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.

Example

Try it.

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.
	One number is nine less than the other.
The system is:
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.
Solve for m.
Substitute \(m=\frac{9}{2}\) into the second equation and then solve for n.
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 \(\frac{9}{2}\) and \(-\frac{9}{2}.\)

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

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.

Example

Try it.

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 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.	\(\begin{array}{l}\text{Let}\ x=\ \text{the measure of the first angle.} \\ y=\ \text{the measure of the second angle}\end{array}\)
Step 4. Translate into a system of equations.	The angles are complementary.
	\(\ x+y=90\)
	The difference of the two angles is 26 degrees.
	\(\ x-y=26\)
The system is shown.	\(\ \{\begin{array}{l}x+y=90 \\ x-y=26\end{array}\)
Step 5. Solve the system of equations by elimination.	\(\ \begin{array}{l}\underset{___________}{\{\begin{array}{l}x+y=90 \\ x-y=26\end{array}} \\ 2x\ =116\end{array}\)
Substitute \(x=58\) into the first equation.	\(\ \begin{array}{l}x=58 \\ \\ \\ x+y=90 \\ 58+y=90 \\ y=32\end{array}\)
Step 6. Check the answer in the problem.
\(\ \begin{array}{l}58+32=90✓ \\ 58-32=26✓\end{array}\)
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.

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

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

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.

Example

Try it.

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.

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.

Since \(D=r\cdot t\) 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 \(\frac{1}{2}\) hour less than Joni’s time. So,

	\(\ k=j-\frac{1}{2}\)
Now we have the system.	\(\ \{\begin{array}{l}\ k=j-\frac{1}{2} \\ 65j=78k\end{array}\)
Solve the system of equations by substitution.
Substitute \(k=j-\frac{1}{2}\) into the second equation, then solve for \(j\).
	\(\ \begin{array}{lll}65j & = & 78k \\ 65j & = & 78(j-\frac{1}{2}) \\ 65j & = & 78j-39 \\ -13j & = & -39 \\ j & = & 3\end{array}\)
To find Kelly’s time, substitute \(j=3\) into the first equation, then solve for \(k\).	\(\ k=j-\frac{1}{2}\)
	\(\ k=3-\frac{1}{2}\)
	\(\ k=\frac{5}{2}\ \text{or}\ k=2\frac{1}{2}\)
Check the answer in the problem.
Joni \(\ 3\ \text{hours}\ (65\ \text{mph})\ =195\ \text{miles}\)
Kelly \(\ 2\frac{1}{2}\ \text{hours}\ (78\ \text{mph})=195\ \text{miles}\)
Yes, they will have traveled the same distance when they meet.
Answer the question.	Kelly will catch up to Joni in \(2\frac{1}{2}\) hours. By then, Joni will have traveled 3 hours.

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

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.

Solve Applications with Systems of Equations

Direct Translation Applications

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

Try it.

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

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

Solution

13 and 17

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

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The sum of two number is \(-26.\) One number is 12 less than the other. Find the numbers.

Solution

\(-7\) and \(-19\)

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

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

Solution

14 and 23

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The sum of two numbers is \(-27.\) Their difference is \(-59.\) Find the numbers.

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The sum of two numbers is \(-45.\) Their difference is \(-89.\) Find the numbers.

Solution

22 and \(-67\)

Try it.

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?

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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.

Try it.

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?

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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.

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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?

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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

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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?

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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

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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.

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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

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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?

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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

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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?

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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

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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?

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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.

Try it.

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

Try it.

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

Solution

\(53.5\) degrees and \(36.5\) degrees

Try it.

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.

Try it.

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

Try it.

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

Try it.

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

Solution

134 degrees and 46 degrees

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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.

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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

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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.

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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

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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.

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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

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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.

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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.

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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.

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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.

Try it.

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?

Try it.

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

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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?

Try it.

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

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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.

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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.

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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.

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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.

Try it.

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.

Try it.

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.

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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.

Try it.

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.

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

Translate to a System of Equations

Many of the problems we solved in earlier applications related two quantities. Here are two of the examples from the chapter on Math Models.

• The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.
• 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?

In that chapter we translated each situation into one equation using only one variable. Sometimes it was a bit of a challenge figuring out how to name the two quantities, wasn’t it?

Let’s see how we can translate these two problems into a system of equations with two variables. We’ll focus on Steps 1 through 4 of our Problem Solving Strategy.

How to Translate to a System of Equations

Try it.

Translate to a system of equations:

The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.

Solution

We’ll do another example where we stop after we write the system of equations.

Example

Try it.

Translate to a system of equations:

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?

Solution

We are looking for the amount that the husband and wife each earn.	Let \(h=\) the amount the husband earns. \(w=\) the amount the wife earns.
Translate.	A married couple together earns $110,000.
	\(w+h=110,000\)
	The wife earns $16,000 less than twice what husband earns.
	\(w=2h-16,000\)
The system of equations is:	\(\{\begin{array}{l}w+h=110,000 \\ w=2h-16,000\end{array}\)

Solve Direct Translation Applications

We set up, but did not solve, the systems of equations in and Now we’ll translate a situation to a system of equations and then solve it.

Example

Try it.

Translate to a system of equations and then solve:

Devon is 26 years older than his son Cooper. The sum of their ages is 50. Find their ages.

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.	We are looking for the ages of Devon and Cooper.
Step 3. Name what we are looking for.	Let \(d=\) Devon’s age. \(\ c=\) Cooper’s age
Step 4. Translate into a system of equations.	Devon is 26 years older than Cooper.
	The sum of their ages is 50.
The system is:
Step 5. Solve the system of equations. Solve by substitution.
Substitute c + 26 into the second equation.
Solve for c.
Substitute c = 12 into the first equation and then solve for d.
Step 6. Check the answer in the problem.	Is Devon’s age 26 more than Cooper’s? Yes, 38 is 26 more than 12. Is the sum of their ages 50? Yes, 38 plus 12 is 50.
Step 7. Answer the question.	Devon is 38 and Cooper is 12 years old.

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

Solve Geometry Applications

When we learned about Math Models, we solved geometry applications using properties of triangles and rectangles. Now we’ll add 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.

Example

Try it.

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 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.	Let \(x=\) the measure of the first angle \(x=\). \(m=\) 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. \(x-y=26\)
The system is	\(\{\begin{array}{l}x+y=90 \\ x-y=26\end{array}\)
Step 5. Solve the system of equations by elimination.	\(\begin{array}{l}\underset{\text{_________}}{\{\begin{array}{l}x+y=90 \\ x-y=26\end{array}} \\ 2x\ =116\end{array}\)
Substitute \(x=58\) into the first equation.	\(\begin{array}{lll}x+y & = & 90 \\ 58+y & = & 90 \\ y & = & 32\end{array}\)
Step 6. Check the answer in the problem. \(\begin{array}{l}58+32\ =\ 90✓ \\ 58-32\ =\ 26✓\end{array}\)
Step 7. Answer the question.	The angle measures are 58 degrees and 32 degrees.

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

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 travelled, 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.

Example

Try it.

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.

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. Since \(D=r\cdot t\) 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 \(\frac{1}{2}\) hour less than Joni’s time. So, \(k=j-\frac{1}{2}.\)
Now we have the system.
Solve the system of equations by substitution.
Substitute \(k=j-\frac{1}{2}\) into the second equation, then solve for j.
To find Kelly’s time, substitute j = 3 into the first equation, then solve for k.
Check the answer in the problem.   Joni 3 hours (65 mph) = 195 miles.   Kelly \(2\frac{1}{2}\) hours (78 mph) = 195 miles.   Yes, they will have traveled the same distance when they meet.
Answer the question.	Kelly will catch up to Joni in \(2\frac{1}{2}\) hours. By then, Joni will have traveled 3 hours.

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

Solve Applications with Systems of Equations

Translate to a System of Equations

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

Try it.

The sum of two numbers is fifteen. One number is three less than the other. Find the numbers.

Solution

The numbers are 6 and 9.

Try it.

The sum of two numbers is twenty-five. One number is five less than the other. Find the numbers.

Try it.

The sum of two numbers is negative thirty. One number is five times the other. Find the numbers.

Solution

The numbers are −5 and −25.

Try it.

The sum of two numbers is negative sixteen. One number is seven times the other. Find the numbers.

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Twice a number plus three times a second number is twenty-two. Three times the first number plus four times the second is thirty-one. Find the numbers.

Solution

The numbers are 5 and 4.

Try it.

Six times a number plus twice a second number is four. Twice the first number plus four times the second number is eighteen. Find the numbers.

Try it.

Three times a number plus three times a second number is fifteen. Four times the first plus twice the second number is fourteen. Find the numbers.

Solution

The numbers are 2 and 3.

Try it.

Twice a number plus three times a second number is negative one. The first number plus four times the second number is two. Find the numbers.

Try it.

A married couple together earn $75,000. The husband earns $15,000 more than five times what his wife earns. What does the wife earn?

Solution

$10,000

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During two years in college, a student earned $9,500. The second year she earned $500 more than twice the amount she earned the first year. How much did she earn the first year?

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Daniela invested a total of $50,000, some in a certificate of deposit (CD) and the remainder in bonds. The amount invested in bonds was $5000 more than twice the amount she put into the CD. How much did she invest in each account?

Solution

She put $15,000 into a CD and $35,000 in bonds.

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Jorge invested $28,000 into two accounts. The amount he put in his money market account was $2,000 less than twice what he put into a CD. How much did he invest in each account?

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In her last two years in college, Marlene received $42,000 in loans. The first year she received a loan that was $6,000 less than three times the amount of the second year’s loan. What was the amount of her loan for each year?

Solution

The amount of the first year’s loan was $30,000 and the amount of the second year’s loan was $12,000.

Try it.

Jen and David owe $22,000 in loans for their two cars. The amount of the loan for Jen’s car is $2000 less than twice the amount of the loan for David’s car. How much is each car loan?

Solve Direct Translation Applications

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

Try it.

Alyssa is twelve years older than her sister, Bethany. The sum of their ages is forty-four. Find their ages.

Solution

Bethany is 16 years old and Alyssa is 28 years old.

Try it.

Robert is 15 years older than his sister, Helen. The sum of their ages is sixty-three. Find their ages.

Try it.

The age of Noelle’s dad is six less than three times Noelle’s age. The sum of their ages is seventy-four. Find their ages.

Solution

Noelle is 20 years old and her dad is 54 years old.

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The age of Mark’s dad is 4 less than twice Marks’s age. The sum of their ages is ninety-five. Find their ages.

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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?

Solution

The small container holds 20 gallons and the large container holds 30 gallons.

Try it.

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?

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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?

Solution

There were 10 calories burned jogging and 10 calories burned cycling.

Try it.

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?

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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.

Solution

Notebooks are $4 and thumb drives are $20.

Try it.

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?

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

Try it.

The difference of two complementary angles is 30 degrees. Find the measures of the angles.

Solution

The measures are 60 degrees and 30 degrees.

Try it.

The difference of two complementary angles is 68 degrees. Find the measures of the angles.

Try it.

The difference of two supplementary angles is 70 degrees. Find the measures of the angles.

Solution

The measures are 125 degrees and 55 degrees.

Try it.

The difference of two supplementary angles is 24 degrees. Find the measure of the angles.

Try it.

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

Solution

94 degrees and 86 degrees

Try it.

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

Try it.

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

Solution

72.5 degrees and 17.5 degrees

Try it.

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

Try it.

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.

Solution

The measures are 44 degrees and 136 degrees.

Try it.

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.

Try it.

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.

Solution

The measures are 34 degrees and 56 degrees.

Try it.

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.

Try it.

Wayne is hanging a string of lights 45 feet long around the three sides of his rectangular 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.

Solution

The width is 10 feet and the length is 25 feet.

Try it.

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

Try it.

A frame around a rectangular family portrait has a perimeter of 60 inches. The length is fifteen less than twice the width. Find the length and width of the frame.

Solution

The width is 15 feet and the length is 15 feet.

Try it.

The perimeter of a rectangular 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.

Solve Uniform Motion Applications In the following exercises, translate to a system of equations and solve.

Try it.

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?

Solution

It took Sarah’s sister 12 hours.

Try it.

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?

Try it.

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?

Solution

It took Lucy’s friend 2 hours.

Try it.

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?

Try it.

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.

Solution

The canoe rate is 5 mph and the current rate is 1 mph.

Try it.

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.

Try it.

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.

Solution

The boat rate is 6.60 mph and the current rate is 2.50 mph.

Try it.

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.

Try it.

A small jet can fly 1,072 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.

Solution

The jet rate is 240 mph and the wind speed is 28 mph.

Try it.

A small jet can fly 1,435 miles in 5 hours with a tailwind but only 1215 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Try it.

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.

Solution

The jet rate is 415 mph and the wind speed is 19 mph.

Try it.

A commercial jet can fly 1,320 miles in 3 hours with a tailwind but only 1,170 miles in 3 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

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

Practice (40)

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

1. The sum of twice a number and nine is 31. Find the number.
   If you missed this problem, review .

   Lafunua jibu

   \(11\)

2. Twins Jon and Ron together earned $96,000 last year. Ron earned $8000 more than three times what Jon earned. How much did each of the twins earn?
   If you missed this problem, review .

   Lafunua jibu

   Jon earned $22,000 and Ron earned $74,000.

3. An express train and a local train leave Pittsburgh to travel to Washington, D.C. The express train can make the trip in four hours and the local train takes five hours for the trip. The speed of the express train is 12 miles per hour faster than the speed of the local train. Find the speed of both trains.
   If you missed this problem, review .

   Lafunua jibu

   The speed of the local train is 48 mph and the speed of the express train is 60 mph.

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

   Lafunua jibu

   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.
   	One number is nine less than the other.
   The system is:
   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.
   Solve for m.
   Substitute \(m=\frac{9}{2}\) into the second equation and then solve for n.
   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 \(\frac{9}{2}\) and \(-\frac{9}{2}.\)

5. The sum of two numbers is 10. One number is 4 less than the other. Find the numbers.

   Lafunua jibu

   3, 7

6. The sum of two numbers is \(-6.\) One number is 10 less than the other. Find the numbers.

   Lafunua jibu

   2, \(-8\)

7. 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 \(\text{\$}10,000+\text{\$}40\) for each training session. How many training sessions would make the salary options equal?

   Lafunua jibu

   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.
   	Option B would pay her $10,000 + $40 for each training session.
   The system is shown.
   Step 5. Solve the system of equations. We will use substitution.
   Substitute 25,000 +15n for s in the second equation.
   Solve for n.
   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.

8. Geraldine has been offered positions by two insurance companies. The first company pays a salary of $12,000 plus a commission of $100 for each policy sold. The second pays a salary of $20,000 plus a commission of $50 for each policy sold. How many policies would need to be sold to make the total pay the same?

   Lafunua jibu

   160 policies

9. Kenneth currently sells suits for company A at a salary of $22,000 plus a $10 commission for each suit sold. Company B offers him a position with a salary of $28,000 plus a $4 commission for each suit sold. How many suits would Kenneth need to sell for the options to be equal?

   Lafunua jibu

   1000 suits

10. 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?

    Lafunua jibu

    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
    	20 minutes on the elliptical and 30 minutes of circuit training burned 473 calories
    The system is:
    Step 5. Solve the system of equations.
    Multiply the first equation by −2 to get opposite coefficients of e.
    Simplify and add the equations. Solve for c.
    Substitute c = 8.3 into one of the original equations to solve for e.
    Step 6. Check the answer in the problem.	Check the math on your own.
    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.

11. Translate to a system of equations and then solve:

    Mark went to the gym and did 40 minutes of Bikram hot yoga and 10 minutes of jumping jacks. He burned 510 calories. The next time he went to the gym, he did 30 minutes of Bikram hot yoga and 20 minutes of jumping jacks burning 470 calories. How many calories were burned for each minute of yoga? How many calories were burned for each minute of jumping jacks?

    Lafunua jibu

    Mark burned 11 calories for each minute of yoga and 7 calories for each minute of jumping jacks.

12. Translate to a system of equations and then solve:

    Erin spent 30 minutes on the rowing machine and 20 minutes lifting weights at the gym and burned 430 calories. During her next visit to the gym she spent 50 minutes on the rowing machine and 10 minutes lifting weights and burned 600 calories. How many calories did she burn for each minutes on the rowing machine? How many calories did she burn for each minute of weight lifting?

    Lafunua jibu

    Erin burned 11 calories for each minute on the rowing machine and 5 calories for each minute of weight lifting.

13. Translate to a system of equations and then solve.

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

    Lafunua jibu

    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.	\(\begin{array}{l}\text{Let}\ x=\ \text{the measure of the first angle.} \\ y=\ \text{the measure of the second angle}\end{array}\)
    Step 4. Translate into a system of equations.	The angles are complementary.
    	\(\ x+y=90\)
    	The difference of the two angles is 26 degrees.
    	\(\ x-y=26\)
    The system is shown.	\(\ \{\begin{array}{l}x+y=90 \\ x-y=26\end{array}\)
    Step 5. Solve the system of equations by elimination.	\(\ \begin{array}{l}\underset{___________}{\{\begin{array}{l}x+y=90 \\ x-y=26\end{array}} \\ 2x\ =116\end{array}\)
    Substitute \(x=58\) into the first equation.	\(\ \begin{array}{l}x=58 \\ \\ \\ x+y=90 \\ 58+y=90 \\ y=32\end{array}\)
    Step 6. Check the answer in the problem.
    \(\ \begin{array}{l}58+32=90✓ \\ 58-32=26✓\end{array}\)
    Step 7. Answer the question.	The angle measures are 58 and 32 degrees.

14. Translate to a system of equations and then solve:

    The difference of two complementary angles is 20 degrees. Find the measures of the angles.

    Lafunua jibu

    The angle measures are 55 and 35.

15. Translate to a system of equations and then solve:

    The difference of two complementary angles is 80 degrees. Find the measures of the angles.

    Lafunua jibu

    The angle measures are 5 and 85.

16. 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.

    Lafunua jibu

    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 larger angle is twelve less than five times the smaller angle.
    The system is shown: Step 5. Solve the system of equations substitution.
    Substitute 5x − 12 for y in the first equation. Solve for x.
    Substitute 32 for x in the second equation, then solve for y.
    Step 6. Check the answer in the problem.
    Step 7. Answer the question.	The angle measures are 148 and 32 degrees.

17. Translate to a system of equations and then solve:

    Two angles are supplementary. The measure of the larger angle is 12 degrees more than three times the smaller angle. Find the measures of the angles.

    Lafunua jibu

    The angle measures are 42 and 138.

18. Translate to a system of equations and then solve:

    Two angles are supplementary. The measure of the larger angle is 18 less than twice the measure of the smaller angle. Find the measures of the angles.

    Lafunua jibu

    The angle measures are 66 and 114.

19. 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.

    Lafunua jibu

    We will draw and label a figure.

    Step 1. Read the problem.
    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.
    	The sum of the measures of the angles of a triangle is 180.
    The system is shown.
    Step 5. Solve the system of equations. We will use substitution since the first equation is solved for a.
    Substitute \(3b+10\) for a in the second equation.
    Solve for b.
    Substitute \(b=20\) into the first equation and then solve for a.
    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.

20. The measure of one of the small angles of a right triangle is 2 more than 3 times the measure of the other small angle. Find the measure of both angles.

    Lafunua jibu

    22, 68

21. The measure of one of the small angles of a right triangle is 18 less than twice the measure of the other small angle. Find the measure of both angles.

    Lafunua jibu

    36, 54

22. 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.

    Lafunua jibu

    Step 1. Read the problem.
    Step 2. Identify what you are looking for.	We are looking for the length and width.
    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.
    	The length will be 5 feet more than four times the width.
    The system is shown. Step 5. Solve The system of equations by substitution.
    Substitute L = 4W + 5 into the first equation, then solve for W.
    Substitute 20 for W in the second equation, then solve for L.
    Step 6. Check the answer in the problem.
    Step 7. Answer the equation.	The length is 85 feet and the width is 20 feet.

23. Translate to a system of equations and then solve:

    Mario wants to put a fence around the pool in his backyard. Since one side is adjacent to the house, he will only need to fence three sides. There are two long sides and the one shorter side is parallel to the house. He needs 155 feet of fencing to enclose the pool. The length of the long side is 10 feet less than twice the width. Find the length and width of the pool area to be enclosed.

    Lafunua jibu

    The length is 60 feet and the width is 35 feet.

24. Translate to a system of equations and then solve:

    Alexis wants to build a rectangular dog run in her yard adjacent to her neighbor’s fence. She will use 136 feet of fencing to completely enclose the rectangular dog run. The length of the dog run along the neighbor’s fence will be 16 feet less than twice the width. Find the length and width of the dog run.

    Lafunua jibu

    The length is 60 feet and the width is 38 feet.

25. 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?

    Lafunua jibu

    A diagram is useful in helping us visualize the situation.

    
    

    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.

    
    

    Since \(D=r\cdot t\) 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 \(\frac{1}{2}\) hour less than Joni’s time. So,

    	\(\ k=j-\frac{1}{2}\)
    Now we have the system.	\(\ \{\begin{array}{l}\ k=j-\frac{1}{2} \\ 65j=78k\end{array}\)
    Solve the system of equations by substitution.
    Substitute \(k=j-\frac{1}{2}\) into the second equation, then solve for \(j\).
    	\(\ \begin{array}{lll}65j & = & 78k \\ 65j & = & 78(j-\frac{1}{2}) \\ 65j & = & 78j-39 \\ -13j & = & -39 \\ j & = & 3\end{array}\)
    To find Kelly’s time, substitute \(j=3\) into the first equation, then solve for \(k\).	\(\ k=j-\frac{1}{2}\)
    	\(\ k=3-\frac{1}{2}\)
    	\(\ k=\frac{5}{2}\ \text{or}\ k=2\frac{1}{2}\)
    Check the answer in the problem.
    Joni \(\ 3\ \text{hours}\ (65\ \text{mph})\ =195\ \text{miles}\)
    Kelly \(\ 2\frac{1}{2}\ \text{hours}\ (78\ \text{mph})=195\ \text{miles}\)
    Yes, they will have traveled the same distance when they meet.
    Answer the question.	Kelly will catch up to Joni in \(2\frac{1}{2}\) hours. By then, Joni will have traveled 3 hours.

26. Translate to a system of equations and then solve:

    Mitchell left Detroit on the interstate driving south towards Orlando at a speed of 60 miles per hour. Clark left Detroit 1 hour later traveling at a speed of 75 miles per hour, following the same route as Mitchell. How long will it take Clark to catch Mitchell?

    Lafunua jibu

    It will take Clark 4 hours to catch Mitchell.

27. Translate to a system of equations and then solve:

    Charlie left his mother’s house traveling at an average speed of 36 miles per hour. His sister Sally left 15 minutes \((\frac{1}{4}\ \text{hour})\) later traveling the same route at an average speed of 42 miles per hour. How long before Sally catches up to Charlie?

    Lafunua jibu

    It will take Sally \(1\frac{1}{2}\) hours to catch up to Charlie.

28. 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.

    Lafunua jibu

    Read the problem.	This is a uniform motion problem and a picture will help us visualize the situation.
    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 s − c.
    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.
    Solve the system of equations. Distribute to put both equations in standard form, then solve by elimination.
    Multiply the top equation by 5 and the bottom equation by 4. Add the equations, then solve for s.
    Substitute s = 13.5 into of the original equations.
    Check the answer in the problem. The downstream rate would be  \(13.5+1.5=15\) mph. In 4 hours the ship would travel   \(15\cdot 4=60\) miles. The upstream rate would be  \(13.5-1.5=12\) mph. In 5 hours the ship would travel   \(12\cdot 5=60\) miles.
    Answer the question.	The rate of the ship is 13.5 mph and the rate of the current is 1.5 mph.

29. Translate to a system of equations and then solve:

    A Mississippi river boat cruise sailed 120 miles upstream for 12 hours and then took 10 hours to return to the dock. Find the speed of the river boat in still water and the speed of the river current.

    Lafunua jibu

    The rate of the boat is 11 mph and the rate of the current is 1 mph.

30. Translate to a system of equations and then solve:

    Jason paddled his canoe 24 miles upstream for 4 hours. It took him 3 hours to paddle back. Find the speed of the canoe in still water and the speed of the river current.

    Lafunua jibu

    The speed of the canoe is 7 mph and the speed of the current is 1 mph.

31. 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.

    Lafunua jibu

    Read the problem.	This is a uniform motion problem and a picture will help us visualize.
    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 j ‒ w.
    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.
    Solve the system of equations. Distribute, then solve by elimination. Add, and solve for j.
    Substitute j = 347 into one of the original equations, then solve for w.
    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   \(365\cdot 3=1,095\) miles Going into the headwind, the jet’s actual rate would be  \(347-18=329\) mph. In 3 hours the jet would travel   \(329\cdot 3=987\) miles.
    Answer the question.	The rate of the jet is 347 mph and the rate of the wind is 18 mph.

32. Translate to a system of equations and then solve:

    A small jet can fly 1,325 miles in 5 hours with a tailwind but only 1,035 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

    Lafunua jibu

    The speed of the jet is 236 mph and the speed of the wind is 29 mph.

33. Translate to a system of equations and then solve:

    A commercial jet can fly 1,728 miles in 4 hours with a tailwind but only 1,536 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

    Lafunua jibu

    The speed of the jet is 408 mph and the speed of the wind is 24 mph.

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

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

    Lafunua jibu

    13 and 17

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

37. The sum of two number is \(-26.\) One number is 12 less than the other. Find the numbers.

    Lafunua jibu

    \(-7\) and \(-19\)

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

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

    Lafunua jibu

    14 and 23

40. The sum of two numbers is \(-27.\) Their difference is \(-59.\) Find the numbers.

Symbols used here

\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.

i

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: Solve Applications with Systems of Equations

1. Solve direct translation applications
2. Solve geometry applications
3. Solve uniform motion applications

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.

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⌨ Type ✎ Draw

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Utasuluhisha

x²√÷×π( )=∫d/dxlim[ ]sinlogeˣ

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

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