Solve for A
A=-\frac{2B}{3}+C+4
Solve for B
B=\frac{3\left(4+C-A\right)}{2}
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3A=3C+12-2B
Subtract 2B from both sides.
3A=12+3C-2B
The equation is in standard form.
\frac{3A}{3}=\frac{12+3C-2B}{3}
Divide both sides by 3.
A=\frac{12+3C-2B}{3}
Dividing by 3 undoes the multiplication by 3.
A=-\frac{2B}{3}+C+4
Divide 3C+12-2B by 3.
2B=3C+12-3A
Subtract 3A from both sides.
2B=12+3C-3A
The equation is in standard form.
\frac{2B}{2}=\frac{12+3C-3A}{2}
Divide both sides by 2.
B=\frac{12+3C-3A}{2}
Dividing by 2 undoes the multiplication by 2.
B=\frac{3C}{2}-\frac{3A}{2}+6
Divide 12+3C-3A by 2.
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