# Problem Solving with Linear Graphs

## Solve story problems using graphs

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Practice Problem Solving with Linear Graphs

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Problem Solving with Linear Relations and Multiple Representations

Learning Goal

By the end of this lesson you will be able to determine other representations given one specific representation and solve problems and compare equations and 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? What about an equation or table of values to model this situation?  In this Concept, you'll learn how to construct an equation, a table and 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:

1. Understand the problem and underline or highlight key information.
2. Translate the problem and devise a method to solve the problem.
3. Carry out the plan and solve the problem.

#### 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:

1. Algebraic Model

Begin by translating the sentence into an algebraic equation.

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

Let m\begin{align*}m\end{align*} represent the number of months and T\begin{align*}T \end{align*} represent the total cost.

The equation becomes:

T=60+40m\begin{align*}T = 60 + 40m\end{align*}

We want to know how much this deal will cost after 9 months.

Since m\begin{align*}m\end{align*} represents the number of months we substitute 9 in for the months.

T=60+40(9)\begin{align*}T = 60 + 40(9)\end{align*}

T=60+360\begin{align*}T = 60 + 360\end{align*}

T=420\begin{align*}T = 420\end{align*}

Therefore the total cost for this plan after 9 months is $420. 2. Graphical Model For the graph we start at$60 and then for every month we increase by $40. 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 m\begin{align*}m\end{align*}.

PhoneCalling planTotal cost=$60=$40×9=$360=$420.\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.

Check out the video of this solution.

3.  Numerical Model

We can build a table of values to see how much it will cost for 9 months.

#### 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:

1.  Algebraic Model

We know that if Christine had never picked up the book, she would have read zero pages. So it takes Christine 0 hours to read 0 pages

We know Christine reads 22 pages per hour.

Let N\begin{align*}N\end{align*} represent the number of pages Christine has read and h\begin{align*}h\end{align*} the number of hours Christine has read.

Number of pages read = 22 times the number of hours Christine has read

N=22h\begin{align*}N = 22h\end{align*}

This time we are looking for how long she read for or h\begin{align*}h\end{align*} when she has read 100 pages.  Therefore N\begin{align*}N\end{align*} is 100.

100=22h\begin{align*}100 = 22h\end{align*}

To get h\begin{align*}h\end{align*} by itself we divide both sides by 22.

10022=22h22\begin{align*}\frac{100}{22} = \frac{22h}{22}\end{align*}

4.54545=h\begin{align*}4.54545 = h\end{align*}

or

h=4.5\begin{align*}h = 4.5\end{align*} hours.

To check we can multiply 4.5 hours times the 22 pages per hour to get 99 pages read which is very close to 100 pages read.

2. Graphical Model

To graph this situation we have two points we can graph. Again we know that she did not read any pages at the beginning. We also know it took Christine one hour to read 22 pages. The two coordinates we can graph are (0, 0) 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.

22 pages1 hour=100 pagesh hours\begin{align*}\frac{22 \ pages}{1 \ hour}=\frac{100 \ pages}{h \ hours}\end{align*}

By cross multiplying you can find out h4.55\begin{align*}h \approx 4.55\end{align*}. It will take Christine about 4.55 hours to read 100 pages, which is very close to your original estimate of 4.5 hours.

#### Again we can approximate how long it takes Christine to read 100 pages by creating a table of values.

 Number of Hours Reading Number of Pages Read 0 0 1 22 2 44 3 66 4 88 5 108

### 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)

1. Using the following graph, determine these values:
1. The slope or rate of change of the line and what it represents

2. The initial value of the line and what it represents

3. An equation to represent the Earnings for n\begin{align*}n\end{align*} hours.

4. A table of values to represent the Earnings over 40 hours (use intervals of 10 hours)

5. The amount of earnings after 40 hours

6. How many hours it takes to earn $250.00 2. 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? 1. 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? 2. 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? 3. 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? 4. 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?

Solution to #2

Step 1: We know: the length of the spring = 12 inches when weight = 2 lbs

the length of the spring = 18 inches when weight = 5 lbs

We want: the length of the spring when weight = 0 lbs

Let  x\begin{align*}x\end{align*} represent the weight attached to the spring.

Let y\begin{align*}y \end{align*} represent the length of the spring.

Step 2: We can solve this problem by making a graph that shows the weight on the horizontal axis and the length of the spring on the vertical axis.

We have two points we can graph:

When the weight is 2 lbs, the length of the spring is 12 inches. This gives point (2, 12).

When the weight is 5 lbs, the length of the spring is 18 inches. This gives point (5, 18).

Graphing those two points and connecting them gives us our line.

Step 3: The question was: “What is the length of the spring when no weights are attached?

We can answer this question by reading the graph we just made. When there is no weight on the spring, the value equals zero, so we are just looking for the intercept of the graph. On the graph, the intercept appears to be approximately 8 inches.

Step 4: To check if this correct, let’s think of the problem again.

You can see that the length of the spring goes up by 6 inches when the weight is increased by 3 lbs, so the slope of the line is 2 inches .

To find the length of the spring when there is no weight attached, we can look at the spring when there are 2 lbs attached. For each pound we take off, the spring will shorten by 2 inches. If we take off 2 lbs, the spring will be shorter by 4 inches. So, the length of the spring with no weights is 12 inches - 4 inches = 8 inches.