Solve for b
b=\frac{2c}{3}+11
Solve for c
c=\frac{3\left(b-11\right)}{2}
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3b=33+2c
Add 2c to both sides.
3b=2c+33
The equation is in standard form.
\frac{3b}{3}=\frac{2c+33}{3}
Divide both sides by 3.
b=\frac{2c+33}{3}
Dividing by 3 undoes the multiplication by 3.
b=\frac{2c}{3}+11
Divide 33+2c by 3.
-2c=33-3b
Subtract 3b from both sides.
\frac{-2c}{-2}=\frac{33-3b}{-2}
Divide both sides by -2.
c=\frac{33-3b}{-2}
Dividing by -2 undoes the multiplication by -2.
c=\frac{3b-33}{2}
Divide 33-3b by -2.
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