Further Applications of Simultaneous Equations
Every question with a fully worked solution.
- Further Applications of Simultaneous Equations - Video - Simultaneous equations word problems Watch
Theory
This Year 10 maths page shows how to turn a worded problem into two equations and solve them: define the pronumerals, write one equation for each fact, solve by substitution or elimination, then answer in words and check. It covers counting problems, money and coins, percentages, ages, speeds with a current or wind, mixtures and break-even points.
A problem needs simultaneous equations when there are two unknown amounts and two separate facts about them.
Start by defining the pronumerals: say exactly what each one stands for, with units, such as "let \(a\) be the number of adult tickets".
Many problems have a count equation (how many things altogether) and a value equation (how much they are worth, how many legs, how many seats). For \(20\) tickets worth $350 with adults at $20 and children at $15: \(a + c = 20\) and \(20a + 15c = 350\).
The break-even point is where two costs (or a cost and an income) are equal. On a graph it is where the two lines cross.
Always interpret the answer: a number of people must be a whole number, and a price should be given in dollars and cents.
Common sentences and the equations they give:
| Words | Equation |
|---|---|
| two numbers add to \(30\) | \(x + y = 30\) |
| they differ by \(8\) | \(x - y = 8\) |
| half their difference is \(4\) | \(\dfrac{x - y}{2} = 4\) |
| \(A\) is \(3\) times \(B\) | \(A = 3B\) |
| \(y\) is \(20\%\) more than \(x\) | \(y = 1.2x\) |
| in \(5\) years (ages) | \(m + 5\) and \(s + 5\) |
| with / against a current | \(b + c\) and \(b - c\) |
How to solve a worded problem
- Read the problem and find the two unknown amounts.
- Define a pronumeral for each, with units.
- Write two equations, one for each fact. Use the same units in both (for example all cents, or all km/h).
- Solve: substitution if a pronumeral is the subject, elimination otherwise.
- Check the values in the original words, not just your equations.
- Answer in a sentence with units, rounding sensibly.
Let there be \(f\) four-seat tables and \(s\) six-seat tables. One equation counts tables, the other counts seats.
| \(f + s\) | \(=\) | \(18 \quad (1)\) |
| \(4f + 6s\) | \(=\) | \(88 \quad (2)\) |
\((1) \times 4\):
| \(4f + 4s\) | \(=\) | \(72 \quad (3)\) |
\((2) - (3)\):
| \(2s\) | \(=\) | \(16\) |
| \(s\) | \(=\) | \(8\) |
| \(f\) | \(=\) | \(18 - 8 = 10\) |
There are \(10\) four-seat tables and \(8\) six-seat tables. Check: \(40 + 48 = 88\) ✓
Let the mother be \(m\) years old and her son \(s\) years old now. Add \(4\) to both ages for the future and subtract \(6\) from both for the past.
| \(m + 4\) | \(=\) | \(3(s + 4)\) |
| \(m\) | \(=\) | \(3s + 8 \quad (1)\) |
| \(m - 6\) | \(=\) | \(8(s - 6)\) |
| \(m\) | \(=\) | \(8s - 42 \quad (2)\) |
Both expressions equal \(m\), so make them equal:
| \(3s + 8\) | \(=\) | \(8s - 42\) |
| \(50\) | \(=\) | \(5s\) |
| \(s\) | \(=\) | \(10\) |
| \(m\) | \(=\) | \(3(10) + 8 = 38\) |
The mother is \(38\) and her son is \(10\). Check: in \(4\) years, \(42 = 3 \times 14\) ✓; six years ago, \(32 = 8 \times 4\) ✓
Let the plane's speed in still air be \(p\) km/h and the wind speed \(w\) km/h. With the wind the speeds add; against it they subtract. Speed \(=\) distance \(\div\) time.
| \(p + w\) | \(=\) | \(1800 \div 3\) |
| \(\) | \(=\) | \(600 \quad (1)\) |
| \(p - w\) | \(=\) | \(1800 \div 4\) |
| \(\) | \(=\) | \(450 \quad (2)\) |
\((1) + (2)\):
| \(2p\) | \(=\) | \(1050\) |
| \(p\) | \(=\) | \(525\) |
| \(w\) | \(=\) | \(600 - 525 = 75\) |
The plane flies at \(525\) km/h in still air and the wind blows at \(75\) km/h.
Let \(c\) kg of cashews and \(n\) kg of peanuts be used. One equation counts kilograms; the other counts dollars. The mix is worth \(10 \times 15 = 150\) dollars.
| \(c + n\) | \(=\) | \(10 \quad (1)\) |
| \(24c + 9n\) | \(=\) | \(150 \quad (2)\) |
From (1), \(n = 10 - c\). Substitute into (2):
| \(24c + 9(10 - c)\) | \(=\) | \(150\) |
| \(15c + 90\) | \(=\) | \(150\) |
| \(c\) | \(=\) | \(4\) |
| \(n\) | \(=\) | \(6\) |
The shop uses \(4~\text{kg}\) of cashews and \(6~\text{kg}\) of peanuts. Check: \(96 + 54 = 150\) ✓
Common pitfalls
Frequently asked questions
How do you write simultaneous equations from a word problem?
Define a pronumeral for each unknown amount. Then write one equation for each fact in the problem, often one that counts the items and one for their total value.
What is a break-even point?
It is where two costs, or a cost and an income, are equal. Solve the two equations together, or read where the two lines cross on a graph.
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, and for the past subtract it from both.
How do current or wind problems work?
Going with the current, the speeds add. Going against it, they subtract. Find each speed as distance divided by time, then solve the two equations.
Should I use substitution or elimination for word problems?
Either works. Substitution is quicker when one equation is like \(A = 3B\). Elimination is quicker when both equations look like \(ax + by = k\).