# 5.1: Linear Equations in Slope-Intercept Form

**At Grade**Created by: CK-12

Previously, you learned how to graph solutions to two-variable equations in slope-intercept form. This lesson focuses on how to write an equation for a graphed line. There are two things you will need from the graph to write the equation in slope-intercept form:

- The \begin{align*}y-\end{align*}
y− intercept of the graph - The slope of the line

Having these two pieces of information will allow you to make the appropriate substitutions in the slope-intercept formula. Recall from the last chapter,

Slope-intercept form: \begin{align*}y=(slope)x+ (y-intercept)\end{align*}

**Example 1:** Write the equation for a line with a slope of 4 and a \begin{align*}y-\end{align*}

**Solution:** Slope-intercept form requires two things: the slope and \begin{align*}y-\end{align*}

\begin{align*}y& =(slope)x+ (y-intercept)\\ y& =4x+(-3)\\ y& =4x-3\end{align*}

You can also use a graphed line to determine the slope and \begin{align*}y-\end{align*}

**Example 2:** *Use the graph below to write its equation in slope-intercept form.*

**Solution:** The \begin{align*}y-\end{align*}

## Writing an Equation Given the Slope and a Point

Sometimes it may be difficult to determine the \begin{align*}y-\end{align*}

**Step 1:** Begin by writing the formula for slope-intercept form \begin{align*}y=mx+b\end{align*}

**Step 2:** Substitute the given slope for \begin{align*}m\end{align*}

**Step 3:** Use the ordered pair you are given \begin{align*}(x, y)\end{align*}

**Step 4:** Solve for \begin{align*}b\end{align*}

**Step 5:** Rewrite the original equation in Step 1, substituting the slope for \begin{align*}m\end{align*}

**Example 3:** *Write an equation for a line with slope of 4 that contains the ordered pair (–1, 5).*

**Solution:**

**Step 1:** Begin by writing the formula for slope-intercept form.

\begin{align*}y=mx+b\end{align*}

**Step 2:** Substitute the given slope for \begin{align*}m\end{align*}

\begin{align*}y=4x+b\end{align*}

**Step 3:** Use the ordered pair you are given (–1, 5) and substitute these values for the variables \begin{align*}x\end{align*}

\begin{align*}5=(4)(-1)+b\end{align*}

**Step 4:** Solve for \begin{align*}b\end{align*}

\begin{align*}5& =-4+b\\ 5+4& =-4+4+b\\ 9& =b\end{align*}

**Step 5:** Rewrite \begin{align*}y=mx+b\end{align*}

\begin{align*}y=4x+9\end{align*}

**Example 4:** Write the equation for a line with a slope of –3 containing the point (3, –5).

**Solution:** Using the five-steps from above:

\begin{align*}y& =(slope)x+(y-intercept)\\ y& =-3x+b\\ -5& =-3(3)+b\\ -5& =-9+b\\ 4& =b\\ y& =-3x+4\end{align*}

## Writing an Equation Given Two Points

In many cases, especially real-world situations, you are given neither the slope nor the \begin{align*}y-\end{align*}intercept. You might have only two points to use to determine the equation of the line.

To find an equation for a line between two points, you need two things:

- The \begin{align*}y-\end{align*}intercept of the graph
- The slope of the line

Previously, you learned how to determine the slope between two points. Let’s repeat the formula here.

The slope between any two points \begin{align*}(x_1, y_1 )\end{align*} and \begin{align*}(x_2, y_2)\end{align*} is: \begin{align*}slope=\frac{y_2-y_1}{x_2-x_1}\end{align*}.

The procedure for determining a line given two points is the same five-step process as writing an equation given the slope and a point.

**Example 5:** *Write the equation for the line containing the points (3, 2) and (–2, 4).*

**Solution:** You need the slope of the line. Find the line's slope by using the formula. Choose one ordered pair to represent \begin{align*}(x_1,y_1)\end{align*} and the other ordered pair to represent \begin{align*}(x_2,y_2)\end{align*}.

\begin{align*}slope=\frac{y_2-y_1}{x_2-x_1}=\frac{4-2}{-2-3}=-\frac{2}{5}\end{align*}

Now use the five-step process to find the equation for this line.

**Step 1:** Begin by writing the formula for slope-intercept form.

\begin{align*}y=mx+b\end{align*}

**Step 2:** Substitute the given slope for \begin{align*}m\end{align*}.

\begin{align*}y=\frac{-2}{5} x+b\end{align*}

**Step 3:** Use one of the ordered pairs you are given (–2, 4) and substitute these values for the variables \begin{align*}x\end{align*} and \begin{align*}y\end{align*} in the equation.

\begin{align*}4=\left (\frac{-2}{5}\right )(-2)+b\end{align*}

**Step 4:** Solve for \begin{align*}b\end{align*} (the \begin{align*}y-\end{align*}intercept of the graph).

\begin{align*}4& =\frac{4}{5}+b\\ 4-\frac{4}{5}& =\frac{4}{5}-\frac{4}{5}+b\\ \frac{16}{5}& =b\end{align*}

**Step 5:** Rewrite \begin{align*}y=mx+b\end{align*}, substituting the slope for \begin{align*}m\end{align*} and the \begin{align*}y-\end{align*}intercept for \begin{align*}b\end{align*}.

\begin{align*}y=\frac{-2}{5} x+\frac{16}{5}\end{align*}

**Example 6:** Write the equation for a line containing the points (–4, 1) and (–2, 3).

**Solution:**

- Start with the slope–intercept form of the line \begin{align*}y=mx+b\end{align*}.
- Find the slope of the line: \begin{align*}m=\frac{y_2-y_1}{x_2-x_1}=\frac{3-1}{-2-(-4)}=\frac{2}{2}=1\end{align*}.
- Substitute the value of slope for \begin{align*}m: y=(1)x+b\end{align*}.
- Substitute the coordinate (–2, 3) into the equation for the variables \begin{align*}x\end{align*} and \begin{align*}y : 3=-2+b \Rightarrow b=5\end{align*}.
- Rewrite the equation, substituting the slope for \begin{align*}m\end{align*} and the \begin{align*}y-\end{align*}intercept for \begin{align*}b\end{align*}: \begin{align*}y=x+5\end{align*}.

## Writing a Function in Slope-Intercept Form

Remember that a linear function has the form \begin{align*}f(x)=mx+b\end{align*}. Here \begin{align*}f(x)\end{align*} represents the \begin{align*}y\end{align*} values of the equation or the graph. So \begin{align*}y=f(x)\end{align*} and they are often used interchangeably. Using the functional notation in an equation often provides you with more information.

For instance, the expression \begin{align*}f(x)=mx+b\end{align*} shows clearly that \begin{align*}x\end{align*} is the independent variable because you **substitute** values of \begin{align*}x\end{align*} into the function and perform a series of operations on the value of \begin{align*}x\end{align*} in order to calculate the values of the dependent variable, \begin{align*}y\end{align*}.

In this case when you substitute \begin{align*}x\end{align*} into the function, the function tells you to multiply it by \begin{align*}m\end{align*} and then add \begin{align*}b\end{align*} to the result. This process generates all the values of \begin{align*}y\end{align*} you need.

**Example 7:** Consider the function \begin{align*}f(x)=3x-4.\end{align*} *Find* \begin{align*}f(2), f(0),\end{align*} *and* \begin{align*}f(-1)\end{align*}.

**Solution:** Each number in parentheses is a value of \begin{align*}x\end{align*} that you need to substitute into the equation of the function.

\begin{align*}f(2)=2; f(0)=-4; \ and \ f(-1)=-7\end{align*}

Function notation tells you much more than the value of the independent variable. It also indicates a point on the graph. For example, in the above example, \begin{align*}f(-1)=-7\end{align*}. This means the ordered pair (–1, –7) is a solution to \begin{align*}f(x)=3x-4\end{align*} and appears on the graphed line. You can use this information to write an equation for a function.

**Example 8:** *Write an equation for a line with \begin{align*}m=3.5\end{align*} and \begin{align*}f(-2)=1\end{align*}*.

**Solution:** You know the slope and you know a point on the graph (–2, 1). Using the methods presented in this lesson, write the equation for the line.

Begin with slope-intercept form.

\begin{align*}&& y& =mx+b\\ \text{Substitute the value for the slope.} && y& =3.5x+b\\ \text{Use the ordered pair to solve for} \ b. && 1& =3.5(-2)+b\\ && b& =8\\ \text{Rewrite the equation.} && y& =3.5x+8 \\ \text{or} && f(x)& =3.5x+8\end{align*}

## Solve Real-World Problems Using Linear Models

Let’s apply the methods we just learned to a few application problems that can be modeled using a linear relationship.

**Example 9:** *Nadia has $200 in her savings account. She gets a job that pays $7.50 per hour and she deposits all her earnings in her savings account. Write the equation describing this problem in slope–intercept form. How many hours would Nadia need to work to have $500 in her account?*

**Solution:** Begin by defining the variables:

\begin{align*}y=\end{align*} amount of money in Nadia’s savings account

\begin{align*}x=\end{align*} number of hours

The problem gives the \begin{align*}y-\end{align*}intercept and the slope of the equation.

We are told that Nadia has $200 in her savings account, so \begin{align*}b=200\end{align*}.

We are told that Nadia has a job that pays $7.50 per hour, so \begin{align*}m=7.50\end{align*}.

By substituting these values in slope–intercept form \begin{align*}y=mx+b\end{align*}, we obtain \begin{align*}y=7.5x+200\end{align*}.

To answer the question, substitute $500 for the value of \begin{align*}y\end{align*} and solve.

\begin{align*}500 = 7.5x+200 \Rightarrow 7.5x=300 \Rightarrow x=40\end{align*}

Nadia must work 40 hours if she is to have $500 in her account.

**Example 10:** *A stalk of bamboo of the family Phyllostachys nigra grows at steady rate of 12 inches per day and achieves its full height of 720 inches in 60 days. Write the equation describing this problem in slope–intercept form. How tall is the bamboo 12 days after it started growing?*

**Solution:** Define the variables.

\begin{align*}y=\end{align*} the height of the bamboo plant in inches

\begin{align*}x=\end{align*} number of days

The problem gives the slope of the equation and a point on the line.

The bamboo grows at a rate of 12 inches per day, so \begin{align*}m=12\end{align*}.

We are told that the plant grows to 720 inches in 60 days, so we have the point (60, 720).

\begin{align*}\text{Start with the slope-intercept form of the line.} && y& =mx+b\\ \text{Substitute 12 for the slope.} && y& =12x+b\\ \text{Substitute the point} \ (60,720). && 720& =12(60)+b \Rightarrow b=0\\ \text{Substitute the value of} \ b \ \text{back into the equation.} && y& =12x\end{align*}

To answer the question, substitute the value \begin{align*}x=12\end{align*} to obtain \begin{align*}y=12(12)=144\end{align*} inches.

## Practice Set

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both.

CK-12 Basic Algebra: Linear Equations in Slope-Intercept Form (14:58)

- What is the formula for slope-intercept form? What do the variables \begin{align*}m\end{align*} and \begin{align*}b\end{align*} represent?
- What are the five steps needed to determine the equation of a line given the slope and a point on the graph (not the \begin{align*}y-\end{align*}intercept)?
- What is the first step in finding the equation of a line given two points?

In 4 – 20, find the equation of the line in slope–intercept form.

- The line has slope of 7 and \begin{align*}y-\end{align*}intercept of –2.
- The line has slope of –5 and \begin{align*}y-\end{align*}intercept of 6.
- The line has \begin{align*}\text{slope}=-2\end{align*} and a \begin{align*}y-\end{align*}intercept = 7.
- The line has \begin{align*}\text{slope}=\frac{2}{3}\end{align*} and a \begin{align*}y-\end{align*}intercept = \begin{align*}\frac{4}{5}\end{align*}.
- The line has slope of \begin{align*}-\frac{1}{4}\end{align*} and contains point (4, –1).
- The line has slope of \begin{align*}\frac{2}{3}\end{align*} and contains point \begin{align*}\left(\frac{1}{2},1\right )\end{align*}.
- The line has slope of –1 and contains point \begin{align*}\left (\frac{4}{5},0\right )\end{align*}.
- The line contains points (2, 6) and (5, 0).
- The line contains points (5, –2) and (8, 4).
- The line contains points (3, 5) and (–3, 0).
- The slope of the line is \begin{align*}-\frac{2}{3}\end{align*} and the line contains point (2, –2).
- The slope of the line is –3 and the line contains point (3, –5).
- The line contains points (10, 15) and (12, 20).

In 21 – 28, find the equation of the linear function in slope–intercept form.

- \begin{align*}m=5, f(0)=-3\end{align*}
- \begin{align*}m=-2\end{align*} and \begin{align*}f(0)=5\end{align*}
- \begin{align*}m=-7, f(2)=-1\end{align*}
- \begin{align*}m=\frac{1}{3}, f(-1)=\frac{2}{3}\end{align*}
- \begin{align*}m=4.2, f(-3)=7.1\end{align*}
- \begin{align*}f\left (\frac{1}{4}\right )=\frac{3}{4}, f(0)=\frac{5}{4}\end{align*}
- \begin{align*}f(1.5)=-3, f(-1)=2\end{align*}
- \begin{align*}f(-1)=1\end{align*} and \begin{align*}f(1)=-1\end{align*}
- To buy a car, Andrew puts in a down payment of $1500 and pays $350 per month in installments. Write an equation describing this problem in slope-intercept form. How much money has Andrew paid at the end of one year?
- Anne transplants a rose seedling in her garden. She wants to track the growth of the rose, so she measures its height every week. In the third week, she finds that the rose is 10 inches tall and in the eleventh week she finds that the rose is 14 inches tall. Assuming the rose grows linearly with time, write an equation describing this problem in slope-intercept form. What was the height of the rose when Anne planted it?
- Ravi hangs from a giant exercise spring whose length is 5 m. When his child Nimi hangs from the spring, its length is 2 m. Ravi weighs 160 lbs. and Nimi weighs 40 lbs. Write the equation for this problem in slope-intercept form. What should we expect the length of the spring to be when his wife Amardeep, who weighs 140 lbs., hangs from it?
- Petra is testing a bungee cord. She ties one end of the bungee cord to the top of a bridge and to the other end she ties different weights. She then measures how far the bungee stretches. She finds that for a weight of 100 lbs., the bungee stretches to 265 feet and for a weight of 120 lbs., the bungee stretches to 275 feet. Physics tells us that in a certain range of values, including the ones given here, the amount of stretch is a linear function of the weight. Write the equation describing this problem in slope–intercept form. What should we expect the stretched length of the cord to be for a weight of 150 lbs?

**Mixed Review**

- Translate into an algebraic sentence:
*One-third of a number is seven less than that number.* - The perimeter of a square is 67 cm. What is the length of its side?
- A hockey team played 17 games. They won two more than they lost. They lost 3 more than they tied. How many games did they win, lose, and tie?
- Simplify \begin{align*}\frac{(30-4+4 \div 2) \div (21 \div 3)}{2}\end{align*}.
- What is the opposite of 16.76?
- Graph the following on a number line: \begin{align*}\left \{6,\frac{11}{3},-5.65,\frac{21}{7}\right \}\end{align*}.
- Simplify: \begin{align*}[(-4+4.5)+(18-|-13|)+(-3.3)]\end{align*}.