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# Intercepts by Substitution

## Graphing equations that form lines by finding x and y intercepts

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Using Intercepts

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 her60.00. 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 cost5.00 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 \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 \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 \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 \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 \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 \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*}

#### Example 3

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

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 \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 \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.

12.

13.

14.

15.

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### Vocabulary Language: English

TermDefinition
$x-$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.

1. [1]^ License: CC BY-NC 3.0
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