Equation Problems
Equation Problems turns word problems into equations: choose a pronumeral for the unknown, write an equation from the information, solve it and answer in words. This NSW Year 9 Advanced maths subtopic practises that process (NESA outcome MA5‑EQU‑C‑01).
You will solve problems about numbers and consecutive integers, angles and perimeters, ages, shared money, taxi fares and ticket sales. You will find the distance to school from a total travel time, how much juice to add to a drink and the area of a fenced paddock, and explain what a negative answer to an age problem means. It is part of Year 9 Advanced Equations.
Theory
An equation problem (word problem) is solved by choosing a pronumeral for one unknown, writing every other quantity in terms of it, forming an equation and solving it. This Year 9 Advanced maths page covers consecutive numbers, perimeters, ages, money, ratios, mixtures and speed, distance and time, and how to interpret the answer in context.
Every word problem has the same shape. Choose a pronumeral for one unknown and write down exactly what it stands for, with units. Write the other unknowns as expressions in that pronumeral. Find the sentence that says two things are equal, and write it as an equation.
Choose the pronumeral for the quantity the others are described from. If Max has \(\$6\) more than Kim and Lee has twice as much as Kim, let Kim's amount be \(k\).
Some facts are not written in the question but are needed: the angles of a triangle add to \(180^\circ\), a perimeter adds all the sides, time \(= \dfrac{\text{distance}}{\text{speed}}\), and consecutive integers differ by \(1\).
Solving the equation is not the end. Interpret the solution: answer the question that was asked, with units, and check it against the story. A negative time can mean "in the past"; a non-whole number of people means a mistake.
Facts that often give the equation:
| Situation | Use |
|---|---|
| perimeter of a rectangle | \(P = 2(l + w)\) |
| angles of a triangle | sum \(= 180^\circ\) |
| speed, distance, time | \(d = st\), \(t = \dfrac{d}{s}\) |
| ages in \(n\) years | add \(n\) to every age |
| items of two kinds, \(N\) in total | \(x\) of one kind, \(N - x\) of the other |
| a mixture | amount of pure ingredient \(=\) percentage \(\times\) total amount |
How to solve a word problem
- Define the pronumeral in a sentence, with units: "Let \(c\) be the number of coffees."
- Express every other unknown in terms of it. A sketch or bar model helps.
- Write the equation from the fact that links them (a total, an equal amount, a known formula).
- Solve it.
- Answer the question in a sentence, with units, then check the numbers in the original story.
Common set-ups
- Ratio \(a : b\): let the amounts be \(ak\) and \(bk\).
- Ages: write a table of now and later (or earlier); every person's age changes by the same amount.
- Mixtures: track the pure ingredient. If \(x\) L of pure juice is added to a drink, the juice goes up by \(x\) and so does the total volume.
- Costs from lengths: write each length in terms of the pronumeral first, then multiply by the price per metre.
- Change and totals: money paid \(=\) cost \(+\) change.
Let the smallest be \(n\). Consecutive odd numbers go up by \(2\): \(n\), \(n + 2\), \(n + 4\).
| \(n + (n + 2) + (n + 4)\) | \(=\) | \(159\) |
| \(3n + 6\) | \(=\) | \(159\) |
| \(3n\) | \(=\) | \(153\) |
| \(n\) | \(=\) | \(51\) |
The numbers are \(\mathbf{51}\), \(\mathbf{53}\) and \(\mathbf{55}\). Check: \(51 + 53 + 55 = 159\).
Let the brother be \(b\) years old now, so Sara is \(3b\). In \(6\) years they will be \(b + 6\) and \(3b + 6\). Both ages increase by \(6\).
| \(3b + 6\) | \(=\) | \(2(b + 6)\) |
| \(3b + 6\) | \(=\) | \(2b + 12\) |
| \(b\) | \(=\) | \(6\) |
The brother is \(\mathbf{6}\) and Sara is \(\mathbf{18}\). Check: in \(6\) years they are \(12\) and \(24\), and \(24 = 2 \times 12\).
Let \(c\) be the number of coffees. The rest are teas: \(64 - c\).
| \(4.5c + 3.5(64 - c)\) | \(=\) | \(262\) |
| \(4.5c + 224 - 3.5c\) | \(=\) | \(262\) |
| \(c + 224\) | \(=\) | \(262\) |
| \(c\) | \(=\) | \(38\) |
\(\mathbf{38}\) coffees (and \(26\) teas). Check: \(38 \times 4.50 + 26 \times 3.50\) \(= 171 + 91\) \(= \$262\).
Let the distance be \(d\) km. Time \(= \dfrac{\text{distance}}{\text{speed}}\), and \(4\) h \(30\) min \(= 4.5\) h.
| \(\dfrac{d}{80} + \dfrac{d}{100}\) | \(=\) | \(4.5\) |
| \(5d + 4d\) | \(=\) | \(1800 \quad (\times 400)\) |
| \(9d\) | \(=\) | \(1800\) |
| \(d\) | \(=\) | \(200\) |
They live \(\mathbf{200}\) km away. Check: \(\dfrac{200}{80} + \dfrac{200}{100} = 2.5 + 2 = 4.5\) hours.
Common pitfalls
Frequently asked questions
How do you turn a word problem into an equation?
Choose a letter for one unknown and say what it stands for. Write every other quantity in terms of that letter. Then find the fact that says two things are equal, such as a total, and write it as an equation.
Which unknown should be the pronumeral?
Usually the one the others are described from. If Jo has twice as much as Mia and Sam has 5 more than Mia, let Mia's amount be the pronumeral, so no fractions appear.
How do you write consecutive numbers in algebra?
Consecutive integers are \(n\), \(n + 1\), \(n + 2\). Consecutive even or odd numbers go up by \(2\): \(n\), \(n + 2\), \(n + 4\).
How do you solve age problems?
Let one person's age now be the pronumeral and write the others from it. For a future age add the same number of years to everyone, then use the sentence that compares them to write the equation.
How do you check the answer to a word problem?
Put the answer back into the words of the problem, not just into your equation. If every condition in the story is true, the answer is right, even if the equation had an error you did not notice.