Question 3

We need to find 'a' and 'b' in the given linear relationship.

Step 1 — Set up equations

Let's use the first given condition. Ice melts at 0 degrees Celsius and 32 degrees Fahrenheit. We substitute these values into the equation $^\circ\text{C} = a ^\circ\text{F} + b$.

$0 = a(32) + b$

$0 = 32a + b \quad \text{... (1)}$

Now, let's use the second condition. Water boils at 100 degrees Celsius and 212 degrees Fahrenheit. We substitute these values into the equation $^\circ\text{C} = a ^\circ\text{F} + b$.

$100 = a(212) + b$

$100 = 212a + b \quad \text{... (2)}$

Step 2 — Solve for 'a'

We have two equations with 'a' and 'b'. Let's subtract Equation (1) from Equation (2). This will help us eliminate 'b'.

$(212a + b) - (32a + b) = 100 - 0$

$212a - 32a = 100$

$180a = 100$

$a = \frac{100}{180}$

$a = \frac{10}{18}$

$\boxed{a = \frac{5}{9}}$

Step 3 — Solve for 'b'

Now we know the value of 'a'. Let's substitute $a = \frac{5}{9}$ into Equation (1). Equation (1) is $0 = 32a + b$.

$0 = 32\left(\frac{5}{9}\right) + b$

$0 = \frac{160}{9} + b$

$b = -\frac{160}{9}$

$\boxed{b = -\frac{160}{9}}$

Step 4 — State the relationship

We found the values for 'a' and 'b'. The relationship is $^\circ\text{C} = a ^\circ\text{F} + b$. Let's substitute $a = \frac{5}{9}$ and $b = -\frac{160}{9}$.

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

This can also be written by factoring out $\frac{5}{9}$.

$^\circ\text{C} = \frac{5}{9}\left(^\circ\text{F} - \frac{160}{5}\right)$

$^\circ\text{C} = \frac{5}{9}(^\circ\text{F} - 32)$

Answer

(i) The value of $a$ is $\frac{5}{9}$. (ii) The value of $b$ is $-\frac{160}{9}$. (iii) The linear relationship is $^\circ\text{C} = \frac{5}{9}^\circ\text{F} - \frac{160}{9}$ or $^\circ\text{C} = \frac{5}{9}(^\circ\text{F} - 32)$.