Turning words into equations · Linear & simultaneous equations

Many WAEC theory questions are word problems about ages, money, tickets or journeys. The algebra in them is usually easy. The marks are lost earlier: forming the equation. This lesson is about that step.

A routine that works every time

Name the unknowns. Write “Let … be $x$ ” and say exactly what $x$ is, with units: “Let Manya’s age now be $x$ years.”
Use a different letter for each different unknown, or write the others in terms of the first.
Turn each sentence of information into an equation. One fact, one equation.
Solve.
Answer the question asked, with units, and check it makes sense.

Translating phrases

Words	Algebra
5 more than $x$	$x + 5$
5 less than $x$	$x - 5$
twice the sum of $x$ and 3	$2(x + 3)$
$x$ is three times $y$	$x = 3y$
the sum of two numbers is three times their difference	$x + y = 3(x - y)$
7 years ago	take 7 from every person’s age
in 4 years’ time	add 4 to every person’s age

Ages

Worked example · WAEC 2018

WAEC 2018 · Paper 2 · Q2

Musa is three years older than Manya. Seven years ago, Musa was twice as old as Manya. How old are they now?

In how many years will the sum of their ages be 45?

Name the unknown
Let Manya’s age now be $x$ years. Musa is three years older: $x + 3$ years.
Think first. If Manya is $x$ years old now, how old is Musa?
Seven years ago
Take 7 from both ages: Manya was $x - 7$ and Musa was $x + 3 - 7 = x - 4$ .
Think first. Write their ages seven years ago, then the sentence as an equation.
Form the equation
Musa was twice as old as Manya:
$x - 4 = 2(x - 7)$
Solve
$x - 4 = 2x - 14 \quad\Rightarrow\quad x = 10$
Manya is 10 and Musa is 13. Check: seven years ago they were 3 and 6, and 6 is twice 3 ✓.
(b) In n years' time
$(10 + n) + (13 + n) = 45 \Rightarrow 23 + 2n = 45 \Rightarrow n = 11$ . In 11 years.
Think first. Both ages go up by $n$ . What's the equation?

More: ages

Two kinds of thing: tickets, prices, coins

When there are two unknown quantities, you usually get one equation from how many and one from how much:

how many: children $+$ adults $= 250$
how much: $2 \times$ children $+ 4 \times$ adults $= 700$

Then solve them as simultaneous equations.

More: money, tickets and prices

Journeys: distance, speed, time

Use $\text{time} = \dfrac{\text{distance}}{\text{speed}}$ for each part of the journey, then add the times.

Speed, distance, timeCover the one you want: T = D ÷ S for each part

More: journeys

More: numbers, fractions and shapes

Your turn

WAEC 2023 · Paper 2 · Q2

In a football match, the tickets for children and adults were sold at D3.00 and D5.00 respectively. 400 people attended the match and D1,700.00 was collected in ticket sales.

(a)
How many tickets were sold to adults?
(b)
Mr. Sonko sold 250 tickets. If 175 of the tickets were for adults, how much sales (in D) did he make altogether?

Worked solution (try it first)

(a)

Let $c$ children's tickets and $a$ adults' tickets be sold.
How many: $c + a = 400$ (1).
How much: $3c + 5a = 1700$ (2).
Multiply (1) by 3: $3c + 3a = 1200$ (3).
Take (3) from (2): $2a = 500$ , so $a = 250$ . 250 tickets were sold to adults.

(b)

Of his 250 tickets, 175 were for adults, so $250 - 175 = 75$ were for children.
Sales $= 75 \times 3 + 175 \times 5$
$= 225 + 875$
$=$ D1,100.00.

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