Constructing Simultaneous Linear Equations
Constructing linear simultaneous equations is a foundational skill assumed for Queensland Year 11 Mathematical Methods (QCAA). Many real situations link two unknowns through two conditions, and writing one equation for each lets you pin both values down.
You will define two pronumerals, form an equation from each condition, solve the pair by substitution or elimination, and interpret the answer in context — prices, counts, speeds or dimensions.
Every question with a fully worked solution.
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Theory
In Year 11 Mathematical Methods (QCAA), many real situations involve two unknowns linked by two conditions. This page shows how to construct simultaneous equations — define two pronumerals, write one linear equation for each condition — then solve the pair by substitution or elimination and interpret the answer in context (prices, counts, speeds, dimensions).
When a problem has two unknowns, we choose two pronumerals and look for two conditions in the words. Each condition becomes one linear equation, giving a pair of simultaneous equations that can be solved together.
For example, ‘\(3\) pies and \(2\) drinks cost \(\$23\)’ becomes \(3p+2d=23\), and ‘\(1\) pie and \(2\) drinks cost \(\$13\)’ becomes \(p+2d=13\). Solving the pair by substitution or elimination gives both unknowns.
Finally we interpret the solution back in the situation — a price, a number of items, a speed, or a length — with the correct units.
Choose pronumerals, then write one equation per condition. A ‘total cost’ condition for items priced \(p\) and \(d\) with counts \(m\) and \(n\) is
A ‘sum’ and a ‘difference’ condition give
For speed with a current \(c\) and still-water speed \(b\), downstream and upstream give
How to construct and solve a system
- Define two pronumerals for the two unknowns, with units.
- Form one linear equation for each condition in the problem.
- Solve the pair by substitution or elimination.
- Interpret and check both values against the words.
Define — let \(p\) be the pie price and \(d\) the drink price (dollars). Form two equations:
| \(3p+2d\) | \(=\) | \(23\) |
| \(p+2d\) | \(=\) | \(13\) |
Eliminate \(d\) — subtract the second from the first:
| \((3p+2d)-(p+2d)\) | \(=\) | \(23-13\) |
| \(2p\) | \(=\) | \(10\) |
| \(p\) | \(=\) | \(5\) |
Back-substitute \(p=5\) into \(p+2d=13\):
| \(5+2d\) | \(=\) | \(13\) |
| \(2d\) | \(=\) | \(8\) |
| \(d\) | \(=\) | \(4\) |
Check in \(3p+2d=23\):
| \(3(5)+2(4)\) | \(=\) | \(15+8=23\;\checkmark\) |
A pie costs \(\$5\) and a drink costs \(\$4\).
Define — let the numbers be \(x\) and \(y\) with \(x\) the larger:
| \(x+y\) | \(=\) | \(40\) |
| \(x-y\) | \(=\) | \(8\) |
Add the equations — \(y\) cancels:
| \(2x\) | \(=\) | \(48\) |
| \(x\) | \(=\) | \(24\) |
Back-substitute into \(x+y=40\):
| \(24+y\) | \(=\) | \(40\) |
| \(y\) | \(=\) | \(16\) |
Check the difference:
| \(24-16\) | \(=\) | \(8\;\checkmark\) |
The numbers are \(24\) and \(16\).
Define — let \(a\) be adult tickets and \(c\) child tickets. Form two equations:
| \(a+c\) | \(=\) | \(200\) |
| \(12a+8c\) | \(=\) | \(1960\) |
From the first, \(a=200-c\). Substitute into the second:
| \(12(200-c)+8c\) | \(=\) | \(1960\) |
| \(2400-12c+8c\) | \(=\) | \(1960\) |
| \(2400-4c\) | \(=\) | \(1960\) |
Solve for \(c\):
| \(-4c\) | \(=\) | \(1960-2400\) |
| \(-4c\) | \(=\) | \(-440\) |
| \(c\) | \(=\) | \(110\) |
Back-substitute \(c=110\): \(a=200-110\):
| \(a\) | \(=\) | \(90\) |
Check the takings:
| \(12(90)+8(110)\) | \(=\) | \(1080+880=1960\;\checkmark\) |
\(90\) adult tickets and \(110\) child tickets.
Define — let \(b\) be the still-water speed and \(c\) the current (km/h). Speed \(=\dfrac{\text{distance}}{\text{time}}\):
| \(b+c\) | \(=\) | \(\dfrac{24}{2}=12\) |
| \(b-c\) | \(=\) | \(\dfrac{24}{3}=8\) |
Add the equations — \(c\) cancels:
| \(2b\) | \(=\) | \(20\) |
| \(b\) | \(=\) | \(10\) |
Back-substitute into \(b+c=12\):
| \(10+c\) | \(=\) | \(12\) |
| \(c\) | \(=\) | \(2\) |
Check upstream:
| \(b-c\) | \(=\) | \(10-2=8\;\checkmark\) |
Still-water speed \(10\) km/h and current \(2\) km/h.
Common pitfalls
Frequently asked questions
How do you construct simultaneous equations from a word problem?
Define a pronumeral for each of the two unknowns, then write one linear equation for each condition in the problem, giving a pair to solve together.
How many equations do I need for two unknowns?
Two. Each unknown needs its own equation, so look for two separate conditions in the wording.
How do I set up a purchase problem?
Let the prices be two pronumerals, then each ‘quantity times price = total’ statement becomes one equation, such as \(3p+2d=23\).
How do I handle a speed-and-current problem?
Downstream the speeds add and upstream they subtract: \(b+c\) and \(b-c\). Turn each trip into a speed first (distance over time), then equate.
Which method should I use to solve the pair?
Substitution if one variable is easy to isolate, elimination if the equations line up neatly; both give the same answer.
Do I need to state units in the final answer?
Yes. Interpret the solution in context with units, for example ‘a pie costs \(\$5\) and a drink \(\$4\)’.