Chapter 7.4, Problem 3QR

Solution

To calculate: The $x$ and $y$ intercept of a line $\frac{x}{20}+\frac{y}{50}=1$ and also draw its graph.

The intercept point are $\left(20,0\right)$; $\left(0,50\right)$ .

  

Given Information:

The equation of a line is $\frac{x}{20}+\frac{y}{50}=1$ .

Calculation:

Consider the equation of line.

The $x$ -intercept of a line is in the form of $\left(a,0\right)$ . To obtain the $x$ -intercept of a line, put $y=0$ in the given equation of a line and solve for $x$ .

  $\begin{array}{c}\frac{x}{20}+\frac{0}{50}=1\\ \frac{x}{20}=1\end{array}$

Now, multiply both the sides of the obtained equation by 20.

  $\begin{array}{c}20\left(\frac{x}{20}\right)=20\left(1\right)\\ x=20\end{array}$

The $x$ -intercept of a line is $\left(20,0\right)$ .

The $y$ -intercept of a line is in the form of $\left(0,b\right)$ . To obtain the $y$ -intercept of a line, put $x=0$ in the given equation of a line and solve for $y$ .

  $\begin{array}{c}\frac{0}{20}+\frac{y}{50}=1\\ \frac{y}{50}=1\end{array}$

Now, multiply both the sides of the obtained equation by 50.

  $\begin{array}{c}50\left(\frac{y}{50}\right)=50\left(1\right)\\ y=50\end{array}$

The $y$ -intercept of a line is $\left(0,50\right)$ .

Draw the graph of a line $\frac{x}{20}+\frac{y}{50}=1$ .