# 4.15: Problem Solving with Linear Graphs

**Basic**Created by: CK-12

**Practice**Problem Solving with Linear Graphs

Suppose that you have a part-time job delivering newspapers and that you can deliver 9 newspapers every 15 minutes. If you deliver at a constant rate, how many newspapers could you deliver in 2 hours? Could you construct a graph to represent this situation? How would the graph help you to solve the problem? In this Concept, you'll learn how to construct a graph to solve real-world linear problems such as this one.

### Guidance

Graphing is a very useful tool when analyzing a situation. This Concept will focus on using graphs to help solve linear problems that occur in real life.

Remember the 4-Step Problem-Solving Plan:

- Understand the problem and underline or highlight key information.
- Translate the problem and devise a method to solve the problem.
- Carry out the plan and solve the problem.
- Check and interpret your answer. Does it make sense?

#### Example A

*A cell phone company is offering its custumers the following deal. You can buy a new cell phone for $60 and pay a monthly flat rate of $40 per month for unlimited calls. How much money will this deal cost you after 9 months?*

**Solution:**

Begin by translating the sentence into an algebraic equation.

\begin{align*}\text{cell phone} = \$60 + \$40 \ \text{per month}\end{align*}

Let \begin{align*}m=\end{align*}*the number of months* and \begin{align*}t=\end{align*}*total cost*. The equation becomes:

\begin{align*}t(m)=60+40m\end{align*}

You could use guess and check or solve this equation. However, this Concept focuses on using a graph to problem-solve. This equation is in slope-intercept form. By graphing the line of this equation, you will find all the ordered pairs that are solutions to the cell phone problem.

Finding the cost at month 9, you can see the cost is approximately $425.00. To check if this is approximately correct, substitute 9 in for the variable \begin{align*}m\end{align*}

\begin{align*}\text{Phone} & = \$60\\
\text{Calling plan} & = \$40 \times 9 = \$360\\
\text{Total cost} & = \$420.\end{align*}

Our answer, $425.00, is approximately equal to the exact solution, $420.00.

#### Example B

*Christine took one hour to read 22 pages of* “Harry Potter and the Order of the Phoenix.” *She has 100 pages left to read in order to finish the book. Assuming that she reads at a constant rate of pages per hour, how much time should she expect to spend reading in order to finish the book?*

**Solution:**

We do not have enough information to write an equation. We do not know the slope or the \begin{align*}y-\end{align*}*and* (1, 22).

Using the graph and finding 100 pages, you can determine it will take Christine about 4.5 hours to read 100 pages.

You can also think of this as a direct variation situation and solve it by writing a proportion.

\begin{align*}\frac{22 \ pages}{1 \ hour}=\frac{100 \ pages}{h \ hours}\end{align*}

By using the Cross Products Theorem, you can find out \begin{align*}h \approx 4.55\end{align*}

#### Example C

*A cell phone monthly messaging plan costs $10 for the first 100 texts and then $0.25 for each additional text. Make a graph of this relation. Use the graph to determine the cost of sending 123 text messages in a month.*

**Solution:**

The plan costs $10 no matter how many texts you send under 100. Even if you send zero texts, the cost is $10. So, $10 is the \begin{align*}y\end{align*}

The coordinate pair for when \begin{align*}x=23\end{align*} is between 15 and 16. In situations like this, it is helpful to write the equation to solve for the exact value, which in this case is 15.75.

It would cost $15.75 to send 123 texts on this plan.

### Guided Practice

Jerome drove 20 miles an hour for 15 minutes on city streets, and then 40 miles an hour for 30 minutes on the highway, 60 miles an hour for an hour on the freeway, and finally 20 miles an hour for 15 minutes on city streets. Use a graph to determine how far Jerome drove.

**Solution:**

Jerome's different speeds are his different rates, in miles per hour. Since his times are given in minutes, convert the rates to miles per minute:

\begin{align*}\frac{20 \text{ miles}}{\text{hour}}=\frac{20 \text{ miles}}{60 \text{ minutes}}=\frac{1 \text{ mile}}{3 \text{ minutes}}\end{align*}

If the change in time is 15 minutes, then the change in distance is \begin{align*}\frac{1 \text{ mile}}{3 \text{ minutes}} \cdot 15 \text{ minutes} = 5 \text{ miles}\end{align*}. So the rate can be written as \begin{align*}\frac{5 \text{ miles}}{15 \text{ minutes}}.\end{align*}

\begin{align*}\frac{40 \text{ miles}}{\text{hour}}=\frac{40 \text{ miles}}{60 \text{ minutes}}=\frac{2 \text{ mile}}{3 \text{ minutes}}\end{align*}

If the change in time is 30 minutes, then the change in distance is \begin{align*}\frac{2 \text{ mile}}{3 \text{ minutes}} \cdot 30 \text{ minutes} = 20 \text{ miles}\end{align*}. So the rate can be written as \begin{align*}\frac{20 \text{ miles}}{30 \text{ minutes}}.\end{align*}

\begin{align*}\frac{60 \text{ miles}}{\text{hour}}=\frac{60 \text{ miles}}{1 \text{ minutes}}=\frac{1 \text{ mile}}{1 \text{ minutes}}\end{align*}

If the change in time is 60 minutes, then the change in distance is \begin{align*}\frac{1 \text{ mile}}{1 \text{ minutes}} \cdot 60 \text{ minutes} = 60 \text{ miles}\end{align*}. So the rate can be written as \begin{align*}\frac{60 \text{ miles}}{60 \text{ minutes}}.\end{align*}

The last interval has the same rate as the first: \begin{align*}\frac{5 \text{ miles}}{15 \text{ minutes}}.\end{align*}

Use these rates to draw a graph of distance in miles as a function of time in minutes.

From the graph you can see that Jerome's trip takes 120 minutes, or two hours, and that he drives 90 miles.

### Practice

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: Word Problem Solving 4 (10:05)

(http://www.youtube.com/watch?v=5EdbPz1ZVn0)

- Using the following graph, determine these values:
- The amount of earnings after 40 hours
- How many hours it takes to earn $250.00
- The slope of the line and what it represents
- The \begin{align*}y-\end{align*}intercept of the line and what it represents

- A stretched spring has a length of 12 inches when a weight of 2 lbs is attached to it. The same spring has a length of 18 inches when a weight of 5 lbs is attached to it. It is known from physics that within certain weight limits, the function that describes how much a spring stretches with different weights is a linear function. What is the length of the spring when no weights are attached?

- A gym is offering a deal to new members. Customers can sign up by paying a registration fee of $200 and a monthly fee of $39. How much will this membership cost a member by the end of one year?
- A candle is burning at a linear rate. The candle measures five inches two minutes after it was lit. It measures three inches eight minutes after it was lit. What was the original length of the candle?
- Tali is trying to find the thickness of a page of his telephone book. To do this, he takes a measurement and finds out that 550 pages measure 1.25 inches. What is the thickness of one page of the phone book?
- Bobby and Petra are running a lemonade stand and they charge 45 cents for each glass of lemonade. To break even, they must make $25. How many glasses of lemonade must they sell to break even?
- The tip for a $78.00 restaurant bill is $9.20. What is the tip for a $21.50 meal?
- Karen left her house and walked at a rate of 4
*miles*/*hour*for 30 minutes. She realized she was late for school and began to jog at a rate of 5.5*miles*/*hour*for 25 minutes. Using a graph, determine how far she is from her house at the end of 45 minutes.

**Mixed Review**

- Simplify \begin{align*}-4|-21-11|+16\end{align*}.
- Give an example of a direct variation equation and label its constant of variation.
- Identify the slope and \begin{align*}y-\end{align*}intercept: \begin{align*}\frac{5}{3} x=y-4\end{align*}.
- Suppose \begin{align*}A=(x,y)\end{align*} is located in quadrant I. Write a rule that would move \begin{align*}A\end{align*} into quadrant III.
- Find the intercepts of \begin{align*}0.04x+0.06y=18\end{align*}.
- Evaluate \begin{align*}f(4)\end{align*} when \begin{align*}f(x)=\frac{3x^2}{8}\end{align*}.

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### Image Attributions

Here you'll learn how to solve real-world linear problems by constructing and interpreting graphs.