We've updated our

TEXT

# Problem Solving

### 2.3 Learning Objectives

• Translate words into algebraic expressions and equations
• Define a process for solving word problems
• Solve distance, rate, and time problems
• Solve area, volume, and perimeter problems
• Solve temperature conversion problems
• Rearrange formulas to isolate specific variables
• Identify an unknown given a formula
Many real-world applications can be modeled by linear equations. For example, a cell phone package may include a monthly service fee plus an additional charge if you exceed your data plan; a car rental company charges a daily fee plus an amount per mile driven; Chipotle offers a base price for your burrito plus additional charges for extra toppings, like guacamole and sour cream. These are examples of applications we come across every day that are modeled by linear equations. In this section, we will set up and use linear equations to solve such problems.

## 2.3.1 Set Up a Linear Equation to Solve an Application

To set up or model a linear equation to fit a real-world application, we must first determine the known quantities and define the unknown quantity as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. Let use an example of a car rental company. The company charges $0.10/mi in addition to a flat rate. In this case, a known cost, such as$0.10/mi, is multiplied by an unknown quantity, the number of miles driven. Therefore, we can write $0.10x$. This expression represents a variable cost because it changes according to the number of miles driven. If a quantity is independent of a variable, we usually just add or subtract it, according to the problem. As these amounts do not change, we call them fixed costs. Consider a car rental agency that charges $0.10/mi plus a daily fee of$50. We can use these quantities to model an equation that can be used to find the daily car rental cost $C$.
$C=0.10x+50$
When dealing with real-world applications, there are certain expressions that we can translate directly into math. The table lists some common verbal expressions and their equivalent mathematical expressions.
Verbal Translation to Math Operations
One number exceeds another by a $x,\text{ }x+a$
Twice a number $2x$
One number is a more than another number $x,\text{ }x+a$
One number is a less than twice another number $x,2x-a$
The product of a number and a, decreased by b $ax-b$
The quotient of a number and the number plus a is three times the number $\frac{x}{x+a}=3x$
The product of three times a number and the number decreased by b is c $3x\left(x-b\right)=c$

### How To: Given a real-world problem, model a linear equation to fit it.

1. Identify known quantities.
2. Assign a variable to represent the unknown quantity.
3. If there is more than one unknown quantity, find a way to write the second unknown in terms of the first.
4. Write an equation interpreting the words as mathematical operations.
5. Solve the equation. Be sure the solution can be explained in words, including the units of measure.

### Example 2.3.A

Find a linear equation to solve for the following unknown quantities: One number exceeds another number by $17$ and their sum is $31$. Find the two numbers.

Answer: Let $x$ equal the first number. Then, as the second number exceeds the first by 17, we can write the second number as $x+17$. The sum of the two numbers is 31. We usually interpret the word is as an equal sign.

$\begin{array}{l}x+\left(x+17\right)\hfill&=31\hfill \\ 2x+17\hfill&=31\hfill&\text{Simplify and solve}.\hfill \\ 2x\hfill&=14\hfill \\ x\hfill&=7\hfill \\ \hfill \\ x+17\hfill&=7+17\hfill \\ \hfill&=24\hfill \end{array}$
The two numbers are $7$ and $24$.

In the following video, we show another example of how to translate an expression in english into a mathematical equation that can then be solved. https://youtu.be/CGC_SwHfwMk In the next example we will write equations that will help us compare cell phone plans.

### Example 2.3.B

There are two cell phone companies that offer different packages. Company A charges a monthly service fee of $34 plus$.05/min talk-time. Company B charges a monthly service fee of $40 plus$.04/min talk-time.
1. Write a linear equation that models the packages offered by both companies.
2. If the average number of minutes used each month is 1,160, which company offers the better plan?
3. If the average number of minutes used each month is 420, which company offers the better plan?
4. How many minutes of talk-time would yield equal monthly statements from both companies?

1. The model for Company A can be written as $A=0.05x+34$. This includes the variable cost of $0.05x$ plus the monthly service charge of $34. Company B’s package charges a higher monthly fee of$40, but a lower variable cost of $0.04x$. Company B’s model can be written as $B=0.04x+\40$.
2. If the average number of minutes used each month is 1,160, we have the following:
$\begin{array}{l}\text{Company }A\hfill&=0.05\left(1,160\right)+34\hfill \\ \hfill&=58+34\hfill \\ \hfill&=92\hfill \\ \hfill \\ \text{Company }B\hfill&=0.04\left(1,160\right)+40\hfill \\ \hfill&=46.4+40\hfill \\ \hfill&=86.4\hfill \end{array}$
So, Company B offers the lower monthly cost of $86.40 as compared with the$92 monthly cost offered by Company A when the average number of minutes used each month is 1,160.
3. If the average number of minutes used each month is 420, we have the following:
$\begin{array}{l}\text{Company }A\hfill&=0.05\left(420\right)+34\hfill \\ \hfill&=21+34\hfill \\ \hfill&=55\hfill \\ \hfill \\ \text{Company }B\hfill&=0.04\left(420\right)+40\hfill \\ \hfill&=16.8+40\hfill \\ \hfill&=56.8\hfill \end{array}$
If the average number of minutes used each month is 420, then Company A offers a lower monthly cost of $55 compared to Company B’s monthly cost of$56.80.
4. To answer the question of how many talk-time minutes would yield the same bill from both companies, we should think about the problem in terms of $\left(x,y\right)$ coordinates: At what point are both the x-value and the y-value equal? We can find this point by setting the equations equal to each other and solving for x.
$\begin{array}{l}0.05x+34=0.04x+40\hfill \\ 0.01x=6\hfill \\ x=600\hfill \end{array}$
Check the x-value in each equation.
$\begin{array}{l}0.05\left(600\right)+34=64\hfill \\ 0.04\left(600\right)+40=64\hfill \end{array}$
Therefore, a monthly average of 600 talk-time minutes renders the plans equal.
Figure 2

The following video shows another example of writing two equations that will allow you to compare two different cell phone plans. https://youtu.be/Q5hlC_VPKGM

## 2.3.2 Solve an Application Using a Formula

Many applications are solved using known formulas. The problem is stated, a formula is identified, the known quantities are substituted into the formula, the equation is solved for the unknown, and the problem’s question is answered. Typically, these problems involve two equations representing two trips, two investments, two areas, and so on. Examples of formulas include the area of a rectangular region, $A=LW$; the perimeter of a rectangle, $P=2L+2W$; and the volume of a rectangular solid, $V=LWH$. When there are two unknowns, we find a way to write one in terms of the other because we can solve for only one variable at a time.

## 2.3.3 Distance, Rate, and Time

### Example 2.3.C

It takes Andrew 30 min to drive to work in the morning. He drives home using the same route, but it takes 10 min longer, and he averages 10 mi/h less than in the morning. How far does Andrew drive to work?

Answer: This is a distance problem, so we can use the formula $d=rt$, where distance equals rate multiplied by time. Note that when rate is given in mi/h, time must be expressed in hours. Consistent units of measurement are key to obtaining a correct solution. First, we identify the known and unknown quantities. Andrew’s morning drive to work takes 30 min, or $\frac{1}{2}$ h at rate $r$. His drive home takes 40 min, or $\frac{2}{3}$ h, and his speed averages 10 mi/h less than the morning drive. Both trips cover distance $d$. A table, such as the one below, is often helpful for keeping track of information in these types of problems.

$d$ $r$ $t$
To Work $d$ $r$ $\frac{1}{2}$
To Home $d$ $r - 10$ $\frac{2}{3}$
Write two equations, one for each trip.
$\begin{array}{ll}d=r\left(\frac{1}{2}\right)\hfill & \text{To work}\hfill \\ d=\left(r - 10\right)\left(\frac{2}{3}\right)\hfill & \text{To home}\hfill \end{array}$
As both equations equal the same distance, we set them equal to each other and solve for r.
$\begin{array}{l}r\left(\frac{1}{2}\right)\hfill&=\left(r - 10\right)\left(\frac{2}{3}\right)\hfill \\ \frac{1}{2}r\hfill&=\frac{2}{3}r-\frac{20}{3}\hfill \\ \frac{1}{2}r-\frac{2}{3}r\hfill&=-\frac{20}{3}\hfill \\ -\frac{1}{6}r\hfill&=-\frac{20}{3}\hfill \\ r\hfill&=-\frac{20}{3}\left(-6\right)\hfill \\ r\hfill&=40\hfill \end{array}$
We have solved for the rate of speed to work, 40 mph. Substituting 40 into the rate on the return trip yields 30 mi/h. Now we can answer the question. Substitute the rate back into either equation and solve for d.
$\begin{array}{l}d\hfill&=40\left(\frac{1}{2}\right)\hfill \\ \hfill&=20\hfill \end{array}$
The distance between home and work is 20 mi.Note that we could have cleared the fractions in the equation by multiplying both sides of the equation by the LCD to solve for $r$.
[latex-display]\begin{array}{l}r\left(\frac{1}{2}\right)\hfill&=\left(r - 10\right)\left(\frac{2}{3}\right)\hfill \\ 6\times r\left(\frac{1}{2}\right)\hfill& =6\times \left(r - 10\right)\left(\frac{2}{3}\right)\hfill \\ 3r\hfill& =4\left(r - 10\right)\hfill \\ 3r\hfill& =4r - 40\hfill \\ -r\hfill& =-40\hfill \\ r\hfill& =40\hfill \end{array}[/latex-display]

In the next example, we will find the lengths of a rectangular field given it's perimeter and a relationship between it's side lengths.

### Example 2.3.D

The perimeter of a rectangular outdoor patio is $54$ ft. The length is $3$ ft greater than the width. What are the dimensions of the patio?

Answer: The perimeter formula is standard: $P=2L+2W$. We have two unknown quantities, length and width. However, we can write the length in terms of the width as $L=W+3$. Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides.

Figure 3
Now we can solve for the width and then calculate the length.
$\begin{array}{l}P=2L+2W\hfill \\ 54=2\left(W+3\right)+2W\hfill \\ 54=2W+6+2W\hfill \\ 54=4W+6\hfill \\ 48=4W\hfill \\ 12=W\hfill \\ \left(12+3\right)=L\hfill \\ 15=L\hfill \end{array}$
The dimensions are $L=15$ ft and $W=12$ ft.

The following video shows an example of finding the dimensions of a rectangular field given it's perimeter. https://youtu.be/VyK-HQr02iQ

### Example 2.3.E

The perimeter of a tablet of graph paper is 48 in. The length is $6$ in. more than the width. Find the area of the graph paper.

Answer: The standard formula for area is $A=LW$; however, we will solve the problem using the perimeter formula. The reason we use the perimeter formula is because we know enough information about the perimeter that the formula will allow us to solve for one of the unknowns. As both perimeter and area use length and width as dimensions, they are often used together to solve a problem such as this one. We know that the length is 6 in. more than the width, so we can write length as $L=W+6$. Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length.

$\begin{array}{l}P\hfill&=2L+2W\hfill \\ 48\hfill&=2\left(W+6\right)+2W\hfill \\ 48\hfill&=2W+12+2W\hfill \\ 48\hfill&=4W+12\hfill \\ 36\hfill&=4W\hfill \\ 9\hfill&=W\hfill \\ \left(9+6\right)\hfill&=L\hfill \\ 15\hfill&=L\hfill \end{array}$
Now, we find the area given the dimensions of $L=15$ in. and $W=9$ in.
$\begin{array}{l}A\hfill&=LW\hfill \\ A\hfill&=15\left(9\right)\hfill \\ \hfill&=135\text{ in}^{2}\hfill \end{array}$ The area is $135$ in2.

The following video shows another example of finding the area of a rectangle given it's perimeter and the relationship between it's side lengths. https://youtu.be/zUlU64Umnq4

## 2.3.4 Volume

### Example 2.3.F

Find the dimensions of a shipping box given that the length is twice the width, the height is $8$ inches, and the volume is 1,600 in.3.

Answer: The formula for the volume of a box is given as $V=LWH$, the product of length, width, and height. We are given that $L=2W$, and $H=8$. The volume is $1,600$ cubic inches.

$\begin{array}{l}V=LWH\hfill \\ 1,600=\left(2W\right)W\left(8\right)\hfill \\ 1,600=16{W}^{2}\hfill \\ 100={W}^{2}\hfill \\ 10=W\hfill \end{array}$
The dimensions are $L=20$ in., $W=10$ in., and $H=8$ in.Note that the square root of ${W}^{2}$ would result in a positive and a negative value. However, because we are describing width, we can use only the positive result.

Express the formula for the surface area of a cylinder, $s=2\pi rh+2\pi r^{2}$, in terms of the height, h. In this example, the variable h is buried pretty deeply in the formula for surface area of a cylinder. Using the order of operations, it can be isolated. Before you look at the solution, use the box below to write down what you think is the best first step to take to isolate h. [practice-area rows="1"][/practice-area]

Answer: Isolate the term containing the variable, h, by subtracting $2\pi r^{2}$from both sides.

$\begin{array}{r}S\,\,=2\pi rh+2\pi r^{2} \\ \underline{-2\pi r^{2}\,\,\,\,\,\,\,\,\,\,\,\,\,-2\pi r^{2}}\\S-2\pi r^{2}\,\,\,\,=\,\,\,\,2\pi rh\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}$

Next, isolate the variable h by dividing both sides of the equation by $2\pi r$.

$\begin{array}{r}\frac{S-2\pi r^{2}}{2\pi r}=\frac{2\pi rh}{2\pi r} \\\\ \frac{S-2\pi r^{2}}{2\pi r}=h\,\,\,\,\,\,\,\,\,\,\end{array}$

You can rewrite the equation so the isolated variable is on the left side.

$h=\frac{S-2\pi r^{2}}{2\pi r}$

In the following video we show how to find the volume of a right circular cylinder formed from a given rectangle. https://youtu.be/KS1pGO_g3vM

## 2.3.5 Isolate Variables in Formulas

Sometimes, it is easier to isolate the variable you you are solving for when you are using a formula. This is especially helpful if you have to perform the same calculation repeatedly, or you are having a computer perform the calculation repeatedly. In the next examples, we will use algebraic properties to isolate a variable in a formula.

### Example 2.3.G

Isolate the term containing the variable, w, from the formula for the perimeter of a rectangle:

${P}=2\left({L}\right)+2\left({W}\right)$.

Answer: First, isolate the term with w by subtracting 2l from both sides of the equation.

$\displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,p\,=\,\,\,\,2l+2w\\\underline{\,\,\,\,\,-2l\,\,\,\,\,-2l\,\,\,\,\,\,\,\,\,\,\,}\\p-2l=\,\,\,\,\,\,\,\,\,\,\,\,\,2w\end{array}$

Next, clear the coefficient of w by dividing both sides of the equation by 2.

$\displaystyle \begin{array}{l}\underline{p-2l}=\underline{2w}\\\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\\ \,\,\,\frac{p-2l}{2}\,\,=\,\,w\\\,\,\,\,\,\,\,\,\,\,\,w=\frac{p-2l}{2}\end{array}$

You can rewrite the equation so the isolated variable is on the left side.

$w=\frac{p-2l}{2}$

### Example 2.3.H

Use the multiplication and division properties of equality to isolate the variable b given $A=\frac{1}{2}bh$

$\begin{array}{l}\,\,\,\,\,\,\,\,A=\frac{1}{2}bh\\\\\left(2\right)A=\left(2\right)\frac{1}{2}bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\frac{2A}{h}=\frac{bh}{h}\\\\\,\,\,\,\,\,\,\,\frac{2A}{h}=\frac{b\cancel{h}}{\cancel{h}}\end{array}$

Write the equation with the desired variable on the left-hand side as a matter of convention:

$b=\frac{2A}{h}$

Use the multiplication and division properties of equality to isolate the variable given $A=\frac{1}{2}bh$

$\begin{array}{l}\,\,\,\,\,\,\,\,A=\frac{1}{2}bh\\\\\left(2\right)A=\left(2\right)\frac{1}{2}bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\frac{2A}{b}=\frac{bh}{b}\\\\\,\,\,\,\,\,\,\,\frac{2A}{b}=\frac{h\cancel{b}}{\cancel{b}}\end{array}$

Write the equation with the desired variable on the left-hand side as a matter of convention:

$h=\frac{2A}{b}$

## 2.3.6 Temperature

Let’s look at another formula that includes parentheses and fractions, the formula for converting from the Fahrenheit temperature scale to the Celsius scale.

$C=\left(F--32\right)\cdot \frac{5}{9}$

### Example 2.3.I

Given a temperature of $12^{\circ}{C}$, find the equivalent in ${}^{\circ}{F}$.

Answer: Substitute the given temperature in${}^{\circ}{C}$ into the conversion formula:

$12=\left(F-32\right)\cdot \frac{5}{9}$

Isolate the variable F to obtain the equivalent temperature.

$\begin{array}{r}12=\left(F-32\right)\cdot \frac{5}{9}\\\\\left(\frac{9}{5}\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\\,\frac{108}{5}=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\21.6=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\underline{+32\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+32}\,\,\,\,\,\,\,\,\,\,\,\,\\\\53.6={}^{\circ}{F}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}$

As with the other formulas we have worked with, we could have isolated the variable F first, then substituted in the given temperature in Celsius.

### Example 2.3.J

Solve the formula shown below for converting from the Fahrenheit scale to the Celsius scale for F. [latex-display]C=\left(F--32\right)\cdot \frac{5}{9}[/latex-display]

Answer: To isolate the variable F, it would be best to clear the fraction involving F first. Multiply both sides of the equation by $\displaystyle \frac{9}{5}$.

$\begin{array}{l}\\\,\,\,\,\left(\frac{9}{5}\right)C=\left(F-32\right)\left(\frac{5}{9}\right)\left(\frac{9}{5}\right)\\\\\,\,\,\,\,\,\,\,\,\,\,\,\frac{9}{5}C=F-32\end{array}$

$\begin{array}{l}\frac{9}{5}\,C+32=F-32+32\\\\\frac{9}{5}\,C+32=F\end{array}$

$F=\frac{9}{5}C+32$