Simultaneous Equations Problems
Simultaneous equations problems turn a worded situation with two unknown quantities into two equations, then solve them together. In NSW Year 10 Advanced maths the situations include prices, ages, digits, fractions, speeds, rates and investments, and answers are checked against the original story (NESA outcome MA5‑EQU‑P‑02).
You will choose two pronumerals, write two equations from the information, and solve them by substitution or elimination. Problems include pizza prices, ute hire, ages, a fraction that changes, investments, a cycling and walking trip, an electricity bill, two pumps filling a tank, changing student numbers, two cyclists and ticket sales. You will also find angles in a parallelogram and decide when a pair has no solution or infinitely many. It is part of Year 10 Advanced Simultaneous Equations.
Every question with a fully worked solution.
- Simultaneous Equations Problems - Video - Simultaneous equations word problems Watch
Theory
A word problem with two unknowns needs two facts. Name each unknown with a pronumeral, turn each fact into an equation, solve the pair by substitution or elimination, and answer in words. This Year 10 Advanced maths page shows how to translate ages, digits, mixtures, percentages, charges, speeds and rates into equations, and how to check the answer against the story.
Modelling a problem means writing equations that say the same thing as the words. Each different fact in the question gives one equation. Two unknowns need two independent equations.
Define each pronumeral fully, with its unit and time: "let \(x\) be the price of an adult ticket, in dollars, on Friday". Vague definitions cause most errors.
After solving, interpret: does the answer make sense (a whole number of people, a positive speed, a digit from \(0\) to \(9\))? Does it fit every fact in the question?
If the two facts turn out to be the same equation in disguise (one is a multiple of the other), there are infinitely many solutions, and the problem does not fix a single answer.
Common translations:
| Words | Algebra |
|---|---|
| two-digit number, tens digit \(t\), units digit \(u\) | \(10t + u\); reversed \(10u + t\) |
| a fraction \(\dfrac{a}{b}\) with \(2\) added to top and bottom | \(\dfrac{a + 2}{b + 2}\) |
| \(15\%\) more than \(x\) | \(1.15x\) |
| \(x\) dollars invested at \(5\%\) p.a. for one year | interest \(0.05x\) |
| fixed charge \(F\) plus \(r\) per unit, \(n\) units | \(F + rn\) |
| time to travel \(d\) km at \(v\) km/h | \(\dfrac{d}{v}\) |
| two machines at \(p\) and \(q\) units per hour for \(h\) hours | \(h(p + q)\) |
Solving a word problem
- Read to the end. Find the two unknowns the question asks about (or that lead to the answer).
- Define a pronumeral for each, with units: "let \(b\) km/h be the boat's speed in still water".
- Write one equation for each fact. A table (people, now, in \(5\) years; or amount, price, value) helps organise the facts.
- Simplify each equation, then solve by substitution (if one is \(x = \ldots\)) or elimination.
- Check in the words of the question, not just in your equations.
- Answer in a sentence with units, and answer the part asked (this year's number, the total bill, the time together).
Tips for common types
- Ages: add or subtract the same years for everyone.
- Mixtures and tickets: one equation for quantities, one for money.
- Before and after a change: let the pronumerals be the original values, then multiply by the percentage factor.
- Working together: find each rate (litres per hour), then time \(=\) amount \(\div\) combined rate.
- Not every question is a story: equations with \(\dfrac{1}{x}\) and \(\dfrac{1}{y}\) (let \(u = \dfrac{1}{x}\), \(v = \dfrac{1}{y}\)), letters as coefficients (eliminate, then factorise and divide) and unknown coefficients from a known point (substitute the point to get two equations) use the same methods as the Substitution and Elimination pages.
- No solution or many: if eliminating gives \(0 = 5\), the facts contradict each other; \(0 = 0\) means one fact repeats the other.
Let Mia be \(a\) years old and her mother \(m\) years old now. Add or subtract the same number of years for both people.
| \(m + 4\) | \(=\) | \(3(a + 4) \quad\Rightarrow\quad m = 3a + 8 \quad (1)\) |
| \(m - 6\) | \(=\) | \(8(a - 6) \quad\Rightarrow\quad m = 8a - 42 \quad (2)\) |
Both equations give \(m\), so set them equal.
| \(8a - 42\) | \(=\) | \(3a + 8\) |
| \(5a\) | \(=\) | \(50\) |
| \(a\) | \(=\) | \(10\) |
| \(m\) | \(=\) | \(3(10) + 8 = 38\) |
Check: in \(4\) years they are \(14\) and \(42 = 3 \times 14\ \checkmark\); six years ago \(4\) and \(32 = 8 \times 4\ \checkmark\)
Mia is \(\mathbf{10}\) and her mother is \(\mathbf{38}\).
Let \(x\) kg be the \(\$24\) beans and \(y\) kg the \(\$36\) beans. One equation counts kilograms, the other counts dollars.
| \(x + y\) | \(=\) | \(30 \quad (1)\) |
| \(24x + 36y\) | \(=\) | \(28 \times 30 = 840 \quad (2)\) |
| \((1) \times 24:\quad 24x + 24y\) | \(=\) | \(720 \quad (3)\) |
| \((2) - (3):\quad 12y\) | \(=\) | \(120\) |
| \(y\) | \(=\) | \(10\) |
| \(x\) | \(=\) | \(20\) |
Check: \(24(20) + 36(10) = 480 + 360 = 840\ \checkmark\)
It uses \(\mathbf{20}\) kg of the \(\$24\) beans and \(\mathbf{10}\) kg of the \(\$36\) beans.
Let the boat's still-water speed be \(b\) km/h and the current \(c\) km/h. Downstream the current helps (\(b + c\)); upstream it slows the boat (\(b - c\)). Use distance \(=\) speed \(\times\) time.
| \(2(b + c)\) | \(=\) | \(36 \quad\Rightarrow\quad b + c = 18 \quad (1)\) |
| \(3(b - c)\) | \(=\) | \(36 \quad\Rightarrow\quad b - c = 12 \quad (2)\) |
| \((1) + (2):\quad 2b\) | \(=\) | \(30\) |
| \(b\) | \(=\) | \(15\) |
| \(c\) | \(=\) | \(18 - 15 = 3\) |
Check upstream: \(3 \times (15 - 3) = 36\ \checkmark\)
Boat \(\mathbf{15}\) km/h, current \(\mathbf{3}\) km/h.
(i) How many of each were sold last month?
(ii) How many coffees were sold this month?
(i) Let \(c\) coffees and \(t\) teas be last month's sales. A \(10\%\) rise multiplies by \(1.1\); a \(20\%\) fall multiplies by \(0.8\).
| \(c + t\) | \(=\) | \(500 \quad (1)\) |
| \(1.1c + 0.8t\) | \(=\) | \(490 \quad (2)\) |
| \((1) \times 0.8:\quad 0.8c + 0.8t\) | \(=\) | \(400 \quad (3)\) |
| \((2) - (3):\quad 0.3c\) | \(=\) | \(90\) |
| \(c\) | \(=\) | \(300\) |
| \(t\) | \(=\) | \(200\) |
Last month: \(\mathbf{300}\) coffees and \(\mathbf{200}\) teas. Check: \(330 + 160 = 490\ \checkmark\)
(ii) This month: \(1.1 \times 300 = \mathbf{330}\) coffees.
Answer the question asked: part (ii) wants this month's number, not \(c\).
Common pitfalls
Frequently asked questions
How do you set up simultaneous equations from a word problem?
Choose a pronumeral for each unknown and say exactly what it stands for. Then turn each separate fact in the question into an equation. Two unknowns need two facts.
Should I use substitution or elimination for word problems?
Use substitution when one equation is already in the form x equals something, such as in age problems. Use elimination when both equations look like ax plus by equals c, such as in mixture or ticket problems.
How do you solve age problems with simultaneous equations?
Let the pronumerals be the ages now. For a time in the future add the same number of years to both ages; for the past subtract them. Write one equation for each statement and solve.
How do you handle percentage increases in these problems?
Let the pronumerals be the original amounts. An increase of 10 percent multiplies by 1.1 and a decrease of 20 percent multiplies by 0.8. Multiply through by 10 or 100 to clear the decimals.
How do I check my answer?
Put your values back into the words of the problem and check every fact, not just the equations, in case an equation was set up incorrectly.