Solve for c
c=\frac{3x}{40}+\frac{1}{4}
Solve for x
x=\frac{40c-10}{3}
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40c=3x+15-5
Subtract 5 from both sides.
40c=3x+10
Subtract 5 from 15 to get 10.
\frac{40c}{40}=\frac{3x+10}{40}
Divide both sides by 40.
c=\frac{3x+10}{40}
Dividing by 40 undoes the multiplication by 40.
c=\frac{3x}{40}+\frac{1}{4}
Divide 3x+10 by 40.
3x+15=40c+5
Swap sides so that all variable terms are on the left hand side.
3x=40c+5-15
Subtract 15 from both sides.
3x=40c-10
Subtract 15 from 5 to get -10.
\frac{3x}{3}=\frac{40c-10}{3}
Divide both sides by 3.
x=\frac{40c-10}{3}
Dividing by 3 undoes the multiplication by 3.
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