Word Problems in Algebra: A Step-by-Step Strategy

Many students can solve equations perfectly but freeze when the same maths is hidden inside a paragraph of words. The difficulty is almost never the algebra; it is translating English into an equation. This guide gives you a reliable strategy and applies it to the word problems that appear most often in Matric, FSc, O Level and entry tests.

The Five-Step Strategy

1. Read and understand. Read the problem twice. Identify what is given and what you are asked to find.
2. Define the variable. Write clearly, for example, "Let \(x\) = the son's present age." Express other unknowns in terms of \(x\).
3. Translate into an equation. Turn each sentence into algebra using key words.
4. Solve the equation. Use normal algebra steps.
5. Check and answer. Substitute back into the original words, not just your equation, and write a sentence with units.

Translating Key Words

Watch the order with "less than": "5 less than \(x\)" is \(x - 5\), not \(5 - x\).

Worked Example 1: Consecutive Integers

The sum of three consecutive integers is 87. Find the integers.

Step 1: We need three integers that follow each other.

Step 2: Let the smallest integer be \(n\). Then the others are \(n + 1\) and \(n + 2\).

Step 3: Translate:

$$ n + (n + 1) + (n + 2) = 87 $$

Step 4: Solve:

$$ 3n + 3 = 87 \quad \Rightarrow \quad 3n = 84 \quad \Rightarrow \quad n = 28 $$

Step 5: The integers are 28, 29 and 30. Check: \(28 + 29 + 30 = 87\). Correct.

Worked Example 2: Age Problem

A father is 4 times as old as his son. In 5 years, he will be 3 times as old as his son. Find their present ages.

Step 1: Two people, two time points (now and in 5 years).

Step 2: Let the son's present age be \(s\). Then the father's present age is \(4s\).

Step 3: In 5 years: son \(= s + 5\), father \(= 4s + 5\). The condition gives:

$$ 4s + 5 = 3(s + 5) $$

Step 4: Solve:

$$ 4s + 5 = 3s + 15 \quad \Rightarrow \quad s = 10 $$

Step 5: The son is 10 and the father is 40. Check: In 5 years they will be 15 and 45, and \(45 = 3 \times 15\). Correct.

Tip: Age problems become much easier if you make a small "now / later" list for each person before writing the equation.

Worked Example 3: Work Rate Problem

Ali can paint a room in 6 hours, and Usman can paint it in 3 hours. How long will they take working together?

Step 1: Work problems use rates: the fraction of the job done per hour.

Step 2: Let \(t\) = hours taken together. Ali's rate \(= \frac{1}{6}\) room per hour; Usman's rate \(= \frac{1}{3}\) room per hour.

Step 3: Combined rate \(\times\) time \(=\) 1 whole job:

$$ \left(\frac{1}{6} + \frac{1}{3}\right)t = 1 $$

Step 4: Simplify:

$$ \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2} \quad \Rightarrow \quad \frac{t}{2} = 1 \quad \Rightarrow \quad t = 2 $$

Step 5: Together they need 2 hours. Check: In 2 hours Ali paints \(\frac{2}{6} = \frac{1}{3}\) and Usman paints \(\frac{2}{3}\) of the room; \(\frac{1}{3} + \frac{2}{3} = 1\). Correct.

Worked Example 4: Speed, Distance and Time

A car leaves Lahore at 60 km/h. Two hours later, a second car leaves from the same place on the same road at 90 km/h. How long after the second car starts will it catch the first?

Step 1: Use distance \(= \text{speed} \times \text{time}\). When the second car catches up, both have travelled the same distance.

Step 2: Let \(t\) = hours travelled by the second car. The first car has then travelled \(t + 2\) hours.

Step 3: Equal distances:

$$ 90t = 60(t + 2) $$

Step 4: Solve:

$$ 90t = 60t + 120 \quad \Rightarrow \quad 30t = 120 \quad \Rightarrow \quad t = 4 $$

Step 5: The second car catches up after 4 hours. Check: second car \(90 \times 4 = 360\) km; first car \(60 \times 6 = 360\) km. Correct.

Worked Example 5: Area Problem Leading to a Quadratic

The length of a rectangle is 3 m more than its width, and its area is 54 m². Find its dimensions.

Step 2: Let width \(= w\). Then length \(= w + 3\).

Step 3: Area \(=\) length \(\times\) width:

$$ w(w + 3) = 54 $$

Step 4: Rearrange and factorise:

$$ w^2 + 3w - 54 = 0 \quad \Rightarrow \quad (w + 9)(w - 6) = 0 $$

So \(w = 6\) or \(w = -9\). A width cannot be negative, so reject \(-9\).

Step 5: Width \(= 6\) m, length \(= 9\) m. Check: \(6 \times 9 = 54\) and \(9 - 6 = 3\). Correct.

Tips

• Draw a diagram for geometry, distance and mixture problems.
• Use a table with columns like "now / later" or "speed / time / distance" to organise information.
• Keep units consistent: convert minutes to hours or grams to kilograms before forming the equation.
• Reject unrealistic answers, such as negative lengths, ages or fractional numbers of people.
• Answer the actual question. If it asks for the length, don't stop at the width.

Common Mistakes

• Defining the variable vaguely, like "Let \(x\) = age" without saying whose age or when.
• Reversing subtraction: "7 less than a number" is \(x - 7\).
• Adding times in work problems instead of adding rates.
• Checking only in the equation, which won't catch an equation you set up wrongly.

Key Takeaways

• Follow five steps: understand, define the variable, translate, solve, check.
• Write every unknown in terms of one variable whenever possible.
• Use standard models: rates for work problems, \(d = st\) for motion, length \(\times\) width for area.
• Always check your answer against the original wording and reject impossible values.

If a word problem in your homework or assignment has you stuck, send it to Amathify on WhatsApp or through the "Ask a Question" form. Our experts and PhD scholars will turn it into equations and solve it step by step for you.