Simultaneous Equations: Substitution and Elimination Explained, with a Solver (Year 10)

Simultaneous equations: substitution and elimination, explained

Simultaneous equations are two equations that share the same unknowns. There are two main ways to solve them, and choosing the right one makes the algebra much shorter. Here’s each method with a worked example, a solver that shows the working, and practice questions with answers.

By Ryan, founder of Bayside Academics, 15 years in private tutoring · Updated 6 November 2026

Quick answer

Simultaneous equations are solved when you find the x and y that make both equations true. Substitution: rearrange one equation for y (or x) and substitute it into the other. Elimination: make the coefficients of one letter match, then add or subtract the equations so it cancels. Always check your answer in the other equation.

What simultaneous equations are

Each equation on its own has endless solutions. Together, there's usually just one pair $(x, y)$ that works in both. That pair is where their graphs cross.

For example, $x + y = 10$ and $x - y = 4$ are both true when $x = 7$ and $y = 3$ .

Method 1: substitution

Best when one equation already says y = … or x = ….

Worked example

Solve $y = 2x + 1$ and $3x + y = 11$ .

Replace $y$ in the second equation with $2x + 1$ :

3x + (2x + 1) = 11

5x = 10 \quad \Rightarrow \quad x = 2

Substitute back: $y = 2(2) + 1 = 5$ . So $x = 2$ , $y = 5$ .

Method 2: elimination

Best when both are in the form ax + by = c. Same signs: subtract. Opposite signs: add.

Coefficients already opposite

Solve $x + y = 10$ and $x - y = 4$ .

The $y$ terms are opposites, so add:

2x = 14 \quad \Rightarrow \quad x = 7

Then $7 + y = 10$ , so $y = 3$ .

Multiply first

Solve $2x + 3y = 13$ (1) and $4x - y = 5$ (2).

Multiply (2) by 3 so the $y$ terms are opposites: $12x - 3y = 15$ .

Add to (1):

14x = 28 \quad \Rightarrow \quad x = 2

Substitute into (1): $4 + 3y = 13$ , so $y = 3$ .

Check in (2): $4(2) - 3 = 5$ ✓

Simultaneous equations solver (with working)

Enter both equations in the form ax + by = c. It solves them by elimination and shows every step, with exact fractions.

Solve $ax + by = c$ and $dx + ey = f$

a (x coefficient)b (y coefficient)c (right side)

d (x coefficient)e (y coefficient)f (right side)

x = 2, y = 3

Label the equations: (1) $2x + 3y = 13$ and (2) $4x - y = 5$ .
Make the $y$ terms match: multiply (1) by $1$ and (2) by $3$ .
(1) becomes $2x + 3y = 13$ and (2) becomes $12x - 3y = 15$ .
The $y$ terms are opposites, so add the equations: $14x = 28$
Solve for $x$ : $x = \dfrac{28}{14} = 2$
Substitute $x = 2$ into (1): $2 \times 2 + 3y = 13$
Solve for $y$ : $y = \dfrac{13 - 4}{3} = 3$
Check in (2): $4 \times 2 + \left(-1\right) \times 3 = 5$ ✓

Write both equations in the form ax + by = c first. If a letter is missing, enter 0.

Method 3: graphing

Draw both lines on the same axes. The solution is the point where they cross. It's a good check, but reading an exact answer off a graph is hard if it isn't a whole number, so the algebra methods are more reliable.

Lines cross once: one solution.
Parallel lines: no solution.
The same line: infinitely many solutions.

Word problems

Choose a letter for each unknown, write one equation for each fact, then solve.

Worked example

Adult tickets cost 15 dollars and child tickets cost 8 dollars. A family buys 7 tickets for 77 dollars. How many of each?

Let $a$ = adults and $c$ = children.

a + c = 7 \qquad 15a + 8c = 77

From the first, $c = 7 - a$ . Substitute:

15a + 8(7 - a) = 77

7a = 21 \quad \Rightarrow \quad a = 3

So 3 adults and 4 children. Check: $15 \times 3 + 8 \times 4 = 77$ ✓

Common mistakes that cost marks

Multiplying only one term. When you multiply an equation, multiply every term, including the right-hand side.
Adding when you should subtract. Same signs: subtract. Opposite signs: add.
Sign errors when subtracting. Subtracting −y is adding y. Write each step.
Stopping after x. The answer needs both values.
Not checking. Substituting into the other equation catches most errors in seconds.

Practice questions with answers

Try each one before you open the answer.

Solve $y = x + 3$ and $2x + y = 12$ .
Show the answer
Substitute: $2x + x + 3 = 12$ , so $3x = 9$ , $x = 3$ and $y = 6$ .
Solve $x + y = 9$ and $x - y = 3$ .
Show the answer
Add: $2x = 12$ , so $x = 6$ and $y = 3$ .
Solve $3x + 2y = 16$ and $x + 2y = 8$ .
Show the answer
Subtract: $2x = 8$ , so $x = 4$ and $y = 2$ .
Solve $5x - 2y = 11$ and $3x + 4y = 4$ .
Show the answer
Multiply the first by 2: $10x - 4y = 22$ . Add: $13x = 26$ , so $x = 2$ . Then $10 - 2y = 11$ , so $y = -\dfrac{1}{2}$ .
Solve $y = 2x - 1$ and $y = -x + 8$ .
Show the answer
Set them equal: $2x - 1 = -x + 8$ , so $3x = 9$ , $x = 3$ and $y = 5$ .
Two numbers add to 25 and their difference is 7. Find them.
Show the answer
$x + y = 25$ and $x - y = 7$ . Add: $2x = 32$ , so the numbers are 16 and 9.

Simultaneous equations by year level

Following the Victorian Curriculum 2.0. Schools sequence topics differently, so check your course.

Year 10

Simultaneous equations

Two linear equations: substitution, elimination and graphing, plus word problems.

Worksheets and topic test · Free PDF with answers

Year 10A

Linear and non-linear simultaneous equations

A line with a parabola, circle or hyperbola: zero, one or two solutions.

Worksheets and topic test · Free PDF with answers

Simultaneous equations: common questions

What are simultaneous equations?

Two (or more) equations with the same unknowns that are true at the same time. The solution is the pair of values, like x = 2 and y = 3, that works in both. On a graph, it's where the two lines cross.

Should I use substitution or elimination?

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

What if there's no solution?

If the lines are parallel (same gradient, different intercepts), they never cross, so there's no solution. If both equations describe the same line, there are infinitely many.

How do I check my answer?

Substitute both values into the equation you didn't use to find the second value. If it works, the answer is right.

What year level learns simultaneous equations?

In the Victorian Curriculum 2.0 they're a Year 10 topic, solved algebraically and graphically. Year 10A extends to one linear and one non-linear equation, such as a line and a parabola.

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