Solve for a
a=\frac{3\left(b+c+180\right)}{2}
Solve for b
b=\frac{2a}{3}-c-180
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-2a+3c=-540-3b
Subtract 3b from both sides.
-2a=-540-3b-3c
Subtract 3c from both sides.
-2a=-3b-3c-540
The equation is in standard form.
\frac{-2a}{-2}=\frac{-3b-3c-540}{-2}
Divide both sides by -2.
a=\frac{-3b-3c-540}{-2}
Dividing by -2 undoes the multiplication by -2.
a=\frac{3b}{2}+\frac{3c}{2}+270
Divide -540-3b-3c by -2.
3b+3c=-540+2a
Add 2a to both sides.
3b=-540+2a-3c
Subtract 3c from both sides.
3b=2a-3c-540
The equation is in standard form.
\frac{3b}{3}=\frac{2a-3c-540}{3}
Divide both sides by 3.
b=\frac{2a-3c-540}{3}
Dividing by 3 undoes the multiplication by 3.
b=\frac{2a}{3}-c-180
Divide -540+2a-3c by 3.
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