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Models and Applications

Set up a linear equation to solve a real-world application.

Models and Applications

Solve a formula for a specified variable (IA 2.3.1)
Use a problem-solving strategy for word problems (IA 2.2.1)

Solve a Formula for a Specific Variable

Try it.

The formula for the perimeter of a rectangle is found using the formula: $P=2l+2w$ . Solve this formula in terms of l.

Solution

$P=2l+2w$	Since we are solving for l we isolate the l term
$P-2w=2l+2w-2w$	Subtract 2w from both sides.
$P-2w=2l$	Combine like terms
$\frac{P-2w}{2}=\frac{2l}{2}$	Divide by 2 to isolate l.
$\frac{P-2w}{2}=l$	Simplify

Solve each formula for the specific variable.

Try it.

Solve for $b$
$P=a+b+c$

Try it.

Solve for $s$
$P=4s$

Try it.

Solve for $r$
$C=2\pi r$

Try it.

Solve for $b$
$A=\frac{1}{2}bh$

Try it.

Solve for $W$
$P=2L+2W$

Try it.

Solve for $m$
$y=mx+b$

Try it.

Solve for $h$
$A=2\pi h+2\pi {r}^{2}$

Try it.

Solve for $r$
$A=\pi {r}^{2}$

Try it.

Solve for $s$
$V=\frac{1}{3}{s}^{2}h$

Try it.

Solve for $L$
$A=2LW+2HW+2LH$

Condensed — the full section is in OpenStax College Algebra 2e.

Setting up a Linear Equation to Solve a Real-World 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 us use the car rental example above. 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. lists some common verbal expressions and their equivalent mathematical expressions.

Verbal	Translation to Math Operations
One number exceeds another by a	$x,\ x+a$
Twice a number	$2x$
One number is a more than another number	$x,\ x+a$
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	$3x(x-b)=c$

Example

Try it.

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.

Solution

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.

\[\begin{array}{lll}x+(x+17) & = & 31 \\ 2x+17 & = & 31\ \text{Simplify and solve}\text{.} \\ 2x & = & 14 \\ x & = & 7 \\ x+17 & = & 7+17 \\ & = & 24\end{array}\]

The two numbers are $7$ and $24.$

Condensed — the full section is in OpenStax College Algebra 2e.

Using a Formula to Solve a Real-World Application

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.

Example

Try it.

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?

Solution

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 , 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.

\[\begin{array}{llll}d & = & r(\frac{1}{2}) & \ \text{To work} \\ d & = & (r-10)(\frac{2}{3}) & \ \text{To home}\end{array}\]

As both equations equal the same distance, we set them equal to each other and solve for r.

\[\begin{array}{lll}r(\frac{1}{2}) & = & (r-10)(\frac{2}{3}) \\ \frac{1}{2}r & = & \frac{2}{3}r-\frac{20}{3} \\ \frac{1}{2}r-\frac{2}{3}r & = & -\frac{20}{3} \\ -\frac{1}{6}r & = & -\frac{20}{3} \\ r & = & -\frac{20}{3}(-6) \\ r & = & 40\end{array}\]

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.

\[\begin{array}{lll}d & = & 40(\frac{1}{2}) \\ & = & 20\end{array}\]

The distance between home and work is 20 mi.

Condensed — the full section is in OpenStax College Algebra 2e.

Key Concepts

A linear equation can be used to solve for an unknown in a number problem. See .
Applications can be written as mathematical problems by identifying known quantities and assigning a variable to unknown quantities. See .
There are many known formulas that can be used to solve applications. Distance problems, for example, are solved using the $d=rt$ formula. See .
Many geometry problems are solved using the perimeter formula $P=2L+2W,$ the area formula $A=LW,$ or the volume formula $V=LWH.$ See , , and .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

The formula for the perimeter of a rectangle is found using the formula: $P=2l+2w$ . Solve this formula in terms of l.

Révèle la réponse

$P=2l+2w$	Since we are solving for l we isolate the l term
$P-2w=2l+2w-2w$	Subtract 2w from both sides.
$P-2w=2l$	Combine like terms
$\frac{P-2w}{2}=\frac{2l}{2}$	Divide by 2 to isolate l.
$\frac{P-2w}{2}=l$	Simplify

Solve for $b$
$P=a+b+c$
Solve for $s$
$P=4s$
Solve for $r$
$C=2\pi r$
Solve for $b$
$A=\frac{1}{2}bh$
Solve for $W$
$P=2L+2W$
Solve for $m$
$y=mx+b$
Solve for $h$
$A=2\pi h+2\pi {r}^{2}$
Solve for $r$
$A=\pi {r}^{2}$
Solve for $s$
$V=\frac{1}{3}{s}^{2}h$
Solve for $L$
$A=2LW+2HW+2LH$

Hang borrowed $7,500 from her parents to pay her tuition. In five years, she paid them $1,500 interest in addition to the $7,500 she borrowed. What was the rate of simple interest?

Révèle la réponse

Write down the given information:	I = $1500
	P = $7500
	r = ?
	t = 5 years
Identify the unknown: let interest rate be represented by r
Write a formula:	$I=Prt$
Substitute in the given information:	$1500=(7500)r(5)$
Solve for r	$1500=37,500r$
	$\frac{1500}{37,500}=r$
	$0.04=r$
	$4\%=r$

The formula for area of a trapezoid is $A=\frac{1}{2}(B+b)h$ where B is the length of the base, b is the length of the other base and h is the height of the trapezoid.

If $B=10\text{cm}$ , $b=8\text{cm}$ and $A=45{\text{cm}}^{2}$ , find the height of the trapezoid.
A married couple together earns $110,000 a year. The wife earns $16,000 less than twice what her husband earns. What does the husband earn?
The label on Audrey’s yogurt said that one serving provided 12 grams of protein, which is 24% of the recommended daily amount. What is the total recommended daily amount of protein?
Recently, the California governor proposed raising community college fees from $36 a unit to $46 a unit. Find the percent change. (Round to the nearest tenth of a percent.)
Sean’s new car loan statement said he would pay $4,866.25 in interest from a simple interest rate of 8.5% over five years. How much did he borrow to buy his new car?
At the campus coffee cart, a medium coffee costs $1.65. MaryAnne brings $2.00 with her when she buys a cup of coffee and leaves the change as a tip. What percent tip does she leave?
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.

Révèle la réponse

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.
\[\begin{array}{lll}x+(x+17) & = & 31 \\ 2x+17 & = & 31\ \text{Simplify and solve}\text{.} \\ 2x & = & 14 \\ x & = & 7 \\ x+17 & = & 7+17 \\ & = & 24\end{array}\]
The two numbers are $7$ and $24.$
Find a linear equation to solve for the following unknown quantities: One number is three more than twice another number. If the sum of the two numbers is $36,$ find the numbers.

Révèle la réponse

11 and 25
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?
Révèle la réponse
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 $B=0.04x+\text{\$}40.$
2. ⓑ
  If the average number of minutes used each month is 1,160, we have the following:
  \[\begin{array}{lll}\text{Company }A & = & 0.05(1,160)+34 \\ & = & 58+34 \\ & = & 92 \\ \text{Company }B & = & 0.04(1,160)+40 \\ & = & 46.4+40 \\ & = & 86.4\end{array}\]
  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:
  \[\begin{array}{lll}\text{Company }A & = & 0.05(420)+34 \\ & = & 21+34 \\ & = & 55 \\ \text{Company }B & = & 0.04(420)+40 \\ & = & 16.8+40 \\ & = & 56.8\end{array}\]
  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 $(x,y)$ 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.
  \[\begin{array}{lll}0.05x+34 & = & 0.04x+40 \\ 0.01x & = & 6 \\ x & = & 600\end{array}\]
  Check the x-value in each equation.
  \[\begin{array}{lll}0.05(600)+34 & = & 64 \\ 0.04(600)+40 & = & 64\end{array}\]
  Therefore, a monthly average of 600 talk-time minutes renders the plans equal. See
Find a linear equation to model this real-world application: It costs ABC electronics company $2.50 per unit to produce a part used in a popular brand of desktop computers. The company has monthly operating expenses of $350 for utilities and $3,300 for salaries. What are the company’s monthly expenses?

Révèle la réponse

$C=2.5x+3,650$
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?

Révèle la réponse

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 , 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.
\[\begin{array}{llll}d & = & r(\frac{1}{2}) & \ \text{To work} \\ d & = & (r-10)(\frac{2}{3}) & \ \text{To home}\end{array}\]
As both equations equal the same distance, we set them equal to each other and solve for r.
\[\begin{array}{lll}r(\frac{1}{2}) & = & (r-10)(\frac{2}{3}) \\ \frac{1}{2}r & = & \frac{2}{3}r-\frac{20}{3} \\ \frac{1}{2}r-\frac{2}{3}r & = & -\frac{20}{3} \\ -\frac{1}{6}r & = & -\frac{20}{3} \\ r & = & -\frac{20}{3}(-6) \\ r & = & 40\end{array}\]
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.
\[\begin{array}{lll}d & = & 40(\frac{1}{2}) \\ & = & 20\end{array}\]
The distance between home and work is 20 mi.
On Saturday morning, it took Jennifer 3.6 h to drive to her mother’s house for the weekend. On Sunday evening, due to heavy traffic, it took Jennifer 4 h to return home. Her speed was 5 mi/h slower on Sunday than on Saturday. What was her speed on Sunday?

Révèle la réponse

45 mi/h
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?

Révèle la réponse

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 as in .
Now we can solve for the width and then calculate the length.
\[\begin{array}{lll}P & = & 2L+2W \\ 54 & = & 2(W+3)+2W \\ 54 & = & 2W+6+2W \\ 54 & = & 4W+6 \\ 48 & = & 4W \\ 12 & = & W \\ (12+3) & = & L \\ 15 & = & L\end{array}\]
The dimensions are $L=15$ ft and $W=12$ ft.
Find the dimensions of a rectangle given that the perimeter is $110$ cm and the length is 1 cm more than twice the width.

Révèle la réponse

$L=37$ cm, $W=18$ cm
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.

Révèle la réponse

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.
\[\begin{array}{lll}P & = & 2L+2W \\ 48 & = & 2(W+6)+2W \\ 48 & = & 2W+12+2W \\ 48 & = & 4W+12 \\ 36 & = & 4W \\ 9 & = & W \\ (9+6) & = & L \\ 15 & = & L\end{array}\]
Now, we find the area given the dimensions of $L=15$ in. and $W=9$ in.
\[\begin{array}{lll}A & = & LW \\ A & = & 15(9) \\ & = & 135\ \text{in}{\text{.}}^{2}\end{array}\]
The area is $135$ in.².
A game room has a perimeter of 70 ft. The length is five more than twice the width. How many ft² of new carpeting should be ordered?

Révèle la réponse

250 ft²
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.

Révèle la réponse

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.
\[\begin{array}{lll}V & = & LWH \\ 1,600 & = & (2W)W(8) \\ 1,600 & = & 16{W}^{2} \\ 100 & = & {W}^{2} \\ 10 & = & W\end{array}\]
The dimensions are $L=20$ in., $W=10$ in., and $H=8$ in.
To set up a model linear equation to fit real-world applications, what should always be the first step?

Révèle la réponse

Answers may vary. Possible answers: We should define in words what our variable is representing. We should declare the variable. A heading.
Use your own words to describe this equation where n is a number: $5(n+3)=2n$
If the total amount of money you had to invest was $2,000 and you deposit $x$ amount in one investment, how can you represent the remaining amount?

Révèle la réponse

$2,000-x$
If a carpenter sawed a 10-ft board into two sections and one section was $n$ ft long, how long would the other section be in terms of $n$ ?
If Bill was traveling $v$ mi/h, how would you represent Daemon’s speed if he was traveling 10 mi/h faster?

Révèle la réponse

$v+10$
Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
Beth and Ann are joking that their combined ages equal Sam’s age. If Beth is twice Ann’s age and Sam is 69 yr old, what are Beth and Ann’s ages?

Révèle la réponse

Ann: $23;$ Beth: $46$
Ruden originally filled out 8 more applications than Hanh. Then each boy filled out 3 additional applications, bringing the total to 28. How many applications did each boy originally fill out?
Find the model of the total cost of Company A’s plan, using $m$ for the minutes.

Révèle la réponse

$20+0.05m$
Find the model of the total cost of Company B’s plan, using $m$ for the minutes.
Find out how many minutes of calling would make the two plans equal.

Révèle la réponse

300 min

Symbols used here

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Models and Applications

Set up a linear equation to solve a real-world application.
Use a formula to solve a real-world application.
Solve a formula for a specified variable (IA 2.3.1)
Use a problem-solving strategy for word problems (IA 2.2.1)
Refer to the appropriate formula and identify the variable you are solving for. Treat the other variable terms as if they were numbers.
Isolate the variable you are solving for using the multiplication/division property of equality.
Identify known quantities.
If there is more than one unknown quantity, find a way to write the second unknown in terms of the first.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Essayez votre propre

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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	\(d\)	\(r\)	\(t\)
To Work	\(d\)	\(r\)	\(\frac{1}{2}\)
To Home	\(d\)	\(r-10\)	\(\frac{2}{3}\)

	\(d\)	\(r\)	\(t\)
To Work	\(d\)	\(r\)	\(\frac{1}{2}\)
To Home	\(d\)	\(r-10\)	\(\frac{2}{3}\)

\(P=2l+2w\)	Since we are solving for l we isolate the l term
\(P-2w=2l+2w-2w\)	Subtract 2w from both sides.
\(P-2w=2l\)	Combine like terms
\(\frac{P-2w}{2}=\frac{2l}{2}\)	Divide by 2 to isolate l.
\(\frac{P-2w}{2}=l\)	Simplify