$2(12 - (4/3)y) + 3y = 7$
$24 - (8/3)y + 3y = 7$
$(1/3)y = 7-24$
$y = z$

Find Z, then plug that in for the value of Y to find your answer for X.

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