Intermediate Algebra 2e — Original English

Solve Applications with Systems of Equations

Solve Direct Translation Applications

Systems of linear equations are very useful for solving applications. Some people find setting up word problems with two variables easier than setting them up with just one variable. To solve an application, we’ll first translate the words into a system of linear equations. Then we will decide the most convenient method to use, and then solve the system.

We solved number problems with one variable earlier. Let’s see how differently it works using two variables.

The sum of two numbers is zero. One number is nine less than the other. Find the numbers.

Solution
Step 1. Read the problem.
Step 2. Identify what we are looking for. We are looking for two numbers.
Step 3. Name what we are looking for. Let n= the first number.
m= the second number
Step 4. Translate into a system of equations. The sum of two numbers is zero.
A clear, concise image displays the algebraic equation 'n + m = 0' written in black characters on a plain white background, presenting a fundamental mathematical concept.
One number is nine less than the other.
A mathematical equation is displayed on a white background, showing 'n = m - 9' in black text, representing a relationship between three variables.
The system is: A system of two linear equations is presented, enclosed by a large left curly brace. The first equation is 'n + m = 0', and the second equation is 'n = m - 9'.
Step 5. Solve the system of
equations. We will use substitution
since the second equation is solved
for n.
Substitute m − 9 for n in the first equation. Two mathematical equations are displayed: n = m - 9 is circled, and an arrow points from it to the second equation, n + m = 0.
Solve for m. A mathematical equation, m - 9 + m = 0, is displayed on a white background. The 'm' and '-9' are in a reddish hue, while the second 'm' and '= 0' are in gray.
A mathematical equation is displayed, reading '2m - 9 = 0'. The text is rendered in a dark gray font against a clean white background.
A simple algebraic equation is displayed, showing '2m = 9' in bold, grey text against a plain white background. This equation involves a variable 'm' being multiplied by 2 and equaling 9.
Substitute m=92 into the second equation
and then solve for n.
A mathematical diagram displays two equations: m = 9/2 and n = m - 9. The value 9/2 for 'm' is circled in red, with an arrow indicating its substitution into the second equation to solve for 'n'.
The image displays two steps of an algebraic equation. The first line is m = 9/2 - 9. The second line converts the integer 9 into a fraction with a common denominator, showing m = 9/2 - 18/2.
A mathematical equation is displayed on a white background, showing 'n = -9/2'.
Step 6. Check the answer in the problem. Do these numbers make sense in
the problem? We will leave this to
you!
Step 7. Answer the question. The numbers are 92 and 92.

Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $25,000 plus $15 for each training session. Option B would pay her $10,000+$40 for each training session. How many training sessions would make the salary options equal?

Solution
Step 1. Read the problem.
Step 2. Identify what we are looking for. We are looking for the number of
training sessions that would make
the pay equal.
Step 3. Name what we are looking for. Let s= Heather’s salary.
n= the number of training sessions
Step 4. Translate into a system of equations. Option A would pay her $25,000
plus $15 for each training
session.
A mathematical equation is displayed on a white background: s = 25,000 + 15n.
Option B would pay her $10,000
+ $40 for each training session.
A mathematical equation is displayed against a white background, reading 's = 10,000 + 40n'. The equation uses a sans-serif font and appears to be a formula for calculating a value 's' based on a base number 10,000 and a variable 'n' multiplied by 40.
The system is shown. A system of two linear equations is displayed. The first equation is s = 25,000 + 15n, and the second equation is s = 10,000 + 40n, where 's' and 'n' are variables.
Step 5. Solve the system of equations.
We will use substitution.
Two linear equations are shown: s = 25,000 + 15n (circled in red) and s = 10,000 + 40n. An arrow points from the first to the second, possibly indicating a transformation or choice between the two.
Substitute 25,000 +15n for s in the second
equation.
A mathematical equation reads '25,000 + 15n = 10,000 + 40n' with the numbers 25,000 and 15n in red, and the rest of the equation in gray.
Solve for n. A step-by-step algebraic solution for 'n' is presented. The initial equation, 25,000 = 10,000 + 25n, is simplified to 15,000 = 25n, and finally solved to show that n = 600.
Step 6. Check the answer. Are 600 training sessions a year reasonable?
Are the two options equal when n = 600?
Step 7. Answer the question. The salary options would be equal for 600 training
sessions.

As you solve each application, remember to analyze which method of solving the system of equations would be most convenient.

Translate to a system of equations and then solve:

When Jenna spent 10 minutes on the elliptical trainer and then did circuit training for 20 minutes, her fitness app says she burned 278 calories. When she spent 20 minutes on the elliptical trainer and 30 minutes circuit training she burned 473 calories. How many calories does she burn for each minute on the elliptical trainer? How many calories for each minute of circuit training?

Solution
Step 1. Read the problem.
Step 2. Identify what we are looking for. We are looking for the number of
calories burned each minute on the
elliptical trainer and each minute of
circuit training.
Step 3. Name what we are looking for. Let e= number of calories burned per
minute on the elliptical trainer.
c= number of calories burned per
minute while circuit training
Step 4. Translate into a system of equations. 10 minutes on the elliptical and circuit
training for 20 minutes, burned
278 calories
A mathematical equation is displayed on a white background, which reads '10e + 20c = 278' in a dark gray sans-serif font.
20 minutes on the elliptical and
30 minutes of circuit training burned
473 calories
A mathematical equation is displayed on a white background, reading 20e + 30c = 473.
The system is: A system of two linear equations is displayed. The first equation is 10e + 20c = 278, and the second equation is 20e + 30c = 473. A curly brace groups the two equations together.
Step 5. Solve the system of equations.
Multiply the first equation by −2 to get
opposite coefficients of e.
The image shows a system of two linear equations. The first equation is given as -2(10e + 20c) = -2(278), indicating that the original equation has been multiplied by -2 on both sides. The second equation in the system is 20e + 30c = 473.
Simplify and add the equations.
Solve for c.
A system of two linear equations, -20e - 40c = -556 and 20e + 30c = 473, is solved using the elimination method, resulting in -10c = -83, which simplifies to c = 8.3.
Substitute c = 8.3 into one of the
original equations to solve for e.
A step-by-step solution to an algebraic equation, starting with 10e + 20c = 278, substituting c with 8.3, and solving for e, resulting in e = 11.2.
Step 6. Check the answer in the problem. Check the math on your own.
Two mathematical expressions are displayed with question marks questioning their equality: 10(11.2) + 20(8.3) ?= 278 and 20(11.2) + 30(8.3) ?= 473. A brace indicates they are grouped together.
Step 7. Answer the question. Jenna burns 8.3 calories per minute
circuit training and 11.2 calories per
minute while on the elliptical trainer.

Solve Geometry Applications

We will now solve geometry applications using systems of linear equations. We will need to add complementary angles and supplementary angles to our list some properties of angles.

The measures of two complementary angles add to 90 degrees. The measures of two supplementary angles add to 180 degrees.

If two angles are complementary, we say that one angle is the complement of the other.

If two angles are supplementary, we say that one angle is the supplement of the other.

Translate to a system of equations and then solve.

The difference of two complementary angles is 26 degrees. Find the measures of the angles.

Solution
Step-by-step solution for finding two complementary angles given their difference, using a system of linear equations.
Step 1. Read the problem.
Step 2. Identify what we are looking for. We are looking for the measure of each angle.
Step 3. Name what we are looking for. Letx=the measure of the first angle.  y=the measure of the second angle
Step 4. Translate into a system of equations. The angles are complementary.
x+y=90
The difference of the two angles is 26 degrees.
xy=26
The system is shown. {x+y=90xy=26
Step 5. Solve the system of equations by elimination. {x+y=90xy=26___________ 2x=116
Substitute x=58 into the first equation. x=58x+y=9058+y=90y=32
Step 6. Check the answer in the problem.
58+32=905832=26
Step 7. Answer the question. The angle measures are 58 and 32 degrees.

In the next example, we remember that the measures of supplementary angles add to 180.

Translate to a system of equations and then solve:

Two angles are supplementary. The measure of the larger angle is twelve degrees less than five times the measure of the smaller angle. Find the measures of both angles.

Solution
Step 1. Read the problem.
Step 2. Identify what we are looking for. We are looking for measure of each
angle.
Step 3. Name what we are looking for. Let x= the measure of the first angle.
y= the measure of the second angle
Step 4. Translate into a system of equations. The angles are supplementary.
The equation 'x + y = 180' is displayed in black text on a plain white background.
The larger angle is twelve less than five
times the smaller angle.
A mathematical equation is displayed on a white background, which reads 'y = 5x - 12'.
The system is shown:
Step 5. Solve the system of equations substitution.
A system of equations: x+y=180 and y=5x-12. An arrow shows substituting '5x-12' for 'y' into the equation x+y=180, illustrating the substitution method for solving linear equations.
Substitute 5x − 12 for y in the first equation.
Solve for x.
Two lines of an algebraic equation are shown. The first line is x + 5x - 12 = 180. The second line, which simplifies the first, reads 6x - 12 = 180.


Substitute 32 for x in the second
equation, then solve for y.
A mathematical problem showing how to solve for 'x' in the equation 6x = 192, yielding x = 32. An arrow indicates this value should be substituted into the equation y = 5x - 12 to find 'y'.
A step-by-step mathematical calculation is shown where y = 5 * 32 - 12 simplifies to y = 160 - 12, resulting in the final answer of y = 148.
Step 6. Check the answer in the problem. Two mathematical equations with checkmarks indicating correctness are shown: 32 + 148 = 180 and 5 * 32 - 12 = 148.
Step 7. Answer the question. The angle measures are 148 and 32 degrees.

Recall that the angles of a triangle add up to 180 degrees. A right triangle has one angle that is 90 degrees. What does that tell us about the other two angles? In the next example we will be finding the measures of the other two angles.

The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles.

Solution

We will draw and label a figure.

Step 1. Read the problem. A right-angled triangle with a vertical side labeled 'a' and an acute angle labeled 'b'. A square symbol indicates the 90-degree angle.
Step 2. Identify what you are looking for. We are looking for the measures of the angles.
Step 3. Name what we are looking for. Let a= the measure of the first angle.
b= the measure of the second angle
Step 4. Translate into a system of equations. The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle.
A mathematical equation is displayed on a white background, which reads 'a = 3b + 10' in black text.
The sum of the measures of the angles of a triangle is 180.
The image displays the algebraic equation a + b + 90 = 180, suggesting a problem involving angles or geometric figures where variables a and b, along with a 90-degree angle, sum up to 180 degrees.

The system is shown. A system of two linear equations is presented. The first equation is a = 3b + 10. The second equation is a + b + 90 = 180.
Step 5. Solve the system of equations. We will use substitution since the first equation is solved for a. An algebraic problem with two equations: a = 3b + 10 and a + b + 90 = 180. The first equation, circled, points to the second, indicating a substitution scenario.
Substitute 3b+10 for a in the second equation. A mathematical equation is displayed: (3b + 10) + b + 90 = 180, featuring variables and numerical values.
Solve for b. A mathematical solution showing the steps to find the value of 'b' from the equation 4b + 100 = 180, resulting in b = 20, and then showing the next equation a = 3b + 10.
Substitute b=20 into the first equation and then solve for a. A mathematical equation is displayed on a white background: a = 3 * 20 + 10, which resolves to a = 70. The number '20' in the first line is highlighted in red.
Step 6. Check the answer in the problem. We will leave this to you!
Step 7. Answer the question. The measures of the small angles are 20 and 70 degrees.

Often it is helpful when solving geometry applications to draw a picture to visualize the situation.

Translate to a system of equations and then solve:

Randall has 125 feet of fencing to enclose the part of his backyard adjacent to his house. He will only need to fence around three sides, because the fourth side will be the wall of the house. He wants the length of the fenced yard (parallel to the house wall) to be 5 feet more than four times as long as the width. Find the length and the width.

Solution
Step 1. Read the problem.
Step 2. Identify what you are looking for. We are looking for the length and width.
A brown paper bag with a textured handle is depicted against a plain white background in a simple, top-down perspective illustration.
Step 3. Name what we are looking for. Let L= the length of the fenced yard.
W= the width of the fenced yard
Step 4. Translate into a system of equations. One length and two widths equal 125.
A mathematical equation is displayed on a white background: L + 2W = 125.
The length will be 5 feet more than
four times the width.
The image shows the mathematical equation L = 4W + 5, which defines the variable L in terms of the variable W.
The system is shown.

Step 5. Solve The system of equations
by substitution.
A system of two linear equations is shown with the second equation, L = 4W + 5, prepared for substitution into the first equation, L + 2W = 125. The expression '4W + 5' is highlighted and indicated to replace 'L'.
Substitute L = 4W + 5 into the first
equation, then solve for W.
A mathematical equation is displayed on a white background: 4W + 5 + 2W = 125. The numbers '4' and '5' are in red, while the rest of the equation is in gray.
Algebraic equation 6W + 5 = 125 is solved step-by-step, resulting in W = 20, which is circled.
Substitute 20 for W in the second
equation, then solve for L.
An algebraic calculation shows L = 4W + 5. Substituting W with 20, the equation becomes L = 4 * 20 + 5, which simplifies to L = 80 + 5, resulting in L = 85.
Step 6. Check the answer in the
problem.
Arithmetic problem: 20 + 85 + 20 = 125. Below, 85 is shown as 4 * 20 + 5, followed by a checkmark confirming 85 = 85. This illustrates a numerical calculation and its verification.
Step 7. Answer the equation. The length is 85 feet and the width is 20 feet.

Solve uniform motion applications

We used a table to organize the information in uniform motion problems when we introduced them earlier. We’ll continue using the table here. The basic equation was D=rt where D is the distance traveled, r is the rate, and t is the time.

Our first example of a uniform motion application will be for a situation similar to some we have already seen, but now we can use two variables and two equations.

Translate to a system of equations and then solve:

Joni left St. Louis on the interstate, driving west towards Denver at a speed of 65 miles per hour. Half an hour later, Kelly left St. Louis on the same route as Joni, driving 78 miles per hour. How long will it take Kelly to catch up to Joni?

Solution

A diagram is useful in helping us visualize the situation.


A diagram illustrating a word problem with two people, Joni and Kelly, traveling from St. Louis towards Denver. Joni travels at 65 mph, while Kelly travels at 78 mph, starting 0.5 hours later.

Identify and name what we are looking for. A chart will help us organize the data. We know the rates of both Joni and Kelly, and so we enter them in the chart. We are looking for the length of time Kelly, k, and Joni, j, will each drive.


A table titled 'Rate * Time = Distance' shows calculations for Joni and Kelly. Joni has a rate of 65, time 'j', and distance '65j'. Kelly has a rate of 78, time 'k', and distance '78k'.

Since D=r·t we can fill in the Distance column.

Translate into a system of equations.

To make the system of equations, we must recognize that Kelly and Joni will drive the same distance. So,

65j=78k

Also, since Kelly left later, her time will be 12 hour less than Joni’s time. So,

Step-by-step solution for a word problem using a system of equations. Calculates when Kelly catches up to Joni, considering their speeds and travel times.
k=j12
Now we have the system. {k=j1265j=78k
Solve the system of equations by substitution.
Substitute k=j12 into the second equation, then solve for j.
65j=78k65j=78(j12)65j=78j39−13j=−39j=3
To find Kelly’s time, substitute j=3 into the first equation, then solve for k. k=j12
k=312
k=52ork=212
Check the answer in the problem.
Joni 3hours(65mph)=195miles
Kelly 212hours(78mph)=195miles
Yes, they will have traveled the same distance when they meet.
Answer the question. Kelly will catch up to Joni in 212 hours. By then, Joni will have traveled 3 hours.

Many real-world applications of uniform motion arise because of the effects of currents—of water or air—on the actual speed of a vehicle. Cross-country airplane flights in the United States generally take longer going west than going east because of the prevailing wind currents.

Let’s take a look at a boat travelling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.

The images below show how a river current affects the speed at which a boat is actually travelling. We’ll call the speed of the boat in still water b and the speed of the river current c.

The boat is going downstream, in the same direction as the river current. The current helps push the boat, so the boat’s actual speed is faster than its speed in still water. The actual speed at which the boat is moving is b+c.

Figure shows a boat and two horizontal arrows, both pointing left. The one to the left of the boat is b and the one to the right is c.

Now, the boat is going upstream, opposite to the river current. The current is going against the boat, so the boat’s actual speed is slower than its speed in still water. The actual speed of the boat is bc.

Figure shows a boat and two horizontal arrows to its left. One, labeled b, points left and the other, labeled c, points right.

We’ll put some numbers to this situation in the next example.

Translate to a system of equations and then solve.

A river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.

Solution
Read the problem. This is a uniform motion problem and a
picture will help us visualize the situation.
A diagram illustrates a 60-mile river journey with a current 'c'. It takes 4 hours to travel downstream (with the current) and 5 hours to travel upstream (against the current).
Identify what we are looking for. We are looking for the speed of the ship
in still water and the speed of the current.
Name what we are looking for. Let s= the rate of the ship in still water.
c= the rate of the current
A chart will help us organize the information.
The ship goes downstream and then upstream.
Going downstream, the current helps the
ship and so the ship's actual rate is s + c.
Going upstream, the current slows the ship
and so the actual rate is sc.
A table showing rate, time, and distance for downstream and upstream travel. Downstream: rate s+c, time 4, distance 60. Upstream: rate s-c, time 5, distance 60.
Downstream it takes 4 hours.
Upstream it takes 5 hours.
Each way the distance is 60 miles.
Translate into a system of equations.
Since rate times time is distance, we can
write the system of equations.
A system of two linear equations is presented. The first equation is 4(s + c) = 60, and the second equation is 5(s - c) = 60. The equations are enclosed within a brace on the left.
Solve the system of equations.
Distribute to put both equations in standard
form, then solve by elimination.
A system of two linear equations is presented. The first equation is 4s + 4c = 60, and the second equation is 5s - 5c = 60. A curly brace indicates they are part of a system.
Multiply the top equation by 5 and the
bottom equation by 4.
Add the equations, then solve for s.
A step-by-step solution demonstrating the elimination method for a system of linear equations: 20s + 20c = 300 and 20s - 20c = 240, yielding s = 13.5, with an arrow pointing to 4(s + c) = 60.
Substitute s = 13.5 into of the original
equations.
An algebraic equation showing the steps to solve for 'c'. The equation 4(13.5 + c) = 60 is solved, leading to c = 1.5 through distribution, subtraction, and division.
Check the answer in the problem.
The downstream rate would be
13.5+1.5=15 mph.
In 4 hours the ship would travel
  15·4=60 miles.
The upstream rate would be
13.51.5=12 mph.
In 5 hours the ship would travel
  12·5=60 miles.
Answer the question. The rate of the ship is 13.5 mph and
the rate of the current is 1.5 mph.

Wind currents affect airplane speeds in the same way as water currents affect boat speeds. We’ll see this in the next example. A wind current in the same direction as the plane is flying is called a tailwind. A wind current blowing against the direction of the plane is called a headwind.

Translate to a system of equations and then solve:

A private jet can fly 1,095 miles in three hours with a tailwind but only 987 miles in three hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Solution
Read the problem. This is a uniform motion problem and a
picture will help us visualize.
This diagram compares distances traveled over 3 hours with and without wind assistance: 1,095 miles with a tailwind vs. 987 miles with a headwind, demonstrating wind's effect on travel.
Identify what we are looking for. We are looking for the speed of the jet
in still air and the speed of the wind.
Name what we are looking for. Let j= the speed of the jet in still air.
w= the speed of the wind.
A chart will help us organize the information.
The jet makes two trips—one in a tailwind
and one in a headwind.
In a tailwind, the wind helps the jet and so
the rate is j + w.
In a headwind, the wind slows the jet and
so the rate is jw.
A table illustrating Rate, Time, and Distance for travel with tailwind and headwind. Tailwind values are (j+w), 3, and 1095. Headwind values are (j-w), 3, and 987, respectively.
Each trip takes 3 hours.
In a tailwind the jet flies 1,095 miles.
In a headwind the jet flies 987 miles.
Translate into a system of equations.
Since rate times time is distance, we get the
system of equations.
A system of two linear equations is presented. The first equation is 3 multiplied by the sum of j and w, which equals 1095. The second equation is 3 multiplied by the difference of j and w, which equals 987.
Solve the system of equations.
Distribute, then solve by elimination.
Add, and solve for j.
A system of linear equations, {3j + 3w = 1095, 3j - 3w = 987}, is solved by adding the equations to eliminate 'w', resulting in 6j = 2082 and j = 347.
Substitute j = 347 into one of the original
equations, then solve for w.
A step-by-step algebraic solution showing how to solve for 'w' in the equation 3(347 + w) = 1095, resulting in w = 18. The steps include distribution, subtraction, and division to isolate the variable.
Check the answer in the problem.
With the tailwind, the actual rate of the
jet would be
347+18=365 mph.
In 3 hours the jet would travel
  365·3=1,095 miles
Going into the headwind, the jet’s actual
rate would be
34718=329 mph.
In 3 hours the jet would travel
  329·3=987 miles.
Answer the question. The rate of the jet is 347 mph and the
rate of the wind is 18 mph.

Key Concepts

  • How To Solve Applications with Systems of Equations
    1. Read the problem. Make sure all the words and ideas are understood.
    2. Identify what we are looking for.
    3. Name what we are looking for. Choose variables to represent those quantities.
    4. Translate into a system of equations.
    5. Solve the system of equations using good algebra techniques.
    6. Check the answer in the problem and make sure it makes sense.
    7. Answer the question with a complete sentence.

Practice Makes Perfect

Direct Translation Applications

In the following exercises, translate to a system of equations and solve.

The sum of two number is 15. One number is 3 less than the other. Find the numbers.

The sum of two number is 30. One number is 4 less than the other. Find the numbers.

Solution

13 and 17

The sum of two number is −16. One number is 20 less than the other. Find the numbers.

The sum of two number is −26. One number is 12 less than the other. Find the numbers.

Solution

−7 and −19

The sum of two numbers is 65. Their difference is 25. Find the numbers.

The sum of two numbers is 37. Their difference is 9. Find the numbers.

Solution

14 and 23

The sum of two numbers is −27. Their difference is −59. Find the numbers.

The sum of two numbers is −45. Their difference is −89. Find the numbers.

Solution

22 and −67

Maxim has been offered positions by two car companies. The first company pays a salary of $10,000 plus a commission of $1000 for each car sold. The second pays a salary of $20,000 plus a commission of $500 for each car sold. How many cars would need to be sold to make the total pay the same?

Jackie has been offered positions by two cable companies. The first company pays a salary of $14,000 plus a commission of $100 for each cable package sold. The second pays a salary of $20,000 plus a commission of $25 for each cable package sold. How many cable packages would need to be sold to make the total pay the same?

Solution

Eighty cable packages would need to be sold to make the total pay the same.

Amara currently sells televisions for company A at a salary of $17,000 plus a $100 commission for each television she sells. Company B offers her a position with a salary of $29,000 plus a $20 commission for each television she sells. How many televisions would Amara need to sell for the options to be equal?

Mitchell currently sells stoves for company A at a salary of $12,000 plus a $150 commission for each stove he sells. Company B offers him a position with a salary of $24,000 plus a $50 commission for each stove he sells. How many stoves would Mitchell need to sell for the options to be equal?

Solution

Mitchell would need to sell 120 stoves for the companies to be equal.

Two containers of gasoline hold a total of fifty gallons. The big container can hold ten gallons less than twice the small container. How many gallons does each container hold?

June needs 48 gallons of punch for a party and has two different coolers to carry it in. The bigger cooler is five times as large as the smaller cooler. How many gallons can each cooler hold?

Solution

8 and 40 gallons

Shelly spent 10 minutes jogging and 20 minutes cycling and burned 300 calories. The next day, Shelly swapped times, doing 20 minutes of jogging and 10 minutes of cycling and burned the same number of calories. How many calories were burned for each minute of jogging and how many for each minute of cycling?

Drew burned 1800 calories Friday playing one hour of basketball and canoeing for two hours. Saturday he spent two hours playing basketball and three hours canoeing and burned 3200 calories. How many calories did he burn per hour when playing basketball? How many calories did he burn per hour when canoeing?

Solution

1000 calories playing basketball and 400 calories canoeing

Troy and Lisa were shopping for school supplies. Each purchased different quantities of the same notebook and thumb drive. Troy bought four notebooks and five thumb drives for $116. Lisa bought two notebooks and three thumb dives for $68. Find the cost of each notebook and each thumb drive.

Nancy bought seven pounds of oranges and three pounds of bananas for $17. Her husband later bought three pounds of oranges and six pounds of bananas for $12. What was the cost per pound of the oranges and the bananas?

Solution

Oranges cost $2 per pound and bananas cost $1 per pound

Andrea is buying some new shirts and sweaters. She is able to buy 3 shirts and 2 sweaters for $114 or she is able to buy 2 shirts and 4 sweaters for $164. How much does a shirt cost? How much does a sweater cost?

Peter is buying office supplies. He is able to buy 3 packages of paper and 4 staplers for $40 or he is able to buy 5 packages of paper and 6 staplers for $62. How much does a package of paper cost? How much does a stapler cost?

Solution

Package of paper $4, stapler $7

The total amount of sodium in 2 hot dogs and 3 cups of cottage cheese is 4720 mg. The total amount of sodium in 5 hot dogs and 2 cups of cottage cheese is 6300 mg. How much sodium is in a hot dog? How much sodium is in a cup of cottage cheese?

The total number of calories in 2 hot dogs and 3 cups of cottage cheese is 960 calories. The total number of calories in 5 hot dogs and 2 cups of cottage cheese is 1190 calories. How many calories are in a hot dog? How many calories are in a cup of cottage cheese?

Solution

Hot dog 150 calories, cup of cottage cheese 220 calories

Molly is making strawberry infused water. For each ounce of strawberry juice, she uses three times as many ounces of water as juice. How many ounces of strawberry juice and how many ounces of water does she need to make 64 ounces of strawberry infused water?

Owen is making lemonade from concentrate. The number of quarts of water he needs is 4 times the number of quarts of concentrate. How many quarts of water and how many quarts of concentrate does Owen need to make 100 quarts of lemonade?

Solution

Owen will need 80 quarts of water and 20 quarts of concentrate to make 100 quarts of lemonade.

Solve Geometry Applications

In the following exercises, translate to a system of equations and solve.

The difference of two complementary angles is 55 degrees. Find the measures of the angles.

The difference of two complementary angles is 17 degrees. Find the measures of the angles.

Solution

53.5 degrees and 36.5 degrees

Two angles are complementary. The measure of the larger angle is twelve less than twice the measure of the smaller angle. Find the measures of both angles.

Two angles are complementary. The measure of the larger angle is ten more than four times the measure of the smaller angle. Find the measures of both angles.

Solution

16 degrees and 74 degrees

The difference of two supplementary angles is 8 degrees. Find the measures of the angles.

The difference of two supplementary angles is 88 degrees. Find the measures of the angles.

Solution

134 degrees and 46 degrees

Two angles are supplementary. The measure of the larger angle is four more than three times the measure of the smaller angle. Find the measures of both angles.

Two angles are supplementary. The measure of the larger angle is five less than four times the measure of the smaller angle. Find the measures of both angles.

Solution

37 degrees and 143 degrees

The measure of one of the small angles of a right triangle is 14 more than 3 times the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 26 more than 3 times the measure of the other small angle. Find the measure of both angles.

Solution

16 degrees and 74 degrees

The measure of one of the small angles of a right triangle is 15 less than twice the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 45 less than twice the measure of the other small angle. Find the measure of both angles.

Solution

45 degrees and 45 degrees

Wayne is hanging a string of lights 45 feet long around the three sides of his patio, which is adjacent to his house. The length of his patio, the side along the house, is five feet longer than twice its width. Find the length and width of the patio.

Darrin is hanging 200 feet of Christmas garland on the three sides of fencing that enclose his front yard. The length is five feet less than three times the width. Find the length and width of the fencing.

Solution

Width is 41 feet and length is 118 feet.

A frame around a family portrait has a perimeter of 90 inches. The length is fifteen less than twice the width. Find the length and width of the frame.

The perimeter of a toddler play area is 100 feet. The length is ten more than three times the width. Find the length and width of the play area.

Solution

Width is 10 feet and length is 40 feet.

Solve Uniform Motion Applications

In the following exercises, translate to a system of equations and solve.

Sarah left Minneapolis heading east on the interstate at a speed of 60 mph. Her sister followed her on the same route, leaving two hours later and driving at a rate of 70 mph. How long will it take for Sarah’s sister to catch up to Sarah?

College roommates John and David were driving home to the same town for the holidays. John drove 55 mph, and David, who left an hour later, drove 60 mph. How long will it take for David to catch up to John?

Solution

12 hours

At the end of spring break, Lucy left the beach and drove back towards home, driving at a rate of 40 mph. Lucy’s friend left the beach for home 30 minutes (half an hour) later, and drove 50 mph. How long did it take Lucy’s friend to catch up to Lucy?

Felecia left her home to visit her daughter driving 45 mph. Her husband waited for the dog sitter to arrive and left home twenty minutes (1/3 hour) later. He drove 55 mph to catch up to Felecia. How long before he reaches her?

Solution

1.83 hour

The Jones family took a 12-mile canoe ride down the Indian River in two hours. After lunch, the return trip back up the river took three hours. Find the rate of the canoe in still water and the rate of the current.

A motor boat travels 60 miles down a river in three hours but takes five hours to return upstream. Find the rate of the boat in still water and the rate of the current.

Solution

Boat rate is 16 mph and current rate is 4 mph.

A motor boat traveled 18 miles down a river in two hours but going back upstream, it took 4.5 hours due to the current. Find the rate of the motor boat in still water and the rate of the current.

A river cruise boat sailed 80 miles down the Mississippi River for four hours. It took five hours to return. Find the rate of the cruise boat in still water and the rate of the current.

Solution

Boat rate is 18 mph and current rate is 2 mph.

A small jet can fly 1072 miles in 4 hours with a tailwind but only 848 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

A small jet can fly 1435 miles in 5 hours with a tailwind but only 1,215 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Solution

Jet rate is 265 mph and wind speed is 22 mph.

A commercial jet can fly 868 miles in 2 hours with a tailwind but only 792 miles in 2 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

A commercial jet can fly 1,320 miles in 3 hours with a tailwind but only 1170 miles in 3 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Solution

Jet rate is 415 mph and wind speed is 25 mph.

Writing Exercises

Write an application problem similar to Example 1. Then translate to a system of equations and solve it.

Write a uniform motion problem similar to Example 2 that relates to where you live with your friends or family members. Then translate to a system of equations and solve it.

Solution

Answers will vary.

Self Check

After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns, 3 rows and a header row. The header row labels each column: I can, confidently, with some help and no, I don’t get it. The first column has the following statements: solve direct translation applications, solve geometry applications, solve uniform motion applications. The remaining columns are empty.

After reviewing this checklist, what will you do to become confident for all objectives?

complementary angles
Two angles are complementary if the sum of the measures of their angles is 90 degrees.
supplementary angles
Two angles are supplementary if the sum of the measures of their angles is 180 degrees.