<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Dismiss
Skip Navigation
Our Terms of Use (click here to view) and Privacy Policy (click here to view) have changed. By continuing to use this site, you are agreeing to our new Terms of Use and Privacy Policy.

Intercepts by Substitution

Graphing equations that form lines by finding x and y intercepts

Atoms Practice
Estimated8 minsto complete
%
Progress
Practice Intercepts by Substitution
Practice
Progress
Estimated8 minsto complete
%
Practice Now
Using Intercepts
License: CC BY-NC 3.0

Mary Ellen has $60 to spend at the craft store and she is very interested in buying either scrapbooks that cost $10 or fashion stickers that cost $5 per package. Before making a purchase, she needs to know how many scrapbooks or how many packages of stickers she can buy with her money. How can Mary Ellen figure this out?

In this concept, you will learn to use intercepts.

Intercepts

Consider the following linear graph.

License: CC BY-NC 3.0

The above graph models the linear function \begin{align*}3x+2y=6\end{align*}. The function is written in standard form \begin{align*}Ax+By=C\end{align*}. Notice that the line passes through the \begin{align*}x\end{align*}-axis at the point \begin{align*}(2,0)\end{align*}. The point where the graph intersects the horizontal \begin{align*}x\end{align*}-axis is called the \begin{align*}x\end{align*}-intercept. The \begin{align*}y\end{align*}-value of any point on the \begin{align*}x\end{align*}-axis is zero. The \begin{align*}x\end{align*}-axis is really the line having the equation \begin{align*}y=0\end{align*}.

The line also passes through the \begin{align*}y\end{align*}-axis at the point \begin{align*}(0,3)\end{align*}. The point where the graph intersects the vertical \begin{align*}y\end{align*}-axis is called the \begin{align*}y\end{align*}-intercept. The \begin{align*}x\end{align*}-value of any point on the \begin{align*}y\end{align*}-axis is zero. The \begin{align*}y\end{align*}-axis is really the line having the equation \begin{align*}x=0\end{align*}.

Remember, only two points are needed to draw a straight line. Therefore, the graph of the linear function \begin{align*}3x+2y=6\end{align*} was drawn by plotting the \begin{align*}x\end{align*}- and \begin{align*}y\end{align*}-intercepts and using a straight edge to join the two plotted points.

From a given function, the intercepts can be determined using algebra. Let’s look at an example.

\begin{align*}5x-3y=15\end{align*}

First, determine the \begin{align*}x\end{align*}-intercept. Substitute \begin{align*}y=0\end{align*} in the equation.

\begin{align*}\begin{array}{rcl} 5x-3y &=& 15 \\ 5x-3(0) &=& 15 \end{array}\end{align*}

Next, perform the multiplication to clear the parenthesis.

\begin{align*}\begin{array}{rcl} 5x-3(0) &=& 15 \\ 5x-0 &=& 15 \\ 5x &=& 15 \end{array}\end{align*}

Next, divide both sides of the equation by ‘5’ to solve for ‘\begin{align*}x\end{align*}’.

\begin{align*}\begin{array}{rcl} 5x &=& 15 \\ \frac{\overset{1}{\cancel{5}}x}{\cancel{5}} &=& \frac{\overset{3}{\cancel{15}}}{\cancel{5}} \\ x &=& 3 \end{array}\end{align*}

The answer is 3.

The \begin{align*}x\end{align*}-intercept is \begin{align*}(3,0)\end{align*}.

Second, determine the \begin{align*}y\end{align*}-intercept. Substitute \begin{align*}x=0\end{align*} in the equation.

\begin{align*}\begin{array}{rcl} 5x-3y &=& 15 \\ 5(0)-3y &=& 15 \end{array}\end{align*}

Next, perform the multiplication to clear the parenthesis.

\begin{align*}\begin{array}{rcl} 5(0)-3y &=& 15 \\ 0-3y &=& 15 \\ -3y &=& 15 \end{array}\end{align*}

Next, divide both sides of the equation by ‘-3’ to solve for ‘\begin{align*}y\end{align*}’.

\begin{align*}\begin{array}{rcl} -3y &=& 15 \\ \frac{\overset{1}{\cancel{-3}}y}{\cancel{-3}} &=& \frac{\overset{-5}{\cancel{15}}}{\cancel{-3}} \\ y &=& -5 \end{array}\end{align*}

The answer is -5.

The \begin{align*}y\end{align*}-intercept is \begin{align*}(0,-5)\end{align*}.

Now, the values of the \begin{align*}x\end{align*}- and \begin{align*}y\end{align*}-intercepts can be used to plot the graph of the linear function \begin{align*}5x-3y=15\end{align*}.

License: CC BY-NC 3.0

Examples

Example 1

Earlier, you were given a problem about Mary Ellen and the scrapbooks or stickers. She needs to figure out how many scrapbooks or how many packages of fashion stickers she can buy with her $60.00. How can she do this?

Mary Ellen can use the \begin{align*}x\end{align*}- and \begin{align*}y\end{align*}-intercepts of a linear function.

First, write a linear function to represent the information given.

Let ‘\begin{align*}x\end{align*}’ represent the number of scrapbooks that cost $10 each and let ‘\begin{align*}y\end{align*}’ represent the number of packages of stickers that cost $5.00 each. She has $60.00 to spend.

The linear function to model the given information is:

\begin{align*}10x+5y=60\end{align*}

Next, determine \begin{align*}x\end{align*}- and \begin{align*}y\end{align*}-intercepts of the linear function.

First, determine the \begin{align*}x\end{align*}-intercept. Substitute \begin{align*}y=0\end{align*} in the equation.

\begin{align*}\begin{array}{rcl} 10x+5y &=& 60 \\ 10x+5(0) &=& 60 \end{array}\end{align*}

Next, perform the multiplication to clear the parenthesis.

\begin{align*}\begin{array}{rcl} 10x+5(0) &=& 60 \\ 10x+0 &=& 60 \\ 10x &=& 60 \end{array}\end{align*}

Next, divide both sides of the equation by ‘10’ to solve for ‘\begin{align*}x\end{align*}’.

\begin{align*}\begin{array}{rcl} 10x &=& 60 \\ \frac{\overset{1}{\cancel{10}}x}{\cancel{10}} &=& \frac{\overset{6}{\cancel{60}}}{\cancel{10}} \\ x &=& 6 \end{array}\end{align*}

The answer is 6.

The \begin{align*}x\end{align*}-intercept is \begin{align*}(6,0)\end{align*}.

Second, determine the \begin{align*}y\end{align*}-intercept. Substitute \begin{align*}x=0\end{align*} in the equation.

\begin{align*}\begin{array}{rcl} 10x+5y &=& 60 \\ 10(0)+5y &=& 60 \end{array}\end{align*}

Next, perform the multiplication to clear the parenthesis.

\begin{align*} \begin{array}{rcl} 10(0)+5y &=& 60 \\ 0+5y &=& 60 \\ 5y &=& 60 \end{array}\end{align*}

Next, divide both sides of the equation by ‘30’ to solve for ‘\begin{align*}y\end{align*}’.

\begin{align*}\begin{array}{rcl} 5y &=& 60 \\ \frac{\overset{1}{\cancel{5}}y}{\cancel{5}} &=& \frac{\overset{12}{\cancel{60}}}{\cancel{5}} \\ y &=& 12 \end{array}\end{align*}

The answer is 12.

The \begin{align*}y\end{align*}-intercept is \begin{align*}(0,12)\end{align*}.

Mary Ellen can buy either 6 scrapbooks and no stickers or 12 packages of stickers and no scrapbooks.

Example 2

For the given linear function, determine the \begin{align*}x\end{align*}- and \begin{align*}y\end{align*}-intercepts.

\begin{align*}4x-3y=-24\end{align*}

First, determine the \begin{align*}x\end{align*}-intercept. Substitute \begin{align*}y=0\end{align*} in the equation.

\begin{align*}\begin{array}{rcl} 4x-3y &=& -24 \\ 4x-3(0) &=& -24 \end{array}\end{align*}

Next, perform the multiplication to clear the parenthesis.

\begin{align*}\begin{array}{rcl} 4x-3(0) &=& -24 \\ 4x-0 &=& -24 \\ 4x &=& -24 \end{array}\end{align*}

Next, divide both sides of the equation by ‘4’ to solve for ‘\begin{align*}x\end{align*}’.

\begin{align*}\begin{array}{rcl} 4x &=& -24 \\ \frac{\overset{1}{\cancel{4}}x}{\cancel{4}} &=& \frac{\overset{-6}{\cancel{-24}}}{\cancel{4}} \\ x &=& -6 \end{array}\end{align*}

The answer is -6.

The \begin{align*}x\end{align*}-intercept is \begin{align*}(-6,0)\end{align*}.

Second, determine the \begin{align*}y\end{align*}-intercept. Substitute \begin{align*}x=0\end{align*} in the equation.

\begin{align*} \begin{array}{rcl} 4x-3y &=& -24 \\ 4(0)-3y &=& -24 \end{array}\end{align*}

Next, perform the multiplication to clear the parenthesis.

\begin{align*}\begin{array}{rcl} 4(0)-3y &=& -24 \\ 0-3y &=& -24 \\ -3y &=& -24 \end{array}\end{align*}

Next, divide both sides of the equation by ‘-3’ to solve for ‘\begin{align*}y\end{align*}’.

\begin{align*}\begin{array}{rcl} -3y &=& -24 \\ \frac{\overset{1}{\cancel{-3}}y}{\cancel{-3}} &=& \frac{\overset{-8}{\cancel{-24}}}{\cancel{-3}} \\ y &=& 8 \end{array}\end{align*}

The answer is 8.

The \begin{align*}y\end{align*}-intercept is \begin{align*}(0, 8)\end{align*}.

Example 2

For the given linear function, use the \begin{align*}x\end{align*}-and \begin{align*}y\end{align*}-intercepts to draw the graph:

\begin{align*}6x-4y=24\end{align*}

First, determine the \begin{align*}x\end{align*}-intercept. Substitute \begin{align*}y=0\end{align*} in the equation.

\begin{align*}\begin{array}{rcl} 6x-4y &=& 24 \\ 6x-4(0) &=& 24 \end{array}\end{align*}

Next, perform the multiplication to clear the parenthesis.

\begin{align*} \begin{array}{rcl} 6x-4(0) &=& 24 \\ 6x-0 &=& 24 \\ 6x &=& 24 \end{array}\end{align*}

Next, divide both sides of the equation by ‘6’ to solve for ‘\begin{align*}x\end{align*}’.

\begin{align*}\begin{array}{rcl} 6x &=& 24 \\ \frac{\overset{1}{\cancel{6}}x}{\cancel{6}} &=& \frac{\overset{4}{\cancel{24}}}{\cancel{6}} \\ x &=& 4 \end{array}\end{align*}

The answer is 4.

The \begin{align*}x\end{align*}-intercept is \begin{align*}(4,0)\end{align*}.

Second, determine the \begin{align*}y\end{align*}-intercept. Substitute \begin{align*}x=0\end{align*} in the equation.

\begin{align*}\begin{array}{rcl} 6x-4y &=& 24 \\ 6(0)-4y &=& 24 \end{array}\end{align*}

Next, perform the multiplication to clear the parenthesis.

\begin{align*}\begin{array}{rcl} 6(0)-4y &=& 24 \\ 0-4y &=& 24 \\ -4y &=& 24 \end{array}\end{align*}

Next, divide both sides of the equation by ‘-4’ to solve for ‘\begin{align*}y\end{align*}’.

\begin{align*}\begin{array}{rcl} -4y &=& 24 \\ \frac{\overset{1}{\cancel{-4}}y}{\cancel{-4}} &=& \frac{\overset{-6}{\cancel{24}}}{\cancel{-4}} \\ y &=& -6 \end{array}\end{align*}

The answer is -6.

The \begin{align*}y\end{align*}-intercept is \begin{align*}(0,-6)\end{align*}.

Now, the values of the \begin{align*}x\end{align*}- and \begin{align*}y\end{align*}-intercepts can be used to plot the graph of the linear function.

\begin{align*}6x-4y=24\end{align*}

License: CC BY-NC 3.0

Example 3

For the following graph, name the \begin{align*}x\end{align*}- and \begin{align*}y\end{align*}-intercepts.

License: CC BY-NC 3.0

The graph crosses the \begin{align*}x\end{align*}-axis at the point \begin{align*}(30,0)\end{align*} and the \begin{align*}y\end{align*}-axis at the point \begin{align*}(0,-50)\end{align*}.

The \begin{align*}x\end{align*}-intercept is \begin{align*}(30,0)\end{align*} and the \begin{align*}y\end{align*}-intercept is \begin{align*}(0,-50)\end{align*}.

Example 4

For the given linear function, determine the \begin{align*}x\end{align*}- and \begin{align*}y\end{align*}-intercepts.

\begin{align*}15x+30y=120\end{align*}

First, determine the \begin{align*}x\end{align*}-intercept. Substitute \begin{align*}y=0\end{align*} in the equation.

\begin{align*}\begin{array}{rcl} 15x+30y &=& 120 \\ 15x+30(0) &=& 120 \end{array}\end{align*}

Next, perform the multiplication to clear the parenthesis.

\begin{align*}\begin{array}{rcl} 15x+30(0) &=& 120 \\ 15x+0 &=& 120 \\ 15x &=& 120 \end{array}\end{align*}

Next, divide both sides of the equation by ‘15’ to solve for ‘\begin{align*}x\end{align*}’.

\begin{align*}\begin{array}{rcl} 15x &=& 120 \\ \frac{\overset{1}{\cancel{15}}x}{\cancel{15}} &=& \frac{\overset{8}{\cancel{120}}}{\cancel{15}} \\ x &=& 8 \end{array}\end{align*}

The answer is 8.

The \begin{align*}x\end{align*}-intercept is \begin{align*}(8,0)\end{align*}.

Second, determine the \begin{align*}y\end{align*}-intercept. Substitute \begin{align*}x=0\end{align*} in the equation.

\begin{align*}\begin{array}{rcl} 15x+30y &=& 120 \\ 15(0)+30y &=& 120 \end{array}\end{align*}

Next, perform the multiplication to clear the parenthesis.

\begin{align*}\begin{array}{rcl} 15(0)+30y &=& 120 \\ 0+30y &=& 120 \\ 30y &=& 120 \end{array}\end{align*}

Next, divide both sides of the equation by ‘30’ to solve for ‘\begin{align*}y\end{align*}’.

\begin{align*}\begin{array}{rcl} 30y &=& 120 \\ \frac{\overset{1}{\cancel{30}}y}{\cancel{30}} &=& \frac{\overset{4}{\cancel{120}}}{\cancel{30}} \\ y &=& 4 \end{array}\end{align*}

The answer is 4.

The \begin{align*}y\end{align*}-intercept is \begin{align*}(0,4)\end{align*}.

Review

Determine the \begin{align*}x\end{align*} and \begin{align*}y\end{align*}-intercepts of each equation. There will be two answers for each equation.

1. \begin{align*}3x + 4y = 12\end{align*}

2. \begin{align*}6x + 2y = 12\end{align*}

3. \begin{align*}4x + 5y = 20\end{align*}

4. \begin{align*}4x + 2y = 8\end{align*}

5. \begin{align*}3x + 5y = 15\end{align*}

6. \begin{align*}-2x + 3y = -6\end{align*}

7. \begin{align*}-3x + y = 9\end{align*}

8. \begin{align*}-2x- 2y = 6\end{align*}

9. \begin{align*}7x + 3y = 21\end{align*}

10. \begin{align*}2x + 9y = 36\end{align*}

Look at each graph and identify the \begin{align*}x\end{align*} and \begin{align*}y\end{align*}-intercept of each equation. Each graph will have two answers.

11. 

License: CC BY-NC 3.0

12. 

License: CC BY-NC 3.0

13. 

License: CC BY-NC 3.0


14. 
License: CC BY-NC 3.0


15. 
License: CC BY-NC 3.0

Review (Answers)

To see the Review answers, open this PDF file and look for section 9.6. 

Vocabulary

x-intercept

An x-intercept is a location where a graph crosses the x-axis. As a coordinate pair, this point will always have the form (x, 0). x-intercepts are also called solutions, roots or zeros.

y-intercept

A y-intercept is a location where a graph crosses the y-axis. As a coordinate pair, this point will always have the form (0, y).

Intercept

The intercepts of a curve are the locations where the curve intersects the x and y axes. An x intercept is a point at which the curve intersects the x-axis. A y intercept is a point at which the curve intersects the y-axis.

Intercept Method

The intercept method is a way of graphing a linear function by using the coordinates of the x and y-intercepts. The graph is drawn by plotting these coordinates on the Cartesian plane and then joining them with a straight line.

Image Attributions

  1. [1]^ License: CC BY-NC 3.0
  2. [2]^ License: CC BY-NC 3.0
  3. [3]^ License: CC BY-NC 3.0
  4. [4]^ License: CC BY-NC 3.0
  5. [5]^ License: CC BY-NC 3.0
  6. [6]^ License: CC BY-NC 3.0
  7. [7]^ License: CC BY-NC 3.0
  8. [8]^ License: CC BY-NC 3.0
  9. [9]^ License: CC BY-NC 3.0
  10. [10]^ License: CC BY-NC 3.0

Explore More

Sign in to explore more, including practice questions and solutions for Intercepts by Substitution.
Please wait...
Please wait...