Solve for a
a=\frac{3b}{2}-2c
Solve for b
b=\frac{2\left(a+2c\right)}{3}
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2a+4c=3b
Add 3b to both sides. Anything plus zero gives itself.
2a=3b-4c
Subtract 4c from both sides.
\frac{2a}{2}=\frac{3b-4c}{2}
Divide both sides by 2.
a=\frac{3b-4c}{2}
Dividing by 2 undoes the multiplication by 2.
a=\frac{3b}{2}-2c
Divide 3b-4c by 2.
-3b+4c=-2a
Subtract 2a from both sides. Anything subtracted from zero gives its negation.
-3b=-2a-4c
Subtract 4c from both sides.
\frac{-3b}{-3}=\frac{-2a-4c}{-3}
Divide both sides by -3.
b=\frac{-2a-4c}{-3}
Dividing by -3 undoes the multiplication by -3.
b=\frac{2a+4c}{3}
Divide -2a-4c by -3.
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Limits
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