Year 11 General Linear Equations And Their Graphs

Problem-Solving With Simultaneous Equations

13 practice questions 2 video lessons Theory + worked examples

Theory

Many word problems involve two unknowns and two pieces of information. Define your variables, write two equations from the wording, solve by substitution or elimination, then state the answer in context with units. Common templates: tickets/coins, mixtures, age, and break-even.

A simultaneous-equation word problem is one where a real-world situation gives you two unknown quantities and two independent facts about them. Each fact becomes one equation; solving the pair gives the two unknowns.

The translation step — turning English into algebra — is the most important part. Once the equations are written correctly, the algebra is the routine substitution or elimination you have already practised.

A break-even point is the value at which two cost (or revenue) expressions are equal — the crossing point of two linear models. It is usually set up as a single equation in one unknown rather than a full simultaneous pair.

The first diagram is a translation table showing how common pieces of word-problem English become algebraic equations. The second is the four-step workflow you should follow for every problem of this type.

Common word-problem phrases and the equations they produce. Define variables first.

The four-step method: define, equations, solve, answer in words.

There are no new formulas. What you need is a small set of templates: for each common problem type, here is the pair of equations to write.

Problem type	Equation 1	Equation 2
Tickets / item-mix	total count of items	total cost or value
Coin tin	total number of coins	total dollar value
Mixture (% solution)	total amount of mixture	total of the active ingredient
Age	relationship between ages NOW	relationship at another time
Break-even (two plans)	cost expression for plan A	cost expression for plan B, set equal

Variable tip. Always define your variables with units — for example "let a = number of adult tickets" or "let d = data used, in MB". A mistake in units is a common source of wrong answers.

The four-step method

Define the variables. Pick the two quantities the question is asking about, give them letters, and state them with units in one short sentence.
Write two equations. Use one piece of information per equation. Translate the wording carefully.
Solve by substitution or elimination — whichever fits the form of the equations.
Answer in context. Write a sentence with units that explicitly says what each number represents. If the question asks for one specific value, state that value clearly.

A quick checklist before you start solving

Did I define both variables with units?
Does each equation come from a distinct fact in the problem?
Are the equations in consistent units (e.g. all dollars, not mixed dollars and cents)?
For age problems — am I using present ages as my variables and adjusting for the other time?
For break-even — am I setting the two cost expressions equal rather than treating them as a system?

EXAMPLE 1 — TICKET MIX

A cinema sells adult tickets for

$ 18

and child tickets for

$ 12

. One evening

13

tickets sold for

$ 186

in total. Find how many of each were sold.

SOLUTION

Let $a$ = number of adult tickets, $c$ = number of child tickets.

From the totals:

$a + c$	$=$	$13$
$18 a + 12 c$	$=$	$186$

Multiply the first equation by $12$ and subtract from the second to eliminate $c$ :

$12 a + 12 c$	$=$	$156$
$6 a$	$=$	$30$
$a$	$=$	$5$

Back-substitute: $c = 13 - 5 = 8$ .

Answer: $5$ adult tickets and $8$ child tickets were sold.

a = 5, c = 8

EXAMPLE 2 — COIN TIN

A money box has

24

coins, all 5-cent or 20-cent, totalling

$ 3.30

. Find the number of each.

SOLUTION

Let $f$ = number of 5-cent coins, $t$ = number of 20-cent coins.

$f + t$	$=$	$24$
$0.05 f + 0.20 t$	$=$	$3.30$

Multiply the first by $0.05$ and subtract from the second to eliminate $f$ :

$0.05 f + 0.05 t$	$=$	$1.20$
$0.15 t$	$=$	$2.10$
$t$	$=$	$14$

Back-substitute: $f = 24 - 14 = 10$ .

Answer: the box contains $10$ five-cent coins and $14$ twenty-cent coins.

f = 10, t = 14

EXAMPLE 3 — AGE PROBLEM

The sum of Anna and Ben's current ages is

50

. In

5

years, Anna will be twice as old as Ben. Find their current ages.

SOLUTION

Let $A$ and $B$ be Anna's and Ben's current ages, in years.

$A + B$	$=$	$50$
$A + 5$	$=$	$2 (B + 5)$

Expand the second equation: $A + 5 = 2 B + 10$ , giving $A = 2 B + 5$ . Substitute into the first:

$(2 B + 5) + B$	$=$	$50$
$3 B + 5$	$=$	$50$
$3 B$	$=$	$45$
$B$	$=$	$15$

Back-substitute: $A = 2 (15) + 5 = 35$ .

Answer: Anna is currently $35$ years old and Ben is currently $15$ years old.

A = 35, B = 15

EXAMPLE 4 — BREAK-EVEN

Plan A costs

$ 40

plus

$ 0.10

per MB. Plan B costs

$ 20

plus

$ 0.20

per MB. At what data usage do the two plans cost the same, and what is that cost?

SOLUTION

Let $d$ = data used, in MB. Set the two cost expressions equal:

$40 + 0.10 d$	$=$	$20 + 0.20 d$
$40 - 20$	$=$	$0.20 d - 0.10 d$
$20$	$=$	$0.10 d$
$d$	$=$	$200$

Substitute $d = 200$ into either cost expression: $40 + 0.10 (200) = 60$ .

Answer: at $200$ MB of data, both plans cost $$ 60$ .

d = 200 MB, cost = $ 60

Common pitfalls

Mixture problems need two equations. One equation for total amount of the mixture, one for the amount of the active ingredient (acid, gold, salt). Don't try to do mixture problems with a single equation.

Age problems use present ages as variables. If the problem says "eight years ago", that means

A - 8

and

B - 8

— you subtract from the present age, you don't make the past age your variable.

Break-even is one equation, not two. "When do plans cost the same?" sets the two cost expressions equal — that's one equation in one unknown. You only need a full simultaneous pair when there are two genuinely unknown quantities.

Don't forget the units. Mixing cents and dollars in the same equation (

5 f + 20 t = 3.30

instead of

0.05 f + 0.20 t = 3.30

) is a classic error. Keep all money in one unit.

Always answer in words. A bare

(5, 8)

isn't a complete answer to a word problem. Write a sentence saying what each number represents, with units, exactly as the question asked.

Frequently asked questions

How do I know to use simultaneous equations for a word problem?

Use simultaneous equations when the problem has two unknown quantities and gives two distinct pieces of information about them. A clue is wording like 'a total of...' or 'altogether...' combined with a second condition such as a price total, an age relationship, or a percentage.

How do I define the variables?

Pick the two unknown quantities the question is actually asking about. Write 'Let x = ... and y = ...' with units, in one short sentence at the very start of the solution. Clear definitions make the equations much easier to write.

Why do mixture problems need two equations, not one?

A mixture has both a total amount (litres, grams) and a total of the active ingredient (acid, gold, salt). Each gives a different equation. One equation alone cannot determine both unknowns.

How do age problems work?

Let the variables be the people's ages NOW. To talk about an earlier or later time, add or subtract the number of years from the present age. For example, 'in 5 years' becomes x + 5 and y + 5, and 'eight years ago' becomes x - 8 and y - 8.

How are break-even problems different from the others?

Break-even problems give you a cost expression for each plan and ask when the two costs are equal. You only need to set the two expressions equal — that's one equation in one unknown, not a full pair of simultaneous equations.

Why does my answer have to include units and context?

A pair like (5, 8) is meaningless on its own — the marker can't tell what the numbers refer to. Always restate the answer in the words of the question: '5 adult tickets and 8 child tickets', '$60 at 200 MB', and so on. Marks are often allocated specifically for the contextual answer.