How do you solve the system $\frac{x}{3} + 3 \frac{y}{4} = - 3$ $x/3 + 3y/4 = -3$ and $\frac{x}{3} - \frac{y}{4} = 5$ $x/3 - y/4 = 5$ ?

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Answer 1 · 2016-06-29T17:07:03

$x = 9$ $x=9$ and $y = - 8$ $y=-8$

The equations are:

$\frac{x}{3} + \frac{3 y}{4} = - 3$ $x/3+(3y)/4=-3$ ...................(A)

$\frac{x}{3} - \frac{y}{4} = 5$ $x/3-y/4=5$ ...................(B)

Now subtracting (B) from (A)

$\frac{3 y}{4} + \frac{y}{4} = - 3 - 5 = - 8$ $(3y)/4+y/4=-3-5=-8$ or $y = - 8$ $y=-8$

Putting this in (B)

$\frac{x}{3} - \frac{- 8}{4} = 5$ $x/3-(-8)/4=5$ or $\frac{x}{3} + 2 = 5$ $x/3+2=5$ or $\frac{x}{3} = 3$ $x/3=3$ or $x = 9$ $x=9$

How do you solve the system x3+3y4=−3x/3 + 3y/4 = -3 and x3−y4=5x/3 - y/4 = 5?