Solve for C
C=\frac{160-5F}{99}
Solve for F
F=-\frac{99C}{5}+32
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\frac{5}{9}F-\frac{160}{9}=-11C
Use the distributive property to multiply \frac{5}{9} by F-32.
-11C=\frac{5}{9}F-\frac{160}{9}
Swap sides so that all variable terms are on the left hand side.
-11C=\frac{5F-160}{9}
The equation is in standard form.
\frac{-11C}{-11}=\frac{5F-160}{-11\times 9}
Divide both sides by -11.
C=\frac{5F-160}{-11\times 9}
Dividing by -11 undoes the multiplication by -11.
C=\frac{160-5F}{99}
Divide \frac{-160+5F}{9} by -11.
\frac{5}{9}F-\frac{160}{9}=-11C
Use the distributive property to multiply \frac{5}{9} by F-32.
\frac{5}{9}F=-11C+\frac{160}{9}
Add \frac{160}{9} to both sides.
\frac{5}{9}F=\frac{160}{9}-11C
The equation is in standard form.
\frac{\frac{5}{9}F}{\frac{5}{9}}=\frac{\frac{160}{9}-11C}{\frac{5}{9}}
Divide both sides of the equation by \frac{5}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.
F=\frac{\frac{160}{9}-11C}{\frac{5}{9}}
Dividing by \frac{5}{9} undoes the multiplication by \frac{5}{9}.
F=-\frac{99C}{5}+32
Divide -11C+\frac{160}{9} by \frac{5}{9} by multiplying -11C+\frac{160}{9} by the reciprocal of \frac{5}{9}.
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