Edexcel GCSE Maths Foundation (1MA1), Foundation tier · Algebra
Practise Simultaneous equations. unlimited generated questions on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Solving two equations with two unknowns at the same time. Both tiers solve two linear equations (by elimination, substitution or from a graph) and form them from worded problems; Higher also solves a linear equation with a quadratic one.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
4
Read a solution from two line graphsThe solution is the coordinates of the point where the two lines cross.
5
Solve linear simultaneous equations by eliminationMake the coefficients of one letter the same, then add or subtract the equations.
5
Form and solve simultaneous equations in contexte.g. 3 teas and 2 coffees cost £8.10 gives \(3t + 2c = 810\) (in pence).
Notes
Elimination
Label the equations (1) and (2).
Make the coefficients of one letter the same size, by multiplying one or both whole equations.
If the signs are the same, subtract the equations; if they are different, add them.
Solve for the letter that is left, then substitute into either original equation to find the other.
\(3x + 2y = 16\) and \(5x - 2y = 16\): adding gives \(8x = 32\), so \(x = 4\). Then \(12 + 2y = 16\), so \(y = 2\).
If nothing matches, multiply both: for \(2x + 3y = 12\) and \(3x + 4y = 17\), do (1) × 3 and (2) × 2 to get \(6x\) in both, then subtract: \(y = 2\) and \(x = 3\).
Substitution and graphs
If one equation is already \(y = \ldots\) (or \(x = \ldots\)), substitute it into the other.
\(y = 2x - 1\) and \(3x + y = 14\): \(3x + 2x - 1 = 14\), so \(x = 3\) and \(y = 5\).
On a graph, the solution is the point where the two lines intersect. Solutions read from a graph are estimates.
Cheatsheet
Same signs: subtract. Different signs: add
Match coefficients by multiplying every term of an equation
Substitute back to find the second letter
Check both values in the other equation
Graph method: the solution is the intersection point
How to answer each type of question
Solve two linear equations
3 marks5
Multiply one or both equations so that one letter has the same coefficient.
Add or subtract to eliminate it, then solve.
Substitute back to find the other letter, and check.
Check your pair in the equation you did not use to find the second value.
Eliminate the letter that is easiest: one that already has opposite signs, or where one coefficient is a multiple of the other.
In money problems, work in pence to avoid decimals.
Higher: a line and a curve usually give two pairs of answers; keep each x with its own y.
Where marks are lost
Multiplying only some terms of an equation when matching coefficients.
Subtracting when the signs are different (or adding when they are the same).
Slipping with negatives when subtracting: \(5 - (-3) = 8\).
Giving x and y when the question asks for the cost of one item.
Higher: mixing up the pairs, e.g. writing \(x = 2, y = 4\) when \(x = 2\) goes with \(y = 2\).
Exam technique
When the question says 'You must show all your working', answers found by trial and improvement (or a calculator's solver alone) may score no marks.
Write the answer clearly as \(x = \ldots\), \(y = \ldots\), or answer the question in context with units.
For the graph method, read both coordinates as accurately as the grid allows.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy3 marks
Solve the simultaneous equations \(x + 4y = 37\) \(5x - 4y = -7\) You must show all your working.[3]
Show the answer and mark scheme
Answer: \(x = 5, \ y = 8\)
M1 for a correct method to eliminate \(x\) or \(y\) (e.g. coefficients of one variable made equal and the equations added or subtracted, allow one arithmetic error)
M1 (dep) for substituting their value into one of the equations or for a correct method to find the second variable
A1 for \(x = 5\) and \(y = 8\)
Worked solution: Add to eliminate \(y\): \(6x = 30\), so \(x = 5\). Substitute into \(x + 4y = 37\): \(4y = 32\), so \(y = 8\).
Question 2Medium3 marks
Solve the simultaneous equations \(4x - y = -5\) \(12x + 5y = 25\) You must show all your working.[3]
Show the answer and mark scheme
Answer: \(x = 0, \ y = 5\)
M1 for a correct method to eliminate \(x\) or \(y\) (e.g. coefficients of one variable made equal and the equations added or subtracted, allow one arithmetic error)
M1 (dep) for substituting their value into one of the equations or for a correct method to find the second variable
A1 for \(x = 0\) and \(y = 5\)
Worked solution: Multiply the first equation by \(5\): \(20x - 5y = -25\) and \(12x + 5y = 25\). Add to eliminate \(y\): \(32x = 0\), so \(x = 0\). Substitute into \(4x - y = -5\): \(-y = -5\), so \(y = 5\).