Systems of Equations Word Problems

Systems of equations word problems describe two unknown amounts with two facts. Name a variable for each unknown and write one equation for each fact. Then solve the system by substitution, elimination or graphing. Last, check that the answer makes sense in the story.

Before this lesson: You should know the substitution method, the elimination method and how to solve systems by graphing.

Extra practice: The Algebra 1 Unit 5 Practice Pack has printable problems, answer keys and a cheat sheet for this unit.

How Word Problems Become Systems

A word problem with two unknowns needs two facts. Each fact becomes one equation. Together, the two equations make a system.

Many problems have a “how many” fact and a “how much” fact. Say a play sold 200 tickets for $1,285. The 200 counts tickets. The $1,285 counts money. Those are two different equations.

Steps for Solving Systems Word Problems

1. Name the unknowns. Write what each variable stands for, with units. For example, “a = number of adult tickets.”
2. Write two equations. Turn each fact into an equation.
3. Pick a method. Use substitution if a variable is already alone. Use elimination if both equations look like Ax + By = C.
4. Solve for both variables.
5. Check and answer. Put the values back into the words, not just the equations. Give units.

Worked Examples

Example 1: Ticket Sales

A school play sold 200 tickets and made $1,285. Adult tickets cost $8 and student tickets cost $5. How many of each were sold?

Step 1. Let a = adult tickets and s = student tickets.

\begin{aligned} a + s &= 200 \\ 8a + 5s &= 1285 \end{aligned}

s = 200 - a

8a + 5(200 - a) = 1285

8a + 1000 - 5a = 1285

3a + 1000 = 1285

3a = 285

a = 95

s = 200 - 95 = 105

Answer: 95 adult tickets and 105 student tickets.

Check: 95 + 105 = 200 tickets and 8(95) + 5(105) = 760 + 525 = 1285 dollars.

Example 2: Rectangle Perimeter

A garden is a rectangle with a perimeter of 58 feet. Its length is 5 feet more than twice its width. Find the length and width.

\begin{aligned} l &= 2w + 5 \\ 2l + 2w &= 58 \end{aligned}

2(2w + 5) + 2w = 58

4w + 10 + 2w = 58

6w + 10 = 58

6w = 48

w = 8

l = 2(8) + 5 = 16 + 5 = 21

Answer: The garden is 21 feet long and 8 feet wide.

Check: 2(21) + 2(8) = 42 + 16 = 58.

Example 3: Comparing Two Plans

Climbing Gym A charges a $50 joining fee plus $25 a month. Gym B charges a $20 joining fee plus $30 a month. After how many months do they cost the same? Which gym is cheaper for a year?

\begin{aligned} c &= 50 + 25m \\ c &= 20 + 30m \end{aligned}

50 + 25m = 20 + 30m

50 = 20 + 5m

30 = 5m

6 = m

c = 50 + 25(6) = 50 + 150 = 200

Step 7. The graph shows the same point, (6, 200), where the lines cross.

50 + 25(12) = 50 + 300 = 350

20 + 30(12) = 20 + 360 = 380

380 - 350 = 30

Answer: They cost the same, $200, at 6 months. For a year, Gym A is cheaper by $30.

Example 4: A Mixture Problem

A cafe mixes coffee that costs $12 per pound with coffee that costs $18 per pound. It wants 30 pounds of a blend worth $14 per pound. How many pounds of each should it use?

Step 1. Let x = pounds of the $12 coffee and y = pounds of the $18 coffee.

14 \cdot 30 = 420

\begin{aligned} x + y &= 30 \\ 12x + 18y &= 420 \end{aligned}

y = 30 - x

12x + 18(30 - x) = 420

12x + 540 - 18x = 420

-6x + 540 = 420

-6x = -120

x = 20

y = 30 - 20 = 10

Answer: Use 20 pounds of the $12 coffee and 10 pounds of the $18 coffee. The answer makes sense: the blend price, $14, is closer to $12, so there is more of the cheaper coffee.

Check: 12(20) + 18(10) = 240 + 180 = 420.

Example 5: Speed with a Current

A kayaker paddles 24 miles down a river in 3 hours. The trip back up the river takes 6 hours. Find the kayaker’s speed in still water and the speed of the current.

Step 1. Let k = kayaker’s speed in still water and c = speed of the current, in miles per hour.

k + c = \frac{24}{3} = 8

k - c = \frac{24}{6} = 4

\begin{array}{rrcl}
 & k + c & = & 8 \\
+ & k - c & = & 4 \\
\hline
 & 2k & = & 12
\end{array}

k = 6

6 + c = 8

c = 2

Answer: The kayaker paddles 6 miles per hour in still water. The current is 2 miles per hour.

Check: Upstream: 6 - 2 = 4 miles per hour, and 4 \cdot 6 = 24 miles.

Common Mistakes

Choosing a Method

Sometimes the lines never cross. If two plans have the same monthly rate but different fees, their lines are parallel. Then they are never equal, and the system has no solution. See number of solutions for more.

Always ask if the answer fits the story. You can’t sell 95.5 tickets or buy −3 pounds of coffee. An answer like that means a setup or math mistake.

Practice Problems

1. Two numbers add to 45. Their difference is 11. Find the numbers.
2. A concert sold 150 tickets for $2,200. Floor seats cost $12 and balcony seats cost $20. How many of each were sold?
3. A jar has 30 dimes and quarters worth $4.80. How many of each coin are there?
4. A rectangle has a perimeter of 70 cm. Its length is 1 cm less than 3 times its width. Find its length and width.
5. Plan A costs $40 a month plus $5 per gigabyte. Plan B costs $60 a month plus $1 per gigabyte. For how many gigabytes do they cost the same? What is that cost?
6. Peanuts cost $3 per pound and cashews cost $9 per pound. How many pounds of each make 12 pounds of a mix worth $5 per pound?
7. A plane flies 1,200 miles with the wind in 3 hours. The return trip against the wind takes 4 hours. Find the plane’s speed in still air and the wind speed.
8. The sum of Sam’s and Lea’s ages is 32. Sam is 3 times as old as Lea. How old is each?
9. A farm has chickens and cows. There are 40 heads and 110 legs. How many of each animal are there?
10. Rental A costs $30 plus $0.20 per mile. Rental B costs $45 plus $0.20 per mile. When do they cost the same?

x + y = 45

x - y = 11

2x = 56

x = 28

28 + y = 45

y = 17

f + b = 150

12f + 20b = 2200

b = 150 - f

12f + 20(150 - f) = 2200

12f + 3000 - 20f = 2200

-8f + 3000 = 2200

-8f = -800

f = 100

b = 150 - 100 = 50

d + q = 30

10d + 25q = 480

d = 30 - q

10(30 - q) + 25q = 480

300 - 10q + 25q = 480

300 + 15q = 480

15q = 180

q = 12

d = 30 - 12 = 18

l = 3w - 1

2l + 2w = 70

2(3w - 1) + 2w = 70

6w - 2 + 2w = 70

8w - 2 = 70

8w = 72

w = 9

l = 3(9) - 1 = 27 - 1 = 26

40 + 5g = 60 + g

40 + 4g = 60

4g = 20

g = 5

40 + 5(5) = 40 + 25 = 65

p + c = 12

5 \cdot 12 = 60

3p + 9c = 60

c = 12 - p

3p + 9(12 - p) = 60

3p + 108 - 9p = 60

-6p + 108 = 60

-6p = -48

p = 8

c = 12 - 8 = 4

p + w = \frac{1200}{3} = 400

p - w = \frac{1200}{4} = 300

2p = 700

p = 350

350 + w = 400

w = 50

s = 3l

s + l = 32

3l + l = 32

4l = 32

l = 8

s = 3(8) = 24

c + w = 40

2c + 4w = 110

w = 40 - c

2c + 4(40 - c) = 110

2c + 160 - 4c = 110

-2c + 160 = 110

-2c = -50

c = 25

w = 40 - 25 = 15

30 + 0.2m = 45 + 0.2m

30 = 45

Want more practice? The Algebra 1 Unit 5 Practice Pack has 55 pages of problems on this and the other lessons in the unit, with step-by-step answers and a cheat sheet.

FAQ

How do you set up a system of equations from a word problem?

Name a variable for each unknown. Then turn each fact into an equation. Most problems give one fact about “how many” and one about “how much.”

Should I use substitution or elimination?

Use substitution when one variable is already alone, like y = 3x. Use elimination when both equations are in Ax + By = C form. Both give the same answer.

How many equations do I need?

You need one equation for each unknown. Two unknowns need two equations that say different things.

What if my answer is a fraction or a negative number?

Check whether that makes sense in the story. A fraction of a ticket or a negative weight means something went wrong. Look again at your equations.

Summary

• Name each unknown with a variable and its units.
• Write one equation for each fact. Keep counts and money in separate equations.
• Solve with substitution, elimination or graphing.
• Check the answer in the words of the problem and give units.
• Parallel lines mean the two amounts are never equal, so there is no solution.

Previous lesson: Number of Solutions

Next lesson: Graphing Linear Inequalities

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