How do you solve for z in $a = x + y + \frac{z}{3}$ $a = x+y+z/3$ ?

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Answer 1 · 2015-04-08T23:07:43

The answer is $z = 3 (a - x - y)$ $z=3(a-x-y)$ .

Beginning Equation:

$a = x + y + \frac{z}{3}$ $a=x+y+z/3$

First multiply both sides of the equation by 3.

$3 a = 3 x + 3 y + z$ $3a=3x+3y+z$

Subract 3x and 3y from both sides.

$3 a - 3 x - 3 y = z$ $3a-3x-3y=z$

Factor out the 3.

$3 (a - x - y) = z$ $3(a-x-y)=z$

How do you solve for z in a = x+y+z/3a=x+y+z3a = x+y+z/3?