Solving Problems with Simultaneous Equations

Turn a word problem with two unknowns into a pair of linear equations, solve them, and check the answer against the words: costs, ages, tickets, coins, heads and legs, digits, fractions, shapes, charges and speeds.

Before you start

A pair of simultaneous equations can be solved by substitution (lesson 2.6) or elimination (lesson 2.7), after tidying (lesson 2.8).
One fact about two unknowns gives one equation with infinitely many solutions (lesson 2.1); a second, different fact pins down the answer.

The Method

Solving a problem with simultaneous equations

(1) Define the two unknowns in words, with their units: "Let one pen cost $ $x$ ."

(2) Write two equations, one from each fact in the question.

(3) Solve them by substitution or elimination.

(4) Check the answer against the words of the question, not only the equations, and make sure it makes sense.

(5) Answer the question that was asked, in a sentence, with units.

Example 1: Pens and pencils

Ali buys 2 pens and 3 pencils for $ $11$ . Ben buys 4 pens and 1 pencil for $ $13$ . The pens are all the same price, and so are the pencils. Find the price of one pen and of one pencil.

Solution

Define. Let one pen cost $ $x$ and one pencil cost $ $y$ .

Equations. Ali: $2x + 3y = 11$ $\quad$ (1). Ben: $4x + y = 13$ $\quad$ (2).

Solve. In (2), $y$ has coefficient $1$ : $y = 13 - 4x$ . Substitute into (1): $2x + 3(13 - 4x) = 11\text{,}$ so $2x + 39 - 12x = 11$ , $-10x = -28$ , $x = 2.8$ . Then $y = 13 - 4(2.8) = 13 - 11.2 = 1.8\text{.}$

Check against the words. Ali: 2 pens at $ $2.80$ and 3 pencils at $ $1.80$ cost $ $5.60$ + $ $5.40$ = $ $11$ . Ben: 4 pens and 1 pencil cost $ $11.20$ + $ $1.80$ = $ $13$ . Both prices are positive.

A pen costs $ $2.80$ and a pencil costs $ $1.80$ .

Does the answer make sense?

Prices, lengths and speeds must be positive. Numbers of people, tickets, coins or animals must be whole numbers that are not negative. A digit must be one of $0$ to $9$ .

If your answer breaks one of these, look for a mistake in the equations or the working. If there is none, the facts in the question cannot all be true.

Numbers and Costs

Example 2: Two numbers

One number is $3$ more than twice another number. The sum of the two numbers is $27$ . Find the numbers.

Solution

Let the first number be $x$ and the other $y$ .

" $3$ more than twice $y$ ": $x = 2y + 3$ $\quad$ (1). "The sum is $27$ ": $x + y = 27$ $\quad$ (2).

Substitute (1) into (2): $2y + 3 + y = 27$ , so $3y = 24$ , $y = 8$ . Then $x = 2(8) + 3 = 19$ .

Check: twice $8$ is $16$ , and $3$ more is $19$ ; $19 + 8 = 27$ .

The numbers are $19$ and $8$ .

Example 3: Books and pens

3 books and 2 pens cost $ $29$ . 2 books and 5 pens cost $ $34$ . Find the cost of one book and of one pen.

Solution

Let a book cost $ $x$ and a pen $ $y$ : $3x + 2y = 29$ $\quad$ (1), $\quad$ $2x + 5y = 34$ $\quad$ (2).

No coefficient is $1$ , so eliminate $y$ . (1) $\times 5$ : $15x + 10y = 145$ . (2) $\times 2$ : $4x + 10y = 68$ .

Subtract: $11x = 77$ , so $x = 7$ . Substitute into (1): $21 + 2y = 29$ , so $y = 4$ .

Check against the words: 3 books and 2 pens cost $ $21$ + $ $8$ = $ $29$ ; 2 books and 5 pens cost $ $14$ + $ $20$ = $ $34$ .

A book costs $ $7$ and a pen costs $ $4$ .

Ages

Let the unknowns be the ages now. Then $n$ years ago each person was $n$ years younger, and in $n$ years' time each will be $n$ years older: a person who is $y$ now will be $y + 5$ in 5 years' time.

Example 4: Five years ago and in five years' time

Five years ago, Mrs Lim was 3 times as old as her daughter. In 5 years' time, she will be twice as old as her daughter. How old are they now?

Solution

Let Mrs Lim be $x$ years old now and her daughter $y$ years old now.

Five years ago they were $x - 5$ and $y - 5$ : $x - 5 = 3(y - 5)$ , so $x = 3y - 10$ $\quad$ (1).

In 5 years they will be $x + 5$ and $y + 5$ : $x + 5 = 2(y + 5)$ , so $x = 2y + 5$ $\quad$ (2).

Both give $x$ , so $3y - 10 = 2y + 5$ , and $y = 15$ . Then $x = 2(15) + 5 = 35$ .

Check against the words: five years ago they were $30$ and $10$ , and $30 = 3 \times 10$ . In five years they will be $40$ and $20$ , and $40 = 2 \times 20$ .

Mrs Lim is 35 years old and her daughter is 15.

Tickets, Coins, and Heads and Legs

These problems give two kinds of total: how many things there are, and how much they are worth (or how many legs they have). Each total gives one equation.

Example 5: Concert tickets

A school sold 250 tickets for a concert. Adult tickets cost $ $8$ and student tickets $ $5$ . The takings were $ $1610$ . How many of each kind were sold?

Solution

Let $x$ adult tickets and $y$ student tickets be sold.

Number of tickets: $x + y = 250$ $\quad$ (1). Takings, in dollars: $8x + 5y = 1610$ $\quad$ (2).

From (1): $y = 250 - x$ . Substitute: $8x + 5(250 - x) = 1610\text{,}$ so $3x + 1250 = 1610$ , $3x = 360$ , $x = 120$ . Then $y = 130$ .

Check: $120 + 130 = 250$ tickets; $120$ tickets at $ $8$ and $130$ at $ $5$ take $ $960$ + $ $650$ = $ $1610$ . Both are whole numbers, as they must be.

120 adult tickets and 130 student tickets were sold.

Example 6: Coins

Jun Hao has 30 coins, all 20-cent and 50-cent coins. Their total value is $ $9.30$ . How many of each coin does he have?

Solution

Let him have $x$ 20-cent coins and $y$ 50-cent coins. Work in cents: $ $9.30$ is $930$ cents.

Number: $x + y = 30$ $\quad$ (1). Value: $20x + 50y = 930$ , or, dividing by $10$ , $2x + 5y = 93$ $\quad$ (2).

(1) $\times 2$ : $2x + 2y = 60$ . Subtract from (2): $3y = 33$ , $y = 11$ . Then $x = 19$ .

Check: $19 + 11 = 30$ coins; $19 \times 20 + 11 \times 50 = 380 + 550 = 930$ cents, which is $ $9.30$ .

He has nineteen 20-cent coins and eleven 50-cent coins.

Example 7: Chickens and rabbits

On a farm there are chickens and rabbits. Altogether they have 35 heads and 94 legs. How many chickens and how many rabbits are there?

Solution

Let there be $x$ chickens and $y$ rabbits. Each animal has one head; a chicken has 2 legs and a rabbit 4.

Heads: $x + y = 35$ $\quad$ (1). Legs: $2x + 4y = 94$ $\quad$ (2).

(1) $\times 2$ : $2x + 2y = 70$ . Subtract from (2): $2y = 24$ , $y = 12$ . Then $x = 23$ .

Check: $23 + 12 = 35$ heads; $23 \times 2 + 12 \times 4 = 46 + 48 = 94$ legs.

There are 23 chickens and 12 rabbits.

Digits and Fractions

A two-digit number with tens digit $x$ and units digit $y$ is worth $10x + y$ , not $xy$ : the number $36$ is $10 \times 3 + 6$ . With its digits reversed it is $10y + x$ .

For a fraction $\frac{x}{y}$ , a fact such as "adding 1 to the numerator and the denominator gives $\frac{1}{2}$ " is an equation with fractions: $\frac{x + 1}{y + 1} = \frac{1}{2}$ . Cross-multiply to clear it: $2(x + 1) = y + 1$ .

Example 8: Digits reversed

The sum of the digits of a two-digit number is 9. When the digits are reversed, the new number is 27 more than the original number. Find the original number.

Solution

Let the tens digit be $x$ and the units digit $y$ . The number is $10x + y$ ; reversed, it is $10y + x$ .

Sum of digits: $x + y = 9$ $\quad$ (1).

Reversed is 27 more: $10y + x = 10x + y + 27\text{,}$ so $9y - 9x = 27$ , and dividing by $9$ , $-x + y = 3$ $\quad$ (2).

Add (1) and (2): $2y = 12$ , $y = 6$ . Then $x = 3$ .

Check: the number is $36$ ; $3 + 6 = 9$ ; reversed it is $63$ , and $63 - 36 = 27$ . Both digits are from $0$ to $9$ .

The original number is 36.

Example 9: A fraction that changes

When 1 is added to both the numerator and the denominator of a fraction, it becomes $\frac{1}{2}$ . When 1 is subtracted from both, it becomes $\frac{1}{3}$ . Find the fraction.

Solution

Let the fraction be $\frac{x}{y}$ .

$\frac{x + 1}{y + 1} = \frac{1}{2}$ : $2(x + 1) = y + 1$ , so $2x - y = -1$ $\quad$ (1).

$\frac{x - 1}{y - 1} = \frac{1}{3}$ : $3(x - 1) = y - 1$ , so $3x - y = 2$ $\quad$ (2).

Subtract (1) from (2): $x = 3$ . Then from (1), $6 - y = -1$ , so $y = 7$ .

Check: $\frac{3 + 1}{7 + 1} = \frac{4}{8} = \frac{1}{2}\text{;}$ $\frac{3 - 1}{7 - 1} = \frac{2}{6} = \frac{1}{3}\text{.}$

The fraction is $\frac{3}{7}$ .

Shapes, Charges and Speeds

Example 10: A rectangle

The perimeter of a rectangle is 46 cm. Its length is 5 cm more than its width. Find its length, its width and its area.

Solution

Let the length be $x$ cm and the width $y$ cm.

Perimeter: $2x + 2y = 46$ , so $x + y = 23$ $\quad$ (1). Length 5 cm more: $x - y = 5$ $\quad$ (2).

Add: $2x = 28$ , $x = 14$ . Then $y = 9$ .

Check: $2(14) + 2(9) = 28 + 18 = 46$ cm; $14 - 9 = 5$ cm.

Area $= 14 \times 9 = 126$ cm $^{2}$ .

The length is 14 cm, the width 9 cm and the area $126$ cm $^{2}$ .

Example 11: A fixed charge and a rate

A plumber charges a fixed call-out fee plus a rate for each hour of work. A 2-hour job costs $ $130$ and a 5-hour job costs $ $250$ . Find the call-out fee and the hourly rate, and the cost of a $3\frac{1}{2}$ -hour job.

Solution

Let the call-out fee be $ $x$ and the rate $ $y$ per hour.

2 hours: $x + 2y = 130$ $\quad$ (1). 5 hours: $x + 5y = 250$ $\quad$ (2).

Subtract (1) from (2): $3y = 120$ , so $y = 40$ . Then $x = 130 - 80 = 50$ .

Check: $50 + 2 \times 40 = 130$ ; $50 + 5 \times 40 = 250$ .

A $3\frac{1}{2}$ -hour job costs $ $50$ + $3.5 \times$ $ $40$ = $ $50$ + $ $140$ = $ $190$ .

The call-out fee is $ $50$ and the rate is $ $40$ an hour; a $3\frac{1}{2}$ -hour job costs $ $190$ .

Example 12: With and against the current

A boat travels 36 km downstream in 2 hours, and the same 36 km back upstream in 3 hours. Find the speed of the boat in still water and the speed of the current.

Solution

Let the speed of the boat in still water be $x$ km/h and the speed of the current $y$ km/h. Downstream the current helps: speed $x + y$ . Upstream it slows the boat: speed $x - y$ .

Speed $=$ distance $\div$ time. Downstream: $x + y = 36 \div 2 = 18$ $\quad$ (1). Upstream: $x - y = 36 \div 3 = 12$ $\quad$ (2).

Add: $2x = 30$ , $x = 15$ . Then $y = 3$ .

Check: downstream at $18$ km/h, 36 km takes 2 hours; upstream at $12$ km/h, 36 km takes 3 hours.

The boat travels at 15 km/h in still water, and the current flows at 3 km/h.

Common Mistakes

Not saying what the letters stand for, or forgetting the units. "Let $x$ be the price of a pen in dollars" makes every later step clear.
Using one fact twice. Two equations must come from two different facts; "2 pens and 2 pencils cost $ $6$ " and "1 pen and 1 pencil cost $ $3$ " are the same fact.
Ages: adding the years to one person only. In 5 years' time both are 5 years older: $x + 5$ and $y + 5$ .
Digits: writing the number as $xy$ . It is $10x + y$ .
Mixing units, such as dollars and cents in one equation. Change everything to the same unit first.
Checking only in your equations. If an equation was written wrongly, the answer still fits it. Check against the words.

Practice Questions

Define your letters, write two equations, solve, and check against the words before you open each solution. The questions get harder as you go.

Question 1

Write a pair of simultaneous equations for each. Do not solve them. (a) 5 apples and 3 pears cost $ $4.10$ ; 2 apples and 4 pears cost $ $3.60$ . (b) A farmer has 50 ducks and goats, with 140 legs altogether. (c) The perimeter of a rectangle is 30 cm and its length is twice its width.

ShowHide solution

(a) Let an apple cost $ $x$ and a pear $ $y$ : $5x + 3y = 4.1$ and $2x + 4y = 3.6$ .

(b) Let there be $x$ ducks and $y$ goats: $x + y = 50$ and $2x + 4y = 140$ .

(a) $5x + 3y = 4.1$ , $2x + 4y = 3.6$ (b) $x + y = 50$ , $2x + 4y = 140$ (c) $2x + 2y = 30$ , $x = 2y$

Question 2

The sum of two numbers is 56. One number is 8 less than three times the other. Find the numbers.

ShowHide solution

Let the numbers be $x$ and $y$ : $x + y = 56$ and $x = 3y - 8$ .

Substitute: $3y - 8 + y = 56$ , so $4y = 64$ , $y = 16$ . Then $x = 40$ .

Check: $40 + 16 = 56$ ; three times $16$ is $48$ , and $8$ less is $40$ .

The numbers are 40 and 16.

Question 3

At a canteen, 4 curry puffs and 3 drinks cost $ $10.10$ , and 2 curry puffs and 5 drinks cost $ $10.30$ . Find the price of a curry puff and of a drink.

ShowHide solution

Let a curry puff cost $ $x$ and a drink $ $y$ : $4x + 3y = 10.1$ (1) and $2x + 5y = 10.3$ (2).

(2) $\times 2$ : $4x + 10y = 20.6$ . Subtract (1): $7y = 10.5$ , $y = 1.5$ . Then $4x = 10.1 - 4.5 = 5.6\text{,}$ $x = 1.4$ .

Check: $4(1.40) + 3(1.50) = 5.60 + 4.50 = 10.10\text{;}$ $2(1.40) + 5(1.50) = 2.80 + 7.50 = 10.30\text{.}$

A curry puff costs $ $1.40$ and a drink $ $1.50$ .

Question 4

A father is 30 years older than his son. In 5 years' time, the father will be 3 times as old as his son. How old are they now?

ShowHide solution

Let the father be $x$ and the son $y$ years old now: $x - y = 30$ and $x + 5 = 3(y + 5)$ , that is $x = 3y + 10$ .

Substitute: $3y + 10 - y = 30$ , so $2y = 20$ , $y = 10$ . Then $x = 40$ .

Check: $40 - 10 = 30$ ; in 5 years they will be $45$ and $15$ , and $45 = 3 \times 15$ .

The father is 40 and the son is 10.

Question 5

A group of 12 people visited a bird park. Adult tickets cost $ $35$ and child tickets $ $25$ , and the group paid $ $370$ . (a) How many adults and how many children were there? (b) Could the group have paid $ $375$ ? Explain.

ShowHide solution

(a) Let there be $x$ adults and $y$ children: $x + y = 12$ and $35x + 25y = 370$ , or dividing by $5$ , $7x + 5y = 74$ .

From the first, $y = 12 - x$ : $7x + 60 - 5x = 74$ , so $2x = 14$ , $x = 7$ , $y = 5$ .

Check: $7 + 5 = 12$ people; $7$ adults at $ $35$ and $5$ children at $ $25$ pay $ $245$ + $ $125$ = $ $370$ .

(b) With $ $375$ : $7x + 5y = 75$ , so $2x = 15$ and $x = 7.5$ . Half a person is not possible, so the group could not have paid $ $375$ .

(a) 7 adults and 5 children. (b) No: it would need $7\frac{1}{2}$ adults.

Question 6

Mr Lee has only $ $2$ notes and $ $10$ notes, 23 notes in all, worth $ $110$ . How many of each does he have?

ShowHide solution

Let him have $x$ $ $2$ notes and $y$ $ $10$ notes: $x + y = 23$ and $2x + 10y = 110$ , or $x + 5y = 55$ .

Subtract the first from $x + 5y = 55$ : $4y = 32$ , $y = 8$ . Then $x = 15$ .

Check: $15 + 8 = 23$ notes; fifteen $ $2$ notes and eight $ $10$ notes make $ $30$ + $ $80$ = $ $110$ .

Fifteen $ $2$ notes and eight $ $10$ notes.

Question 7

A farmer keeps ducks and goats. There are 40 heads and 112 legs. How many ducks and how many goats are there?

ShowHide solution

Let there be $x$ ducks and $y$ goats: $x + y = 40$ and $2x + 4y = 112$ .

First $\times 2$ : $2x + 2y = 80$ . Subtract from the second: $2y = 32$ , $y = 16$ . Then $x = 24$ .

Check: $24 + 16 = 40$ heads; $48 + 64 = 112$ legs.

24 ducks and 16 goats.

Question 8

In a two-digit number, the tens digit is twice the units digit. When the digits are reversed, the number decreases by 36. Find the number.

ShowHide solution

Let the tens digit be $x$ and the units digit $y$ : $x = 2y$ , and $(10x + y) - (10y + x) = 36\text{,}$ so $9x - 9y = 36$ , $x - y = 4$ .

Substitute $x = 2y$ : $2y - y = 4$ , $y = 4$ . Then $x = 8$ .

Check: the number is $84$ ; $8 = 2 \times 4$ ; reversed it is $48$ , and $84 - 48 = 36$ .

The number is 84.

Question 9

If 3 is subtracted from the numerator of a fraction, it becomes $\frac{1}{4}$ . If 2 is added to its denominator instead, it becomes $\frac{1}{2}$ . Find the fraction.

ShowHide solution

Let the fraction be $\frac{x}{y}$ : $\frac{x - 3}{y} = \frac{1}{4}$ gives $4(x - 3) = y$ , so $y = 4x - 12$ ; and $\frac{x}{y + 2} = \frac{1}{2}$ gives $2x = y + 2$ .

Substitute: $2x = 4x - 12 + 2$ , so $-2x = -10$ , $x = 5$ . Then $y = 20 - 12 = 8$ .

Check: $\frac{5 - 3}{8} = \frac{2}{8} = \frac{1}{4}\text{;}$ $\frac{5}{8 + 2} = \frac{5}{10} = \frac{1}{2}\text{.}$

The fraction is $\frac{5}{8}$ .

Question 10

In triangle $ABC$ , $\angle A = (2x + y)^\circ$ , $\angle B = (x + 2y)^\circ$ and $\angle C = 30^\circ$ . $\angle A$ is $10^\circ$ larger than $\angle B$ . Find $x$ , $y$ and the size of each angle.

ShowHide solution

Angle sum of a triangle: $(2x + y) + (x + 2y) + 30 = 180\text{,}$ so $3x + 3y = 150$ and $x + y = 50$ .

$\angle A - \angle B = 10$ : $(2x + y) - (x + 2y) = 10\text{,}$ so $x - y = 10$ .

Add: $2x = 60$ , $x = 30$ . Then $y = 20$ .

$\angle A = 60 + 20 = 80^\circ\text{,}$ $\angle B = 30 + 40 = 70^\circ\text{.}$ Check: $80 + 70 + 30 = 180$ , and $80 - 70 = 10$ .

$x = 30$ , $y = 20$ ; $\angle A = 80^\circ$ , $\angle B = 70^\circ$ , $\angle C = 30^\circ$ .

Question 11

A taxi fare is a fixed flag-down charge plus a charge for each kilometre. A 6 km ride costs $ $7.60$ and a 10 km ride costs $ $10$ . Find the flag-down charge, the charge per kilometre, and the fare for a 15 km ride.

ShowHide solution

Let the flag-down charge be $ $x$ and the charge $ $y$ per km: $x + 6y = 7.6$ and $x + 10y = 10$ .

Subtract: $4y = 2.4$ , $y = 0.6$ . Then $x = 7.6 - 3.6 = 4$ .

Check: $4 + 6(0.60) = 7.60$ ; $4 + 10(0.60) = 10$ .

For 15 km: $ $4$ + $15 \times$ $ $0.60$ = $ $4$ + $ $9$ = $ $13$ .

The flag-down charge is $ $4$ , the charge is 60 cents per km, and a 15 km ride costs $ $13$ .

Question 12

Aisha cycles at a steady speed and walks at a steady speed. Cycling for 2 hours and then walking for 1 hour, she covers 33 km. Cycling for 1 hour and walking for 2 hours, she covers 24 km. Find her cycling speed and her walking speed.

ShowHide solution

Let her cycle at $x$ km/h and walk at $y$ km/h. Distance $=$ speed $\times$ time: $2x + y = 33$ and $x + 2y = 24$ .

Add: $3x + 3y = 57$ , so $x + y = 19$ . Subtract the second from the first: $x - y = 9$ .

Add these: $2x = 28$ , $x = 14$ . Then $y = 5$ .

Check: $2 \times 14 + 5 = 33$ km; $14 + 2 \times 5 = 24$ km. Both speeds are sensible.

She cycles at 14 km/h and walks at 5 km/h.

Key Points

Define both unknowns in words, with units, before writing any equation.
Two different facts give two equations; solve them by substitution or elimination.
Count and value (tickets, coins, notes; heads and legs) give one equation each.
Ages: use the ages now; $n$ years ago subtract $n$ from both, in $n$ years add $n$ to both.
A two-digit number is $10x + y$ ; reversed, $10y + x$ . A fraction is $\frac{x}{y}$ ; clear fractions by cross-multiplying.
Fixed charge plus rate: subtract the two bills to find the rate. Downstream speed is $x + y$ , upstream $x - y$ .
Check against the words, and make sure the answer makes sense: whole numbers of things, positive prices and speeds, digits from 0 to 9.
Answer the question asked, in a sentence, with units.