First, we need to translate word problems into equations with variables. Then, we need to solve the equation(s) to find the solution(s) to the word problem.

### Example 1

**1. **What number when added to 8 gives 15?**2.** Segun thinks of a number, subtracts 7 from it and the result is 11. What is the number?**3.** When a number is multiplied by 4, the result is 24. What is the number?

**Solution**

**1.** Let the number added to 8 that gives 15 be x

i.e. \( \scriptsize x \: + \: 8 = 15 \)

This statement is true if x = 7

The number added is 7

**âˆ´** \( \scriptsize 7 \: + \: 8 = 15 \: is \: true \)

**2. **Let y be the number you subtract 7 from to get 11.

i.e. \( \scriptsize y \: – \: 7 = 11 \)

The statement is true if y = 18

The number is 18

**âˆ´ **\( \scriptsize 18 \: – \: 7 = 11 \)

**3.** Let z be the number you multiply by 4 to get 24.

i.e. \( \scriptsize z \: \times \: 4 = 24 \)

This statement is true if z = 6

The number is 6.

### Example 2

A woman is twice as old as her daughter. The woman is x years old.**a.** How old is the daughter?**b.** How old were they 5 years ago?**c.** How old will the woman be in 15 years’ time?

**Solution**

**a. **If the woman is x years old, then her daughter is \( \frac{x}{2} \: \scriptsize years.\)

**b.** 5 years ago, the woman was (x – 5) years and the daughter was \( \left( \frac{x}{2} \scriptsize \: – \: 5 \right ) \: \scriptsize years\)

**c.** In 15 years’ time the woman will beÂ (xÂ +Â 15) years.

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