Simultaneous equations: revision notes and practice

What you will learn

• Recognise a solution as a pair satisfying both equations.
• Choose elimination or substitution for a linear system.
• Handle negative values and whole-equation operations carefully.
• Check solutions algebraically and understand their graphical meaning.

Key rules

Elimination

Add or subtract whole equations to cancel one variable, then solve the resulting equation.

Substitution

Replace a variable with its complete expression from the other equation.

Check both

Substitute the final x and y values into both original equations.

Worked examples

Example 1: Solve 2x + y = 11 and 3x − y = 9.

1. Add the equations: 5x = 20, so x = 4.
2. Use 2x + y = 11: 8 + y = 11, so y = 3.
3. Check: 3 × 4 − 3 = 9.

x = 4, y = 3

Example 2: Solve y = x + 2 and 2x + y = 11.

1. Substitute x + 2 for y: 2x + x + 2 = 11.
2. Then 3x = 9, so x = 3; y = 5.

x = 3, y = 5

A practice question

Solve 3x + 1y = -40 and 4x − 1y = -37 by elimination. Find x.

1. Understand the shared-solution requirement

A single equation such as 2x + y = 11 has many solutions: choosing an x-value determines a matching y-value. A second equation restricts which pair is allowed. Solving simultaneous equations means finding values that satisfy all the given equations at the same time. The answer therefore normally contains both variables, written clearly as x = something and y = something.

Think of each equation as a condition. A pair that satisfies only the first condition is not a solution of the whole system. For 2x + y = 11 and 3x − y = 9, the pair (4, 3) works in both: 8 + 3 = 11 and 12 − 3 = 9. This double check is more reliable than checking only the equation used to find the second variable.

2. Use elimination when terms cancel

Elimination removes one variable by adding or subtracting equations. When the y-terms have equal sizes and opposite signs, adding the equations cancels them. Add 2x + y = 11 to 3x − y = 9. The result is 5x = 20 because y + (−y) = 0. Dividing both sides by 5 gives x = 4.

If the coefficients have the same sign, subtraction may be more convenient. For 3x + 2y = 16 and x + 2y = 8, subtract the second equation from the first. This gives 2x = 8, so x = 4. Write the operation clearly and include the right-hand sides. Cancelling a variable on the left without performing the matching operation on the right destroys the equality.

3. Make coefficients match when necessary

Sometimes neither variable immediately cancels. Multiply one or both complete equations to create matching coefficients. For x + 2y = 7 and 3x + y = 11, doubling the second equation gives 6x + 2y = 22. Subtract the first equation to obtain 5x = 15, giving x = 3. Then y = 2 from the first equation.

Multiply every term, including the constant on the right. Doubling 3x + y = 11 means 6x + 2y = 22, not 6x + y = 11. Choose the variable that gives manageable multipliers, and leave enough space to show the rewritten equations. This method stays within linear simultaneous equations; a system involving a squared variable needs additional techniques beyond these core practice questions.

4. Find the second variable by substitution

Once you know one variable, put it into a simple original equation. After finding x = 4 for 2x + y = 11, write 2(4) + y = 11. Then 8 + y = 11, so y = 3. Do not stop after the first variable if the question asks you to solve the system. The pair is the complete result.

Use brackets around negative values during substitution. If x = −2 in 3x − y = 5, write 3(−2) − y = 5. This becomes −6 − y = 5, so −y = 11 and y = −11. A negative coefficient on the remaining variable requires careful division; adding 6 and then writing y = 11 would lose that negative sign.

5. Solve directly by substitution

Substitution is useful when one variable is already isolated. For y = x + 2 and 2x + y = 11, replace y in the second equation with the complete expression x + 2. This produces 2x + (x + 2) = 11. Collect the x-terms to obtain 3x + 2 = 11, then solve x = 3 and use the first equation to get y = 5.

The replacement expression needs brackets when it is multiplied or subtracted. If an equation contains 2y and y = x − 4, the replacement is 2(x − 4), not 2x − 4. If it contains −y, write −(x − 4) before expanding. Treat the expression as one complete quantity until the next algebra step; this keeps the original equation’s structure intact.

6. Link a system to its graph

Each linear equation describes a straight line. Their shared solution is the coordinate of their intersection, because that point lies on both lines. A diagram is useful for checking the approximate location and the signs of the answer. If the lines cross to the left of the y-axis and below the x-axis, both coordinates of the solution should be negative.

A graph can provide an estimate, but algebra should supply the exact answer when the question gives exact equations. Thick lines and screen scaling can make an intersection difficult to read precisely. Do not replace an exact fractional answer with a nearby integer just because the picture looks close. The plotted points are givens or representations of the equations; they do not change what the equations require.

7. Check both equations independently

Insert your final pair into each original equation, including every negative sign. Evaluate each left-hand side separately and compare it with the stated right-hand side. For x = 4 and y = 3, the system 2x + y = 11 and 3x − y = 9 gives 11 and 9, exactly as required. Write those equalities to show why the answer works.

If one check fails, return to the step where equations were combined or a value was substituted. Common problems include subtracting only part of an equation, forgetting to multiply the right-hand side, and giving the variables in the wrong order. A solution pair might accidentally fit one equation despite an arithmetic mistake, so checking both is an important safeguard rather than unnecessary repetition.

8. Interpret unusual systems and changed conditions

Two distinct parallel lines have no shared solution. Their equations reduce to incompatible conditions, such as 0 = 5, after elimination. Two equations describing exactly the same line have infinitely many shared solutions and may reduce to 0 = 0. These are different situations from the unique intersection used in the app’s generated questions. Dividing by a zero coefficient cannot create a unique solution.

A changed right-hand side changes the condition, so a previous pair need not remain valid. Rewrite the new system and solve it afresh rather than adjusting one variable by guesswork. In a word problem, define what x and y represent before forming equations, and consider units or realistic values after solving. The practice here uses linear systems only and does not claim complete simultaneous-equation coverage for every course.

Check your understanding

How to earn the marks

• Choose the variable whose coefficients are easiest to cancel.
• Multiply or subtract the entire equation, including its right-hand side.
• Show the value of both variables unless only one is requested.
• Check the final pair in both original equations.

A common mistake

When subtracting an equation, subtract every term and the right-hand side, including negative terms.

Course context and official specifications

This guide covers selected skills. Some extensions are Higher content; choose your official tier separately from practice difficulty in the app. These are original questions and explanations.

• AQA GCSE Mathematics — A19

  These are pairs of linear equations, available at both tiers. Linear–quadratic systems are Higher content and are not in this bank.

• Edexcel GCSE Mathematics — A19

  These are pairs of linear equations, available at both tiers. Linear–quadratic systems are Higher content and are not in this bank.

• OCR GCSE Mathematics — 6.03

  These are pairs of linear equations, available at both tiers. Linear–quadratic systems are Higher content and are not in this bank.

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