Question

I need to solve for C the formula for temperature scale conversion

Answer 1

We need to rewrite the equation

`F=(9C)/(5)+32`

isolating C on the left side of the equation.

First, we can subtract 9C/5 from both sides of the equation. We obtain:

`\begin{gathered} F-(9C)/(5)=(9C)/(5)+32-(9C)/(5) \\ \\ F-(9C)/(5)=32 \end{gathered}`

Then, we can subtract F from both sides of the equation. We obtain:

`\begin{gathered} F-(9C)/(5)-F=32-F \\ \\ -(9C)/(5)=32-F \end{gathered}`

Then, we can multiply both sides by -5/9. We obtain:

`\begin{gathered} -(9C)/(5)\cdot(-(5)/(9))=(32-F)\cdot(-(5)/(9)) \\ \\ (-1)(-1)(9)/(9)\cdot(5)/(5)C=(32-F)\cdot(-(5)/(9)) \\ \\ C=(32-F)\cdot(-(5)/(9)) \end{gathered}`

Then, we can distribute the factor -5/9 over the sum 32-F on the right side:

`\begin{gathered} C=32\cdot(-(5)/(9)_{})-F\cdot(-(5)/(9)) \\ \\ C=-(160)/(9)+(5)/(9)F \end{gathered}`

Therefore, the answer can be written as

`C=-(160)/(9)+(5)/(9)F`

Or as

`C=(5)/(9)(F-32)`