Unit 4 · Lesson 3 of 4

Systems word problems

About 25 minutes · 5 steps · one idea: two facts in the story give two equations

You'll be able to: turn a story about tickets, coins, mixtures, or wind and current into a system, and solve it.

  1. 1 Warm-up
  2. 2 The idea
  3. 3 Examples
  4. 4 Practice
  5. 5 Wrap-up

Step 1 of 5Warm-up

Two quick questions from earlier lessons. Answer from memory first.

Solve by elimination: \(x + 2y = 9\) and \(3x - 2y = 3\).

The \(y\) terms are opposites. Add: \(4x = 12\), so \(x = 3\).

Then \(3 + 2y = 9\), so \(2y = 6\) and \(y = 3\).

Answer: \((3, 3)\).

A jacket costs $80 and is on sale for 25% off. What is the sale price, in dollars?

The discount is \(0.25 \cdot 80 = 20\) dollars.

Sale price: \(80 - 20 = 60\).

Answer: $60.

Step 2 of 5The idea

Many word problems have two unknown amounts and give you two facts about them. Each fact becomes one equation, so you get a system. Then you solve it with substitution or elimination, whichever looks easier.

Four steps for any systems word problem:

  1. Name the two unknowns with letters, and write what each one means ("\(a\) = number of adult tickets").
  2. Write one equation for each fact.
  3. Solve the system.
  4. Answer the question that was asked, with units, and check it against the story.

Common patterns:

  • Tickets, coins, mixtures: a count equation (\(a + s = \) total number) and a value equation (price \(\cdot\, a\) + price \(\cdot\, s\) = total money).
  • Wind or current: with it, speed \(= p + w\); against it, speed \(= p - w\). Find each speed from distance \(\div\) time.

Why this shows up on the placement test: placement tests like to wrap a system inside a short story, and setting up the two equations is the part most people miss.

Step 3 of 5Worked examples

Example 1: Tickets

A school play sold 200 tickets. Adult tickets cost $8 and student tickets cost $5. Ticket sales were $1,360. How many of each were sold?

  1. Name: \(a\) = adult tickets, \(s\) = student tickets.
  2. Count: \(a + s = 200\). Value: \(8a + 5s = 1360\).
  3. Substitute \(s = 200 - a\): \(8a + 5(200 - a) = 1360\), so \(8a + 1000 - 5a = 1360\).
  4. \(3a = 360\), so \(a = 120\). Then \(s = 200 - 120 = 80\).
  5. Check: \(8(120) + 5(80) = 960 + 400 = 1360\). 120 adult and 80 student tickets.

Example 2: Plane and wind

A plane flies 1,200 miles with the wind in 3 hours. The trip back, against the wind, takes 4 hours. Find the plane's speed in still air and the wind speed.

  1. Name: \(p\) = plane speed in still air, \(w\) = wind speed (miles per hour).
  2. With the wind: \(1200 \div 3 = 400\), so \(p + w = 400\). The wind pushes the plane, so the speeds add.
  3. Against the wind: \(1200 \div 4 = 300\), so \(p - w = 300\).
  4. Add the equations: \(2p = 700\), so \(p = 350\). Then \(350 + w = 400\), so \(w = 50\).
  5. Plane: 350 mph. Wind: 50 mph. Check: \(350 - 50 = 300\). True.

Step 4 of 5Practice

Write the two equations on paper first, then solve. Type just the number the question asks for. If you miss twice, open the worked solution.

Two numbers add up to 30. Their difference is 8. What is the larger number?

Let \(x\) = larger, \(y\) = smaller. Then \(x + y = 30\) and \(x - y = 8\).

Add: \(2x = 38\), so \(x = 19\). (The smaller is \(11\).)

Answer: \(19\).

A concert sold 150 tickets. Adult tickets cost $12 and child tickets cost $7. Sales were $1,500. How many child tickets were sold?

Let \(a\) = adult, \(c\) = child. Count: \(a + c = 150\). Value: \(12a + 7c = 1500\).

Substitute \(a = 150 - c\): \(12(150 - c) + 7c = 1500\), so \(1800 - 12c + 7c = 1500\).

\(-5c = -300\), so \(c = 60\). (Adults: 90. Check: \(1080 + 420 = 1500\).)

Answer: \(60\) child tickets.

A jar holds 30 coins, all dimes and quarters. Together they are worth $4.80. How many dimes are in the jar?

Let \(d\) = dimes, \(q\) = quarters. Count: \(d + q = 30\). Value in cents: \(10d + 25q = 480\).

Substitute \(d = 30 - q\): \(10(30 - q) + 25q = 480\), so \(300 + 15q = 480\).

\(15q = 180\), so \(q = 12\). Then \(d = 30 - 12 = 18\).

Answer: \(18\) dimes. (Check: \(180 + 300 = 480\) cents.)

A shop mixes coffee that costs $6 per pound with coffee that costs $10 per pound. It wants 20 pounds of a blend worth $9 per pound. How many pounds of the $6 coffee should it use?

Let \(x\) = pounds at $6, \(y\) = pounds at $10. Count: \(x + y = 20\). Value: \(6x + 10y = 9 \cdot 20 = 180\).

Substitute \(y = 20 - x\): \(6x + 10(20 - x) = 180\), so \(6x + 200 - 10x = 180\).

\(-4x = -20\), so \(x = 5\). (And \(y = 15\). Check: \(30 + 150 = 180\).)

Answer: \(5\) pounds.

A boat goes 36 miles downstream (with the current) in 2 hours. Going back upstream takes 3 hours. What is the speed of the current, in miles per hour?

Let \(b\) = boat speed in still water, \(c\) = current speed.

Downstream: \(36 \div 2 = 18\), so \(b + c = 18\). Upstream: \(36 \div 3 = 12\), so \(b - c = 12\).

Add: \(2b = 30\), so \(b = 15\). Then \(15 + c = 18\), so \(c = 3\).

Answer: \(3\) mph.

3 burgers and 2 orders of fries cost $21. 2 burgers and 4 orders of fries cost $22. What does one burger cost, in dollars?

Let \(b\) = burger price, \(f\) = fries price. Then \(3b + 2f = 21\) and \(2b + 4f = 22\).

Multiply the first equation by \(-2\): \(-6b - 4f = -42\). Add it to the second: \(-4b = -20\), so \(b = 5\).

(Fries: \(3(5) + 2f = 21\), so \(f = 3\). Check: \(10 + 12 = 22\).)

Answer: $5.

A shop has 40 bikes. Some rent for $15 a day and the rest for $25 a day. Renting all of them for one day brings in $760. Let \(x\) = number of $15 bikes and \(y\) = number of $25 bikes. Which system fits the story?

  1. \(x + y = 760\) and \(15x + 25y = 40\)
  2. \(x + y = 40\) and \(15x + 25y = 760\)
  3. \(x + y = 40\) and \(25x + 15y = 760\)
  4. \(x - y = 40\) and \(15x + 25y = 760\)

The count of bikes is 40, so \(x + y = 40\). That rules out (a), which swaps the count and the money, and (d), which subtracts.

The money: each $15 bike earns \(15x\) and each $25 bike earns \(25y\), so \(15x + 25y = 760\). Choice (c) puts the prices on the wrong letters.

Answer: (b). (It solves to 24 bikes at $15 and 16 bikes at $25.)

A plane flies 900 miles with the wind in 3 hours, and 900 miles back against the wind in 4.5 hours. What is the plane's speed in still air, in miles per hour?

With the wind: \(900 \div 3 = 300\), so \(p + w = 300\).

Against the wind: \(900 \div 4.5 = 200\), so \(p - w = 200\).

Add: \(2p = 500\), so \(p = 250\). (The wind is 50 mph.)

Answer: \(250\) mph.

Step 5 of 5Wrap-up

You can now name two unknowns, write a count equation and a value equation (or a with/against pair), solve the system, and answer in words.

Read or watch next: OpenStax Intermediate Algebra 2e, 4.2 Solve Applications with Systems of Equations (free textbook, with more practice at the end of the section).

Also good: Khan Academy: Systems of equations word problems (short videos). Quick rules: Lines and systems cheat sheet.

Something unclear? Ask your teacher bot a follow-up question, for example: "In a mixture problem, how do I know which numbers go in the value equation?" It knows this course and can explain things another way.