Simultaneous equations word problems — Mathematics

Before this

Forming equations from word problems · coming soon
Simultaneous equations — substitution
Simultaneous equations — elimination

This leads to

Finding the equation of a straight line
Finding equilibrium algebraically (Economics)

The algebra is right, and the answer is wrong

Read this sentence: "At a university, there are six times as many students as professors." Now write it as an equation. Use $S$ for the number of students and $P$ for the number of professors.

In 1981, three researchers, John Clement, Jack Lochhead and George Monk, published a study in which university engineering students were asked exactly this. These were students who could solve equations well. More than a third of them still wrote the wrong equation. The most common wrong answer was $6 S = P$ .

It reads almost like the sentence. But test it. If there are 10 professors, it says $6 S = 10$ , so $S \approx 1.67$ . That is fewer than two students for ten professors. The equation says the opposite of the sentence.

For three lectures you have solved pairs of equations: by substitution, by elimination, and with a curve. But in a word problem, nobody gives you the equations. You build them from sentences. And at the end, nobody tells you whether the answer is sensible. You judge that yourself.

So the difficulty is not in the algebra. It lives at the two ends of the problem: words into equations at the start, and the answer back into the world at the end. This lecture is about those two ends.

A letter is a number, not a label

Here is why $6 S = P$ feels right. The writer is reading $S$ as a label, a short way of writing "students". Then "6S" looks like "six students", and the sentence seems to become "six students for each professor". But in algebra, $S$ is not the word "students". It is the number of students, a number we do not know yet.

Once you read the letters as numbers, the sentence becomes a comparison between two numbers. Which number is bigger? The number of students. How much bigger? Six times. So the number of students equals six times the number of professors:

S = 6 P

Test it with a number you can picture. With 10 professors, $S = 60$ . Sixty students and ten professors: six times as many students. The equation now says what the sentence says.

That test is the most useful habit in this lecture. Before you solve an equation, try it with small numbers you can imagine. If the result is strange, the equation is wrong.

Try the same sentence in a new form below. Commit to an answer before you read the explanations.

Predict, then reveal — six students for every teacher

A school has 6 students for every teacher. Let s be the number of students and t the number of teachers. Which equation is correct? Choose one.

The sentence compares two numbers: the number of students is six times the number of teachers. So s = 6t. The tempting answer 6s = t is reversed: with 10 teachers it gives s ≈ 1.67, fewer students than teachers.

Equation	Test: 10 teachers	Verdict
s = 6t	s = 60	correct: 60 students, 10 teachers
6s = t	s ≈ 1.67	reversed
s = t + 6	s = 16	"6 more", not "6 times"
s + t = 6	s = −4	impossible

The test with 10 teachers decides it. Only $s = 6 t$ gives a school that looks like the sentence. This is why Step 1 below asks you to write exactly what each letter means. "Let $s$ = students" invites the reversal. "Let $s$ = the number of students" prevents it.

Four steps, every time

Every word problem in this topic goes through the same four steps. Steps 1, 2 and 4 are the new skills. Step 3 is what the last three lectures taught.

Solving a word problem with two unknowns

1. Name the unknowns. Write what each letter means, as a number with its unit: "let $p$ = the cost of one pen, in cents".

2. Write two equations. Each separate fact in the problem gives one equation. Test each one with small numbers.

3. Solve by substitution or elimination, whichever is cleaner.

4. Interpret and check. Answer the question that was asked, in words and units. Ask whether the answer is possible in the real situation. Check it against the original words, not just your equations.

Notice the last part of Step 4. A check in your own equations only shows that you solved them correctly. If an equation was reversed, that check still passes. The words are the real test.

Each equation counts one thing

Here is a problem that is very old and still very good. A farmer has chickens and cows. She counts 20 heads and 56 legs. How many of each animal does she have?

The problem gives two facts, and each one counts something different: heads and legs. That is where the two equations come from.

Worked Example 1 — Heads and legs

Step 1. Let $c$ = the number of chickens and $w$ = the number of cows.

Step 2. Every animal has one head, so counting heads gives $c + w = 20$ … (1). A chicken has 2 legs and a cow has 4, so counting legs gives $2 c + 4 w = 56$ … (2).

Look at the units. In equation (1), every term counts heads. In equation (2), every term counts legs: $2 c$ is the number of chicken legs, and $4 w$ is the number of cow legs. An equation that adds heads to legs would be meaningless. Checking that every term measures the same thing is a quick test of Step 2.

Step 3. From (1), $c = 20 - w$ . Substitute into (2): $2 (20 - w) + 4 w = 56$ . Expand: $40 - 2 w + 4 w = 56$ , so $2 w = 16$ and $w = 8$ . Then $c = 20 - 8 = 12$ .

Step 4. The farmer has 12 chickens and 8 cows. Both are whole numbers and neither is negative, so the answer is possible. Check against the words: $12 + 8 = 20$ heads ✓, and $12 \times 2 + 8 \times 4 = 24 + 32 = 56$ legs ✓.

Now look at the problem without algebra. Change the number of cows in the explorer below and count the legs. Then change the number of legs she counted, and see what happens to the answer.

Explore — chickens, cows and legs

There are always 20 animals. Change the number of cows and watch the leg count. Find the number of cows that gives the legs she counted. Then try 57 legs and 90 legs.

Cows: 0

Legs she counted:

Filled box: a cow (4 legs). Outlined box: a chicken (2 legs). Below, the dot is the number of legs now, and the dashed mark is the number she counted.

With 20 animals, each cow in place of a chicken adds 2 legs. So the leg count goes up in steps of 2 from 40 (all chickens) to 80 (all cows).

Cows	Chickens	Legs
0	20	40
4	16	48
8	12	56
12	8	64
20	0	80

56 legs: 8 cows. 57 legs: impossible, because every leg count is even (the algebra gives 8.5 cows). 90 legs: impossible, because even 20 cows give only 80 legs (the algebra gives 25 cows and −5 chickens).

What did you see? Starting from 20 chickens, there are 40 legs. Every time one chicken became a cow, the count went up by exactly 2. To reach 56 legs you needed 16 more legs, which is 8 swaps. That is the answer, 8 cows, without writing an equation.

With 57 legs, no position works, because every count is even. The algebra still gives $w = 8.5$ , and it is correct algebra. But half a cow is not possible. With 90 legs, the dot cannot reach the mark at all. The algebra gives $w = 25$ and $c = - 5$ . Again the algebra is right, and the answer is impossible.

That is it. That is the whole idea of Step 4. The equations do not know that cows come in whole numbers, or that there cannot be fewer than zero chickens. Only you know that. When the answer is impossible, the problem is not in the solving. Either the information in the question is impossible, or one of your equations does not say what the words say. Go back to Step 1 and Step 2.

A puzzle from 1 500 years ago

This problem is older than algebra as you know it. A Chinese book called the Sunzi Suanjing, usually dated to around the year 400, asks it with pheasants and rabbits: 35 heads above and 94 feet below. How many of each?

The method usually told with it is beautiful. Imagine every animal lifts half its feet. Each pheasant now stands on one foot, and each rabbit on two. The number of feet on the ground is half of 94, which is 47. There are 35 heads, so if every animal stood on one foot, there would be 35 feet. The extra 12 feet must come from the rabbits, one extra each. So there are 12 rabbits and $35 - 12 = 23$ pheasants.

Now write it in this topic's language. Let $c$ be the number of pheasants and $w$ the number of rabbits. Then $c + w = 35$ and $2 c + 4 w = 94$ . "Lift half the feet" means divide the second equation by 2: $c + 2 w = 47$ . "Compare with one foot each" means subtract the first equation:

(c + 2 w) - (c + w) w = 47 - 35 = 12

The ancient trick is elimination. It is exactly the method of the second lecture in this topic, told as a story. Check: $23 + 12 = 35$ heads ✓ and $46 + 48 = 94$ feet ✓.

The picture below shows why 56 legs has one answer and 57 has none. Each equation is a straight line, and only whole-number points make sense.

Every point on the dark line has 20 heads, but only the dots are whole numbers of animals. The legs line for 56 meets it at the dot (12, 8). A legs line for 57 would meet the dark line at (11.5, 8.5), between two dots, so no real farm fits.

Stop and think

A problem says: "The difference between two numbers is 7." Kofi writes $x - y = 7$ . Is that always right?

Show the answer

No. "The difference is 7" does not say which number is bigger. If $x$ is the bigger one, then $x - y = 7$ . If $y$ is the bigger one, then $y - x = 7$ . These are two different equations, and they give two different answers. Before writing it, look in the problem for a fact that tells you which is bigger, such as "Ana is older than her brother". If there is none, solve both cases and use Step 4 to see which ones make sense.

Worked examples

Worked Example 2 — Pens and notebooks

Aisha buys 3 pens and 2 notebooks and pays $7.20. Ben buys 1 pen and 4 notebooks and pays $8.80. Find the cost of one pen and of one notebook.

Step 1. Decimals make the arithmetic harder, so choose the unit to avoid them. Let $p$ = the cost of one pen, in cents, and $n$ = the cost of one notebook, in cents. Then $7.20 is 720 cents and $8.80 is 880 cents.

Step 2. Aisha: $3 p + 2 n = 720$ … (1). Ben: $p + 4 n = 880$ … (2). Every term is in cents.

Step 3. Elimination is clean here. Multiply (2) by 3 so that both equations contain $3 p$ : $3 p + 12 n = 2640$ … (3). Subtract (1) from (3): $10 n = 1920$ , so $n = 192$ . Substitute into (2): $p + 768 = 880$ , so $p = 112$ .

Step 4. A pen costs 112 cents, which is $1.12, and a notebook costs $1.92. These are sensible prices. Check against the words: Aisha pays $3 \times 1.12 + 2 \times 1.92 = 3.36 + 3.84 = 7.20$ dollars ✓. Ben pays $1.12 + 4 \times 1.92 = 1.12 + 7.68 = 8.80$ dollars ✓.

Worked Example 3 — Ages, six years ago

The ages of Nadia and her son add up to 48. Six years ago, Nadia was five times as old as her son. How old are they now?

Step 1. Let $a$ = Nadia's age now, in years, and $s$ = her son's age now, in years. Name the ages now, because that is what the question asks for.

Step 2. The first fact is about now: $a + s = 48$ … (1). The second fact is about six years ago. Then Nadia was $a - 6$ and her son was $s - 6$ . Both of them were six years younger. "Nadia was five times as old" puts the five with the son's age, the smaller number:

a - 6 = 5 (s - 6) \dots (2)

Test (2) with numbers you can picture. If the son was 2 six years ago, Nadia was 10. That is five times as old ✓. The five is in the right place.

Step 3. Expand (2): $a - 6 = 5 s - 30$ , so $a = 5 s - 24$ . Substitute into (1): $5 s - 24 + s = 48$ , so $6 s = 72$ and $s = 12$ . Then $a = 48 - 12 = 36$ .

Step 4. Nadia is 36 and her son is 12. Check against the words: $36 + 12 = 48$ ✓. Six years ago they were 30 and 6, and $30 = 5 \times 6$ ✓. Both ages are sensible.

Worked Example 4 — A boat on a river

A boat travels 24 km downstream in 2 hours. The return journey of 24 km upstream takes 3 hours. Find the speed of the boat in still water and the speed of the current.

Step 1. Let $b$ = the speed of the boat in still water, in km/h, and $c$ = the speed of the current, in km/h.

Step 2. Here the facts are hidden inside a physical idea. Going downstream, the current pushes the boat, so its speed over the ground is $b + c$ . Going upstream, the current pushes against it, so its speed is $b - c$ . Speed is distance divided by time:

b + c = \frac{24}{2} = 12 \dots (1)

b - c = \frac{24}{3} = 8 \dots (2)

Step 3. Add (1) and (2): $2 b = 20$ , so $b = 10$ . Then $c = 12 - 10 = 2$ .

Step 4. The boat moves at 10 km/h in still water, and the current flows at 2 km/h. The current is slower than the boat, which it must be: otherwise the boat could not go upstream at all. Check: downstream $12 \times 2 = 24$ km ✓, upstream $8 \times 3 = 24$ km ✓.

Three classic mistakes

Trap 1 — Reversing a comparison

"There are three times as many girls as boys" becomes $3 g = b$ when the letters are read as labels. The correct equation is $g = 3 b$ . Put a small number in before you go on: with 5 boys, $g = 15$ girls, which matches the sentence.

Trap 2 — Mixing units in one equation

A problem gives one price in dollars and another in cents, or one time in minutes and another in hours. If you put them into one equation as they stand, the answer is wrong by a factor of 100 or 60. Choose one unit in Step 1, write it down, and convert every number to it before Step 2.

Trap 3 — Stopping at the letters

Solving gives $p = 112$ and $n = 192$ . But the question may ask "How much do 2 pens and a notebook cost?" or want the answer in dollars. Step 4 starts by reading the question again. Answer it in a sentence, with units.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

A theatre sells adult tickets for $12 and child tickets for $7. One evening it sells 250 tickets and takes $2 450. How many of each ticket did it sell?

Hint 1
Two facts, two counts: one counts tickets, the other counts dollars.

Hint 2
With $a$ adult and $k$ child tickets: $a + k = 250$ and $12 a + 7 k = 2450$ . Substitute $k = 250 - a$ .

Full answer
$12 a + 7 (250 - a) = 2450$ gives $12 a + 1750 - 7 a = 2450$ , so $5 a = 700$ and $a = 140$ . Then $k = 110$ . The theatre sold 140 adult tickets and 110 child tickets. Check: $140 + 110 = 250$ ✓ and $140 \times 12 + 110 \times 7 = 1680 + 770 = 2450$ ✓. Both are whole numbers, as tickets must be.
A quiz has 20 questions. Each right answer scores 5 marks, and each wrong answer loses 2 marks. Lin answers every question and scores 58. How many did she get right?

Hint 1
Let $r$ = the number right and $x$ = the number wrong. One equation counts questions, the other counts marks.

Hint 2
$r + x = 20$ and $5 r - 2 x = 58$ . The minus sign is the lost marks.

Full answer
Substitute $x = 20 - r$ : $5 r - 2 (20 - r) = 58$ , so $5 r - 40 + 2 r = 58$ , so $7 r = 98$ and $r = 14$ . Lin got 14 right and 6 wrong. Check: $14 \times 5 - 6 \times 2 = 70 - 12 = 58$ ✓. Step 4: both numbers are whole and between 0 and 20, so the answer is possible.
A laboratory has a salt solution that is 10% salt and another that is 25% salt. How much of each should be mixed to make 600 ml of a solution that is 15% salt?

Hint 1
Count two things: millilitres of solution, and millilitres of salt. The final mixture contains 15% of 600 ml of salt.

Hint 2
With $x$ ml of the 10% solution and $y$ ml of the 25% solution: $x + y = 600$ and $0.10 x + 0.25 y = 90$ .

Full answer
Substitute $y = 600 - x$ : $0.10 x + 0.25 (600 - x) = 90$ , so $0.10 x + 150 - 0.25 x = 90$ , so $0.15 x = 60$ and $x = 400$ . Then $y = 200$ . Mix 400 ml of the 10% solution with 200 ml of the 25% solution. Check the salt: $40 + 50 = 90$ ml, and $90 \div 600 = 0.15$ ✓. Step 4: 15% is closer to 10% than to 25%, so we expect more of the 10% solution. We have twice as much ✓.
In a market for mangoes, the quantity buyers want each week is $Q = 120 - 4 P$ , and the quantity sellers offer is $Q = 2 P + 30$ . Here $P$ is the price in dollars per box and $Q$ is the number of boxes. (a) Find the price and quantity where the two are equal. (b) After a huge harvest, sellers offer $Q = 2 P + 150$ . Solve again and explain what your answer means.

Hint 1
Both equations give $Q$ . At the price where they agree, the two expressions are equal.

Hint 2
(a) $120 - 4 P = 2 P + 30$ . (b) $120 - 4 P = 2 P + 150$ . Then use Step 4 on the result.

Full answer
(a) $120 - 4 P = 2 P + 30$ gives $90 = 6 P$ , so $P = 15$ . Then $Q = 120 - 60 = 60$ . Check in the other equation: $2 \times 15 + 30 = 60$ ✓. At $15 a box, 60 boxes a week are bought and sold. (b) $120 - 4 P = 2 P + 150$ gives $- 30 = 6 P$ , so $P = - 5$ . A negative price would mean sellers pay buyers to take the mangoes. The algebra is correct, but the model does not describe a real market here. Even at a price of zero, sellers offer 150 boxes and buyers want only 120. The straight-line model was built for ordinary prices, and it breaks when it is pushed this far. In real life the price would fall very low, and some mangoes would not be sold or picked. This is Step 4 at work: an impossible answer tells you about the limits of the model.

The bigger idea

This lecture ends the topic, so look back at it. Two lectures gave you methods: substitution and elimination. One showed that the same idea works when one graph is a curve. This one showed where equations come from. That last skill has a name: modelling. You decide what to call things, which facts to turn into equations, and what the answer means.

Modelling is most of applied mathematics. The last question was a small model of a market. In Economics, the price where supply equals demand is found exactly this way. Engineers and scientists do the same four steps with much bigger systems.

And they all do Step 4. When a model predicts a negative price or half a cow, a careful scientist does not report it. They ask which assumption failed. Reality is a check on mathematics. That check is part of the mathematics.

Key concepts in this lesson

In a word problem the difficulty lives at the two ends: turning sentences into equations, and turning the answer back into the real situation.
A letter stands for a number with a unit, such as "the number of students", never for the thing itself, which is why $S = 6 P$ and not $6 S = P$ .
Each separate fact gives one equation, every term in an equation must measure the same thing, and a test with small numbers catches a wrong equation before you solve it.
An answer that is negative, fractional or too large for the situation means the question or an equation is wrong, even when the algebra is correct.
The final check is against the original words, because an equation that was written wrongly will still pass a check in itself.