Is x = 2, y = -1 a solution of 3x + y = 5 and x - 2y = 3?

No. It works in the first, but 2 - 2(-1) = 4, not 3

Solve 3x + y = 11 and x + y = 5.

Subtract: 2x = 6, so x = 3, y = 2

Solve y = 2x - 3 and 4x - y = 9.

4x - 2x + 3 = 9, so x = 3, y = 3

Solve x + 2y = 7 and 3x + y = 11.

Double the second and subtract: 5x = 15, so x = 3, y = 2

Solve 4x + 3y = 18 and 5x - 2y = 11.

8x + 6y = 36 and 15x - 6y = 33, so 23x = 69, x = 3, y = 2

Three pens and two notebooks cost $13. One pen and four notebooks cost $16. Find the cost of each.

3p + 2n = 13 and p + 4n = 16, so a pen is $2 and a notebook is $3.50

Free worksheet · Australian Curriculum

Simultaneous equations worksheets with answers

Simultaneous equations are two equations that share the same solution. Learn substitution and elimination, then make a free worksheet with full solutions.

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Simultaneous equations at each year level, matched to the textbook chapters most Victorian schools use: a free printable worksheet with answers, sample questions, and an auto-marked test teachers can share.

Free Year 9 simultaneous equations worksheet

10 questions, then the answers with worked solutions. Printable PDF, no sign-up.

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Year 9 topics covered

Simultaneous equations using substitution
Simultaneous equations using elimination
Applications of simultaneous equations
Graphical solutions to simultaneous equations

Sample Year 9 simultaneous equations questions

Simultaneous equations using substitution
Use substitution to solve the simultaneous equations $y = 2x + 1$ and $x + y = 10$ .
Show the answer
Key idea: Replace $y$ in the second equation with the expression it equals from the first equation.
1. $x + (2x + 1) = 10$
2. $3x + 1 = 10$ , so $x = 3$
3. $y = 2(3) + 1 = 7$ (check: $3 + 7 = 10$ ✓)
Answer: $x = 3$ , $y = 7$
Simultaneous equations using elimination
Use elimination to solve $x + y = 9$ and $x - y = 3$ .
Show the answer
Key idea: The $y$ terms have opposite signs, so adding the equations eliminates $y$ .
1. Add: $2x = 12$ , so $x = 6$
2. Substitute into $x + y = 9$ : $6 + y = 9$ , so $y = 3$
Answer: $x = 6$ , $y = 3$
Applications of simultaneous equations
A phone plan from Company A costs $20 per month plus $0.10 per minute of calls. Company B charges $35 per month with no call costs. Write an equation for the total monthly cost of each plan in terms of $m$ minutes, then find the number of minutes at which both plans cost the same.
Show the answer
Key idea: Write a linear equation for each plan, then set them equal and solve for $m$ .
1. Company A: $C = 0.10m + 20$ . Company B: $C = 35$ .
2. Set the costs equal: $0.10m + 20 = 35$ .
3. Subtract $20$ from both sides: $0.10m = 15$ .
4. Divide both sides by $0.10$ : $m = 150$ .
5. Check: $0.10(150) + 20 = 15 + 20 = 35$ . ✓
Answer: The plans cost the same at $150$ minutes per month.
Graphical solutions to simultaneous equations
(a) Copy and complete the table of values for the lines $y = x + 1$ and $y = -x + 5$ .

$x$ $0$ $1$ $2$ $3$ $4$
$y = x + 1$ $\Box$ $\Box$ $\Box$ $\Box$ $\Box$
$y = -x + 5$ $\Box$ $\Box$ $\Box$ $\Box$ $\Box$

(b) Use the table to find the point where the two lines intersect.
Show the answer
Key idea: The lines intersect at the point where both rules give the same $y$ -value for the same $x$ -value.
1. (a) $y = x + 1$ gives 1, 2, 3, 4, 5
2. (a) $y = -x + 5$ gives 5, 4, 3, 2, 1
3. (b) At $x = 2$ both rules give $y = 3$ , so the lines intersect at $(2, 3)$
Answer: (a) $y = x + 1$ : 1, 2, 3, 4, 5; $y = -x + 5$ : 5, 4, 3, 2, 1 (b) $(2, 3)$

$x$	$0$	$1$	$2$	$3$	$4$
$y = x + 1$	$\Box$	$\Box$	$\Box$	$\Box$	$\Box$
$y = -x + 5$	$\Box$	$\Box$	$\Box$	$\Box$	$\Box$

Free Year 10 simultaneous equations worksheet

10 questions, then the answers with worked solutions. Printable PDF, no sign-up.

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Auto-marked test for your class

Teachers: share one link and every student is marked straight away, with worked solutions.

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Printable practice exam

Mix in harder questions, with marks and a suggested time.

Build an exam

Year 10 topics covered

Simultaneous equations using substitution
Simultaneous equations using elimination
Further applications of simultaneous equations

Sample Year 10 simultaneous equations questions

Simultaneous equations using substitution
Solve the simultaneous equations:\
$\begin{gathered}y = 2x - 1 \\ 3x + y = 9\end{gathered}$
Show the answer
Key idea: Substitute the expression for $y$ from the first equation directly into the second equation.
1. Substitute $y = 2x - 1$ into $3x + y = 9$ :\
$3x + (2x - 1) = 9$
1. Simplify and solve for $x$ :\
$\begin{aligned}5x - 1 &= 9 \\ 5x &= 10 \\ x &= 2\end{aligned}$
1. Substitute $x = 2$ back into $y = 2x - 1$ :\
$y = 2(2) - 1 = 3$
1. Check in $3x + y = 9$ : $3(2) + 3 = 9$ ✓
Answer: $x = 2,\; y = 3$
Simultaneous equations using elimination
Solve the simultaneous equations:\
$\begin{gathered}3x + 2y = 16 \\ 5x - 2y = 8\end{gathered}$
Show the answer
Key idea: Add the equations to eliminate $y$ , since the $y$ -coefficients are opposites.
1. Add the two equations:\
$\begin{aligned}(3x + 2y) + (5x - 2y) &= 16 + 8 \\ 8x &= 24 \\ x &= 3\end{aligned}$
1. Substitute $x = 3$ into $3x + 2y = 16$ :\
$\begin{aligned}9 + 2y &= 16 \\ 2y &= 7 \\ y &= 3.5\end{aligned}$
1. Check in $5x - 2y = 8$ : $5(3) - 2(3.5) = 15 - 7 = 8$ ✓
Answer: $x = 3,\; y = 3.5$
Further applications of simultaneous equations
Two friends, Aiden and Brianna, buy items at a school canteen. Aiden buys 2 pies and 1 drink and pays $8.70. Brianna buys 1 pie and 1 drink and pays $5.20. Find the cost of one pie and one drink.
Show the answer
Key idea: Define variables, write two equations from the given information, then solve.
1. Let $p$ = cost of one pie (dollars) and $d$ = cost of one drink (dollars). Write the equations:\
$\begin{aligned}2p + d &= 8.70 (1)\\ p + d &= 5.20 (2)\end{aligned}$
1. Subtract equation (2) from equation (1) to eliminate $d$ :\
$\begin{aligned}(2p + d) - (p + d) &= 8.70 - 5.20 \\ p &= 3.50\end{aligned}$
1. Substitute $p = 3.50$ into equation (2):\
$\begin{aligned}3.50 + d &= 5.20 \\ d &= 1.70\end{aligned}$
1. Check in equation (1): $2(3.50) + 1.70 = 7.00 + 1.70 = 8.70$ ✓
Answer: One pie costs $3.50 and one drink costs $1.70.
Simultaneous equations using substitution
Solve using substitution: $y = 3x - 2$ and $2x + y = 13$ . Give your answer as $x$ , then $y$ .
Show the answer
Key idea: Replace $y$ in the second equation with the expression from the first.
1. Substitute $y = 3x - 2$ into $2x + y = 13$ : $2x + 3x - 2 = 13$ .
2. $5x = 15$ , so $x = 3$ .
3. $y = 3(3) - 2 = 7$ .
Answer: $x = 3$ , $y = 7$

Topic guide

Simultaneous equations, explained

Two methods cover every linear pair. Here's how each one works and when to use it.

1. What the solution means

The solution is the pair of values that makes both equations true at once. For

$x + y = 10 x - y = 4$

the solution is $x = 7$ , $y = 3$ , because $7 + 3 = 10$ and $7 - 3 = 4$ . On a graph, it is the point $(7, 3)$ where the two lines cross.

2. Substitution

Use this when one equation already says $y = \ldots$ or $x = \ldots$ . Replace that pronumeral in the other equation:

$y = 2x - 1 3x + y = 14$

$3x + (2x - 1) = 14, 5x = 15, x = 3$

Then $y = 2(3) - 1 = 5$ .

3. Elimination

If one pronumeral has the same coefficient in both equations, subtract the equations to remove it (add them if the signs are opposite):

$2x + 3y = 13 2x + y = 7$

Subtracting gives $2y = 6$ , so $y = 3$ . Then $2x + 3 = 7$ , so $x = 2$ .

4. Multiplying first

If no coefficients match, multiply one or both equations so they do:

$2x + 3y = 12 3x - 2y = 5$

Multiply the first by $2$ and the second by $3$ :

$4x + 6y = 24 9x - 6y = 15$

Adding gives $13x = 39$ , so $x = 3$ . Then $2(3) + 3y = 12$ , so $y = 2$ .

5. Check in both equations

Substitute the answer into both original equations, not just one. For $x = 3$ , $y = 2$ :

$2(3) + 3(2) = 12 \checkmark 3(3) - 2(2) = 5 \checkmark$

A pair that only works in one equation is not a solution.

6. Worded problems

Adult tickets cost $15 and child tickets cost $9. A group buys $10$ tickets for $126. Let $a$ be adult tickets and $c$ child tickets:

$a + c = 10 15a + 9c = 126$

Substitute $c = 10 - a$ : $15a + 90 - 9a = 126$ , so $6a = 36$ , $a = 6$ and $c = 4$ . Define your pronumerals, write two equations, then solve.

Quick check: 10 questions

Try each one, then open the answer.

Solve $x + y = 9$ and $x - y = 3$ .
Answer
$x = 6$ , $y = 3$
Solve $y = x + 2$ and $x + y = 12$ .
Answer
$2x + 2 = 12$ , so $x = 5$ , $y = 7$
Is $x = 2$ , $y = -1$ a solution of $3x + y = 5$ and $x - 2y = 3$ ?
Answer
No. It works in the first, but $2 - 2(-1) = 4$ , not $3$
Solve $y = 3x$ and $2x + y = 20$ .
Answer
$5x = 20$ , so $x = 4$ , $y = 12$
Solve $3x + y = 11$ and $x + y = 5$ .
Answer
Subtract: $2x = 6$ , so $x = 3$ , $y = 2$
Solve $2x + y = 9$ and $3x - y = 11$ .
Answer
Add: $5x = 20$ , so $x = 4$ , $y = 1$
Solve $y = 2x - 3$ and $4x - y = 9$ .
Answer
$4x - 2x + 3 = 9$ , so $x = 3$ , $y = 3$
Solve $x + 2y = 7$ and $3x + y = 11$ .
Answer
Double the second and subtract: $5x = 15$ , so $x = 3$ , $y = 2$
Solve $4x + 3y = 18$ and $5x - 2y = 11$ .
Answer
$8x + 6y = 36$ and $15x - 6y = 33$ , so $23x = 69$ , $x = 3$ , $y = 2$
Three pens and two notebooks cost $13. One pen and four notebooks cost $16. Find the cost of each.
Answer
$3p + 2n = 13$ and $p + 4n = 16$ , so a pen is $2 and a notebook is $3.50

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These topic pages cover this skill, with worked examples, common mistakes and a question to try.

Year 10 worksheetSimultaneous equations Year 9 worksheetLinear equations and graphs Year 9 worksheetCoordinate geometry Year 10A worksheetLinear and non-linear simultaneous equations

Questions

What are simultaneous equations?

Simultaneous equations are two or more equations with the same unknowns that are solved together. The solution is the pair of values that makes every equation true, which is where their graphs cross.

Should I use substitution or elimination?

Use substitution when one equation is already written as y = or x =. Use elimination when both equations are in the form ax + by = c, especially if a coefficient already matches.

Do the worksheets include worked solutions?

Yes. Every question on a generated worksheet comes with a fully worked solution, so students can follow each step of the method.

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