Answer
$C=\displaystyle \frac{5}{9}F-\frac{160}{9}$, (C as a linear function of F)
$F=104,\displaystyle \quad C=\frac{5}{9}(104)-\frac{160}{9}=40$
$F=77,\displaystyle \quad C=\frac{5}{9}(77)-\frac{160}{9}=25$
$F=14,\displaystyle \quad C=\frac{5}{9}(14)-\frac{160}{9}=-10$
$F=-40,\displaystyle \quad C=\frac{5}{9}(-40)-\frac{160}{9}=-40$
Work Step by Step
If $F=1.8C+32 $, as we found in the previous problem,
we solve for $C:$
$F=1.8C+32\qquad/-32$
$F-32=\displaystyle \frac{9}{5}C\qquad/\times\frac{5}{9}$
$\displaystyle \frac{5}{9}F-\frac{5}{9}\cdot 32=C$
$C=\displaystyle \frac{5}{9}F-\frac{160}{9}$, (C as a linear function of F)
$F=104,\displaystyle \quad C=\frac{5}{9}(104)-\frac{160}{9}=40$
$F=77,\displaystyle \quad C=\frac{5}{9}(77)-\frac{160}{9}=25$
$F=14,\displaystyle \quad C=\frac{5}{9}(14)-\frac{160}{9}=-10$
$F=-40,\displaystyle \quad C=\frac{5}{9}(-40)-\frac{160}{9}=-40$
