How do you solve the system $(3/4)x - (1/2)y = -(1/2)$ and $x + y = -(3/2)$ ?

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Answer 1 · 2015-03-30T12:45:51

Simplify your equations by multiplying both sides by a constant to clear the fractions then solve as normal.

$(3/4)x - (1/2)y = -(1/2)$ multiplied by $4$ becomes
$3x - 2y = -2$

$x+y= -(3/2)$ multiplied by $2$ becomes
$2x+2y = -3$

Adding these modified equations together gives
$5x = -5$

So
$x=-1$
and by substitution back into any one of the prior equations
$y=-1/2$

How do you solve the system (3/4)x - (1/2)y = -(1/2)(3/4)x - (1/2)y = -(1/2) and x + y = -(3/2)x + y = -(3/2)?