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.
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.
| Words in the problem | What to write |
|---|---|
| “a total of 200 tickets” | a + s = 200 |
| “adult tickets cost $8 each” | 8a dollars from adult tickets |
| “5 more than twice the width” | l = 2w + 5 |
| “the same cost” | Set the two costs equal |
| “with the current” / “against the current” | Add the speeds / subtract the speeds |
Steps for Solving Systems Word Problems
- Name the unknowns. Write what each variable stands for, with units. For example, “a = number of adult tickets.”
- Write two equations. Turn each fact into an equation.
- Pick a method. Use substitution if a variable is already alone. Use elimination if both equations look like Ax + By = C.
- Solve for both variables.
- 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.
Step 2: Write one equation for tickets and one for money.
\begin{aligned} a + s &= 200 \\ 8a + 5s &= 1285 \end{aligned}Step 3: Solve the first equation for s: s = 200 - a. Substitute it into the second.
8a + 5(200 - a) = 1285 \quad\Rightarrow\quad 3a + 1000 = 1285 \quad\Rightarrow\quad a = 95
Step 4: s = 200 - 95 = 105. Check: 95 + 105 = 200 tickets and 8(95) + 5(105) = 760 + 525 = 1285 dollars.
Answer: 95 adult tickets and 105 student tickets.
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.
Step 1: Let l = length and w = width, in feet.
\begin{aligned} l &= 2w + 5 \\ 2l + 2w &= 58 \end{aligned}Step 2: l is already alone, so substitute.
2(2w + 5) + 2w = 58 \quad\Rightarrow\quad 6w + 10 = 58 \quad\Rightarrow\quad w = 8
Step 3: l = 2(8) + 5 = 21. Check: 2(21) + 2(8) = 42 + 16 = 58.
Answer: The garden is 21 feet long and 8 feet wide.
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?
Step 1: Let m = months and c = total cost in dollars.
\begin{aligned} c &= 50 + 25m \\ c &= 20 + 30m \end{aligned}Step 2: Both equal c, so set them equal.
50 + 25m = 20 + 30m \quad\Rightarrow\quad 30 = 5m \quad\Rightarrow\quad m = 6
Step 3: c = 50 + 25(6) = 200. Both gyms cost $200 after 6 months.
Step 4: For 12 months, Gym A costs 50 + 25(12) = 350 dollars and Gym B costs 20 + 30(12) = 380 dollars.
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.
Step 2: The pounds add to 30. The values add to 14 \cdot 30 = 420 dollars.
\begin{aligned} x + y &= 30 \\ 12x + 18y &= 420 \end{aligned}Step 3: Substitute y = 30 - x.
12x + 18(30 - x) = 420 \quad\Rightarrow\quad -6x + 540 = 420 \quad\Rightarrow\quad x = 20
Step 4: y = 30 - 20 = 10. Check: 12(20) + 18(10) = 240 + 180 = 420.
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.
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.
Step 2: Going down the river, the current helps, so the speed is k + c. Going up, it slows the kayak, so the speed is k - c. Use speed = distance ÷ time.
\begin{aligned} k + c &= \tfrac{24}{3} = 8 \\ k - c &= \tfrac{24}{6} = 4 \end{aligned}Step 3: Add the equations to eliminate c: 2k = 12, so k = 6. Then 6 + c = 8, so c = 2.
Answer: The kayaker paddles 6 miles per hour in still water. The current is 2 miles per hour.
Common Mistakes
| Mistake | Example of the error | Fix |
|---|---|---|
| Mixing a count with money | a + s = 1285 | Count equation: a + s = 200. Money equation: 8a + 5s = 1285 |
| Not saying what each variable means | Finding x = 20 but not knowing which coffee it is | Write “x = pounds of $12 coffee” first |
| Using the blend price as a total | x + y = 14 | The pounds add to 30. Use 14 only in the value: 14 \cdot 30 = 420 |
| Adding the current going upstream | Upstream speed = k + c | Against the current, subtract: k - c |
| Stopping after one variable | “95 adult tickets” and no student count | Find both values and answer every question |
Choosing a Method
| Method | Best when | Example |
|---|---|---|
| Substitution | One variable is alone, like l = 2w + 5 | Examples 1, 2 and 4 |
| Elimination | Both equations are in Ax + By = C form and a variable has matching or opposite coefficients | Example 5 |
| Graphing | You want to see which choice is better before and after a point | Example 3 |
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
- Two numbers add to 45. Their difference is 11. Find the numbers.
- A concert sold 150 tickets for $2,200. Floor seats cost $12 and balcony seats cost $20. How many of each were sold?
- A jar has 30 dimes and quarters worth $4.80. How many of each coin are there?
- A rectangle has a perimeter of 70 cm. Its length is 1 cm less than 3 times its width. Find its length and width.
- 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?
- 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?
- 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.
- The sum of Sam’s and Lea’s ages is 32. Sam is 3 times as old as Lea. How old is each?
- A farm has chickens and cows. There are 40 heads and 110 legs. How many of each animal are there?
- Rental A costs $30 plus $0.20 per mile. Rental B costs $45 plus $0.20 per mile. When do they cost the same?
Show answers
- x + y = 45 and x - y = 11. Add: 2x = 56, so x = 28 and y = 17.
- f + b = 150, 12f + 20b = 2200. 12f + 20(150 - f) = 2200 gives -8f = -800, so 100 floor and 50 balcony.
- d + q = 30 and 10d + 25q = 480 (in cents). 10(30 - q) + 25q = 480 gives 15q = 180, so 12 quarters and 18 dimes.
- l = 3w - 1 and 2l + 2w = 70. 2(3w - 1) + 2w = 70 gives 8w = 72, so w = 9 cm and l = 26 cm.
- 40 + 5g = 60 + g gives 4g = 20, so g = 5. Both cost $65 for 5 gigabytes.
- p + c = 12 and 3p + 9c = 60. 3p + 9(12 - p) = 60 gives -6p = -48, so 8 pounds of peanuts and 4 pounds of cashews.
- p + w = 400 and p - w = 300. Add: 2p = 700. The plane flies 350 mph and the wind is 50 mph.
- s = 3l and s + l = 32. 3l + l = 32, so Lea is 8 and Sam is 24.
- c + w = 40 and 2c + 4w = 110. 2c + 4(40 - c) = 110 gives -2c = -50, so 25 chickens and 15 cows.
- 30 + 0.2m = 45 + 0.2m gives 30 = 45, which is false. There is no solution. The lines are parallel, and Rental B always costs $15 more.
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.
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Next lesson: Graphing Linear Inequalities
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