To form an equation from words, define the unknown, translate each phrase into algebra, then join the pieces with ”=” using the fact the question gives you. Forming equations appears in Core and Extended papers as perimeter, age, cost and number problems.
It uses the solving skills from brackets and fractions, and leads into simultaneous relationships.
What is the idea behind it?
A sentence carries two jobs. Some phrases describe one quantity in terms of another (“3 more than twice the width”). One phrase gives a fact that two expressions must satisfy (“the perimeter is 54 cm”).
The first job defines expressions, the second creates the equation.
Write the relationship in words first, then replace each word with its algebra.
How to do it, step by step
- Define the unknown with a letter and units: “let w = width in cm”.
- Write the other quantities in terms of w.
- Find the fact that links things: a total, a difference or an equal cost.
- Build the equation, using brackets where needed.
- Solve.
- Check against the words and give the answer in context.
Worked example
A rectangle’s length is 3 cm more than twice its width. Its perimeter is 54 cm. Find its width and length.
Step 1, define: let w = width in cm.
Step 2, other quantity: length = 2w + 3.
Step 3, fact: perimeter = 2 × (length + width) = 54.
Step 4, equation: 2(w + 2w + 3) = 54, so 2(3w + 3) = 54.
Step 5, solve: 6w + 6 = 54, so 6w = 48 and w = 8.
Step 6, read back: width 8 cm, length 2 × 8 + 3 = 19 cm. Perimeter 2 × (8 + 19) = 54 cm. ✓
The mistake to watch for
The usual slip is bracketing in the wrong place, so the phrase means something else.
Mistaken working: length = 2(w + 3), which is “twice 3 more than the width”.
That gives 2(w + 2(w + 3)) = 54, so 3w + 6 = 27, w = 7 and length 20.
Reading back against the words shows the problem: twice 7 is 14, and 3 more is 17, not 20. The equation was solved correctly, but it described a different rectangle.
The correction is to build the phrase in order: “twice the width” is 2w, then “3 more” adds 3.
Check yourself
1. Five times a number, minus 8, equals twice the number plus 13. Find the number.
Show answer
Let the number be n. Then 5n − 8 = 2n + 13. Subtract 2n: 3n − 8 = 13. Add 8: 3n = 21. So n = 7.
Check: 5 × 7 − 8 = 27 and 2 × 7 + 13 = 27. ✓
2. An adult ticket costs RM12 and a child ticket costs RM7. A group buys 3 more child tickets than adult tickets and pays RM173. How many adult tickets did they buy?
Show answer
Let a = number of adult tickets, so child tickets = a + 3. Cost: 12a + 7(a + 3) = 173. Expand: 12a + 7a + 21 = 173. So 19a = 152 and a = 8.
Check: 8 adult and 11 child tickets cost 96 + 77 = RM173. ✓
3. Three consecutive integers add up to 87. Find them.
Show answer
Let the smallest be n. The others are n + 1 and n + 2. So n + (n + 1) + (n + 2) = 87, giving 3n + 3 = 87 and n = 28.
The integers are 28, 29 and 30. Check: 28 + 29 + 30 = 87. ✓
Where this leads next
Finish the module with checking a solution in the original relationship, which is the last step of every problem like these. Then try the equations and formulas practice set.
If you can solve equations but freeze on word problems, online one-to-one Mathematics tuition gives you a teacher who can hear how you read the sentence and adjust from there.