Q46. - Algebra 2

Question

For Exercises 46-48, use the information on temperatures at the left.

 

Write and graph the linear equation that gives the number $\mathbf{y}$ of degrees Fahrenheit in terms of the number $\mathbf{x}$ of degrees Celsius.

Step-by-Step Solution

Answer

The required linear equation is $y=\frac{9}{5}x+32$. The graph of this equation is,

Step 1 – State the concept

The equation of a line passing through $\left(a,b\right)$ and $\left(c,d\right)$ is given as $\frac{y-b}{x-a}=\frac{d-b}{c-a}$.

Step 2 – List the given data

It is clear, from the given information, that the temperature of $0{}^{\circ }C$ is equivalent to the temperature of $32{}^{\circ }F$ and the temperature of $100{}^{\circ }C$ corresponds to the temperature of $212{}^{\circ }F$.

 This implies that the required line passes through the points, $\left(0,32\right)$ and $\left(100,212\right)$.

 Then, $\left(a,b\right)=\left(0,32\right)$ and $\left(c,d\right)=\left(100,212\right)$.

It is clear, from the given information, that the temperature of $0{}^{\circ }C$ is equivalent to the temperature of $32{}^{\circ }F$ and the temperature of $100{}^{\circ }C$ corresponds to the temperature of $212{}^{\circ }F$.

 This implies that the required line passes through the points, $\left(0,32\right)$ and$\left(100,212\right)$.

 Then, $\left(a,b\right)=\left(0,32\right)$ and $\left(c,d\right)=\left(100,212\right)$.

Step 3 – Write the equation

Put$\left(a,b\right)=\left(0,32\right)$ and $\left(c,d\right)=\left(100,212\right)$ in $\frac{y-b}{x-a}=\frac{d-b}{c-a}$ to get,

$\frac{y-32}{x-0}=\frac{212-32}{100-0}$

 

$\frac{y-32}{x}=\frac{180}{100}$  (Simplify)

 

$\frac{y-32}{x}=\frac{9}{5}$  (Simplify)

 

$\frac{y-32}{x}\left(x\right)=\frac{9}{5}\left(x\right)$  (Multiply both sides by $x$)

 

$y-32=\frac{9}{5}x$   (Simplify)

 

$y-32+32=\frac{9}{5}x+32$  (Add 32 to both sides)

 

$y=\frac{9}{5}x+32$  (Simplify)

 

So, the required linear equation is $y=\frac{9}{5}x+32$.