Consider the relationship between temperature measured in degrees Celsius (${}^{\circ}\mathrm{C}$) and degrees Fahrenheit (${}^{\circ}\mathrm{F}$), which is given by ${}^{\circ}\mathrm{C} = a{}^{\circ}\mathrm{F} + b$. Find $a$ and $b$, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit. (Hint: When ${}^{\circ}\mathrm{C} = 0$, ${}^{\circ}\mathrm{F} = 32$ and when ${}^{\circ}\mathrm{C} = 100$, ${}^{\circ}\mathrm{F} = 212$. Use this information to find $a$ and $b$, and thus, the linear relationship between ${}^{\circ}\mathrm{C}$ and ${}^{\circ}\mathrm{F}$.) [Page No. 27]

The given relation is $^\circ\mathrm{C} = a ^\circ\mathrm{F} + b$

Given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, then expression is given by

$$0 = 32a + b \quad (\mathrm{i})$$

Also, given that water boils at 100 degrees Celsius and 212 degrees Fahrenheit, then expression becomes

$$100 = 212a + b \quad \dots (ii)$$

Subtract eq. (i) from eq. (ii), we have

100 - 0 = 212a + b - 32a - b

100 = 180a

$\alpha = \frac{100}{180} = \frac{5}{9}$

Substitute in eq. (i), we have

$b = -32a = -32 \times \frac{5}{9} = -\frac{160}{9}$

Therefore, the value of $a$ is $\frac{5}{9}$ and the value of $b$ is $\left(-\frac{160}{9}\right)$. Hence, the linear relationship between $^\circ\mathrm{C}$ and $^\circ\mathrm{F}$ is given by $^\circ\mathrm{C} = \frac{5}{9}{}^\circ\mathrm{F} - \frac{160}{9} = \frac{5}{9}({}^\circ\mathrm{F} - 32)$.