Solve in Terms of the Arbitrary Variable C C=5/9(F-32) F=9/5C+32

$C = \frac{5}{9} (F - 32)$ $F = \frac{9}{5} C + 32$

Step 1

Simplify the right side.

Simplify $\frac{9 (\frac{5 F}{9} - \frac{160}{9})}{5} + 32$ .

Simplify each term.

Simplify the numerator.

Combine the numerators over the common denominator.

$F = \frac{9 (\frac{5 F - 160}{9})}{5} + 32$

$C = \frac{5 F}{9} - \frac{160}{9}$

Factor $5$ out of $5 F - 160$ .

Factor $5$ out of $5 F$ .

$F = \frac{9 (\frac{5 (F) - 160}{9})}{5} + 32$

$C = \frac{5 F}{9} - \frac{160}{9}$

Factor $5$ out of $- 160$ .

$F = \frac{9 (\frac{5 F + 5 \cdot - 32}{9})}{5} + 32$

$C = \frac{5 F}{9} - \frac{160}{9}$

Factor $5$ out of $5 F + 5 \cdot - 32$ .

$F = \frac{9 (\frac{5 (F - 32)}{9})}{5} + 32$

$C = \frac{5 F}{9} - \frac{160}{9}$

$F = \frac{9 (\frac{5 (F - 32)}{9})}{5} + 32$

$C = \frac{5 F}{9} - \frac{160}{9}$

$F = \frac{9 (\frac{5 (F - 32)}{9})}{5} + 32$

$C = \frac{5 F}{9} - \frac{160}{9}$

Combine $9$ and $\frac{5 (F - 32)}{9}$ .

$F = \frac{\frac{9 (5 (F - 32))}{9}}{5} + 32$

$C = \frac{5 F}{9} - \frac{160}{9}$

Multiply $9$ by $5$ .

$F = \frac{\frac{45 (F - 32)}{9}}{5} + 32$

$C = \frac{5 F}{9} - \frac{160}{9}$

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{45 (F - 32)}{9}$ by cancelling the common factors.

Factor $9$ out of $45 (F - 32)$ .

$F = \frac{\frac{9 (5 (F - 32))}{9}}{5} + 32$

$C = \frac{5 F}{9} - \frac{160}{9}$

Factor $9$ out of $9$ .

$F = \frac{\frac{9 (5 (F - 32))}{9 (1)}}{5} + 32$

$C = \frac{5 F}{9} - \frac{160}{9}$

Cancel the common factor.

$F = \frac{\frac{9 (5 (F - 32))}{9 \cdot 1}}{5} + 32$

$C = \frac{5 F}{9} - \frac{160}{9}$

Rewrite the expression.

$F = \frac{\frac{5 (F - 32)}{1}}{5} + 32$

$C = \frac{5 F}{9} - \frac{160}{9}$

$F = \frac{\frac{5 (F - 32)}{1}}{5} + 32$

$C = \frac{5 F}{9} - \frac{160}{9}$

Divide $5 (F - 32)$ by $1$ .

$F = \frac{5 (F - 32)}{5} + 32$

$C = \frac{5 F}{9} - \frac{160}{9}$

$F = \frac{5 (F - 32)}{5} + 32$

$C = \frac{5 F}{9} - \frac{160}{9}$

Cancel the common factor of $5$ .

Cancel the common factor.

$F = \frac{5 (F - 32)}{5} + 32$

$C = \frac{5 F}{9} - \frac{160}{9}$

Divide $F - 32$ by $1$ .

$F = F - 32 + 32$

$C = \frac{5 F}{9} - \frac{160}{9}$

$F = F - 32 + 32$

$C = \frac{5 F}{9} - \frac{160}{9}$

$F = F - 32 + 32$

$C = \frac{5 F}{9} - \frac{160}{9}$

Combine the opposite terms in $F - 32 + 32$ .

Add $- 32$ and $32$ .

$F = F + 0$

$C = \frac{5 F}{9} - \frac{160}{9}$

Add $F$ and $0$ .

$F = F$

$C = \frac{5 F}{9} - \frac{160}{9}$

$F = F$

$C = \frac{5 F}{9} - \frac{160}{9}$

$F = F$

$C = \frac{5 F}{9} - \frac{160}{9}$

$F = F$

$C = \frac{5 F}{9} - \frac{160}{9}$

Step 2

Simplify the right side.

Multiply $5$ by $F$ .

$C = \frac{5 F}{9} - \frac{160}{9}$

$F = F$

$C = \frac{5 F}{9} - \frac{160}{9}$

$F = F$

Step 3

The simplified system is the arbitrary solution of the original system of equations.

$F = F$

$C = \frac{5 F}{9} - \frac{160}{9}$

Step 4

Reduce the system.

$C = \frac{5 F}{9} - \frac{160}{9}$

$F = F$

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