Secondary 4 · Mathematics · Algebra
Simultaneous linear equations
Solve pairs of linear equations graphically and algebraically using substitution and elimination, and apply them to practical problems.
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These lessons were written around Australian Curriculum strands and matched to Singapore levels by age (Primary 1 is the first year of primary school). They are not aligned to MOE syllabuses or the PSLE, so check them against the syllabus your child is working towards.
Learning goals
Learners will
- Interpret the solution of a pair of linear equations as the point of intersection of their graphs.
- Solve simultaneous linear equations algebraically using substitution and elimination.
- Set up and solve simultaneous equations from practical word problems.
Success looks like
- I can find the solution to two linear equations by reading the intersection point of their graphs.
- I can choose between substitution and elimination and solve a pair of equations accurately.
- I can write two equations from a word problem, solve them and check my answer in both equations.
The big idea
One linear equation such as x + y = 10 has infinitely many solutions (1 and 9, 2 and 8, 3.5 and 6.5…). When we have two equations that must both be true, there is usually just one pair of values that works. Finding it is called solving simultaneous equations.
Graphically, each linear equation is a straight line. The solution is the point of intersection, because that point lies on both lines. For y = x + 1 and y = 3x − 3, the lines cross at (2, 3), so x = 2 and y = 3.
Graphs are great for seeing the idea, but reading an exact answer can be hard. Algebraic methods give exact solutions.
Substitution works well when one equation already has a variable on its own, such as y = 2x + 1. Substitute that expression into the other equation, solve for x, then substitute back to find y.
Elimination works well when both equations are in the form ax + by = c. If the coefficients of one variable are the same, subtract the equations; if they are opposites (such as +y and −y), add them. If neither, multiply one or both equations first so that the coefficients match.
Special cases: if the lines are parallel (same gradient, different y-intercept), they never meet and there is no solution. If the two equations describe the same line, there are infinitely many solutions.
Word problems: define your two variables clearly ('let a be the number of adult tickets'), write one equation for each piece of information, solve, and then check your answer makes sense in the original story.
Always check by substituting your x and y values into both original equations.
Worked examples
Substitution
Solve y = 2x + 1 and 3x + y = 11.
- Substitute y = 2x + 1 into the second equation: 3x + (2x + 1) = 11.
- Simplify and solve: 5x + 1 = 11, so 5x = 10 and x = 2.
- Substitute x = 2 into y = 2x + 1: y = 5.
- Check in 3x + y = 11: 6 + 5 = 11. Correct.
Answer: x = 2, y = 5
Elimination after multiplying
Solve 2x + 3y = 12 and x − y = 1.
- Multiply the second equation by 3: 3x − 3y = 3.
- Add it to the first equation so y cancels: (2x + 3y) + (3x − 3y) = 12 + 3, giving 5x = 15, so x = 3.
- Substitute x = 3 into x − y = 1: 3 − y = 1, so y = 2.
- Check in 2x + 3y = 12: 6 + 6 = 12. Correct.
Answer: x = 3, y = 2
A practical problem
A school concert sells adult tickets for $12 and child tickets for $7. Ten tickets are sold for a total of $95. How many of each were sold?
- Let a = number of adult tickets and c = number of child tickets.
- Number of tickets: a + c = 10. Money: 12a + 7c = 95.
- From the first equation, c = 10 − a. Substitute: 12a + 7(10 − a) = 95, so 12a + 70 − 7a = 95, giving 5a = 25 and a = 5.
- Then c = 10 − 5 = 5. Check: 12 × 5 + 7 × 5 = 60 + 35 = 95.
Answer: 5 adult tickets and 5 child tickets
Practice check
Have a go, then check your answers. Each answer comes with an explanation.
Answer guide for parents
The lines y = x + 1 and y = 3x − 3 intersect at which point?
(2, 3) — x + 1 = 3x − 3 gives 4 = 2x, so x = 2 and y = 3. Check: 3 × 2 − 3 = 3.
Solve x + y = 12 and x − y = 2. What is the value of x?
x = 7 — Adding the equations gives 2x = 14, so x = 7 (and y = 5).
Solve y = 2x and x + y = 9. What is the value of y?
y = 6 — Substitute: x + 2x = 9, so x = 3 and y = 2 × 3 = 6.
What is the best first step to solve 2x + 3y = 13 and 2x + y = 7 by elimination?
Subtract the second equation from the first to eliminate x — Both equations contain 2x, so subtracting removes x: 2y = 6, giving y = 3 and then x = 2.
How many solutions does the pair y = 2x + 1 and y = 2x − 4 have?
None — Both lines have gradient 2 but different y-intercepts, so they are parallel and never meet.
Solve 3x + 2y = 16 and x − 2y = 0. What is the value of y?
y = 2 — Adding the equations gives 4x = 16, so x = 4. Then 4 − 2y = 0, so y = 2.
Two numbers add to 20 and their difference is 6. What is the larger number?
13 — x + y = 20 and x − y = 6. Adding gives 2x = 26, so x = 13 and y = 7.
Two pens and one notebook cost $7. One pen and one notebook cost $5. How much does one pen cost?
$2 — Subtracting the equations (2p + n) − (p + n) = 7 − 5 gives p = 2. So a pen costs $2 and a notebook costs $3.
Watch out for
- Finding only one variable and stopping. A solution to simultaneous equations is a pair of values – find both x and y.
- Adding equations when the coefficients are the same (or subtracting when they are opposites), so nothing cancels. Same signs: subtract; opposite signs: add.
- Multiplying only one side of an equation when preparing for elimination. Every term on both sides must be multiplied.
Tips for parents
- Encourage your learner to check the final answer in both original equations – it takes ten seconds and catches most errors.
- Use real situations: comparing two phone plans or two gym memberships gives natural simultaneous equations.
- Let your learner sketch a quick graph to see whether the answer looks reasonable before solving algebraically.
Go further
- Two phone plans: Plan A costs $20 per month plus $0.10 per minute; Plan B costs $35 per month with unlimited calls. Find the number of minutes at which the plans cost the same (150 minutes) and graph both plans.
- Use graphing software to draw pairs of lines with no solution, one solution and infinitely many solutions. Describe what the equations in each case have in common.
- Investigate how to solve three equations in three unknowns, such as x + y + z = 6, x − y = 1 and y + z = 4.
About 55 minutes
You'll need
- Paper and pencil
- Grid paper and a ruler
- A calculator (optional)
- Graphing software (optional)
Key words
- Simultaneous equations
- Two or more equations that are true at the same time for the same values of the variables.
- Point of intersection
- The point where two graphs cross; its coordinates satisfy both equations.
- Substitution method
- Replacing one variable with an equivalent expression from the other equation.
- Elimination method
- Adding or subtracting the equations so that one variable cancels out.
- Parallel lines
- Lines with the same gradient that never meet, so the equations have no solution.
Still tricky?
For parents
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