# The linear equation that converts Fahrenheit

Question:

The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation,

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

(i) If the temperature is 86°F, what is the temperature in Celsius?

(ii) If the temperature is 35°C, what is the temperature in Fahrenheit?

(iii) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?

(iv) What is the numerical value of the temperature which is same in both the scales?

Solution:

Given relation is,

$C=\frac{5 F-160}{9}$  $\ldots$ (i)

$\Rightarrow$ $9 C=5 F-160 \Rightarrow 5 F=9 C+160$

$\Rightarrow$  $F=\frac{9 C+160}{5}$ ...(ii)

(i) Given, $F=86^{\circ} \mathrm{F}$, then from Eq. (i), we get

$C=\frac{5 \times 86-160}{9}=\frac{430-160}{9}=\frac{270}{9}=30^{\circ} \mathrm{C}$

(ii) Given, $C=35^{\circ} \mathrm{C}$, then from Eq. (ii), we get

$F=\frac{9 \times 35+160}{5}=\frac{315+160}{5}=\frac{475}{5}=95^{\circ} \mathrm{F}$

(iii) Given, $C=0^{\circ} \mathrm{C}$, then from Eq. (ii), we get

$F=\frac{9 \times 0+160}{5}=\frac{160}{5}=32^{\circ} \mathrm{F}$

$F=0^{\circ} \mathrm{F}$, then from $\mathrm{Eq}$. (i), we get

$C=\frac{5 \times 0-160}{9}=\frac{-160}{9}=\left(-\frac{160}{9}\right)^{\circ} \mathrm{C}$

(iv) By given condition, $C=F$

Put this value in Eq. (i), we get

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

$\Rightarrow \quad 9 C-5 C=-160$

$\Rightarrow$ $4 C=-160 \Rightarrow C=\frac{-160}{4} \Rightarrow C=-40=F$ [numerical value of the temperature]