## Algebra 1

(a) $F=\dfrac {9}{5}C+32$ (b) $95 ^\circ F$
(a) As we are given that $C=\dfrac {5}{9}(F-32)$ $9C=5(F-32)$ [ Cross multiplication ] $\dfrac {9}{5}C=(F-32)$ $F-32=\dfrac {9}{5}C$ $F=32+\dfrac {9}{5}C$ or, $F=\dfrac {9}{5}C+32$ (b) From part (a), we have : $F=\dfrac {9}{5}C+32$ Thus, at $C=35$, the equation becomes $F=\dfrac {9}{5}(35)+32$ $F=63+32=95 ^\circ F$