IGCSE Mathematics 0580 · Topic 2.5

IGCSE Mathematics: Simultaneous Equations Practice Questions

Simultaneous equations are two equations that are both true for the same pair of values. Solve a linear pair by elimination or substitution, and solve a linear and non-linear pair by substituting the linear equation into the non-linear one.

Cambridge IGCSE Mathematics (0580) · Topic 2.5: Simultaneous Equations

Simultaneous equations appear on every Cambridge IGCSE Mathematics 0580 paper, usually once as a pure algebra question and once hidden inside a word problem. Extended candidates also face a linear pair combined with a quadratic or a circle. The marks are almost entirely method marks, so showing every line matters more than reaching the answer quickly.

What you need to know for Simultaneous Equations

IGCSE Mathematics Simultaneous Equations questions and answers

4 exam-style questions written to the 0580 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.

Question 1[3 marks]

Solve the simultaneous equations 4x + 3y = 25 and 2x - y = 5.

Show the worked answer
Answer: x = 4 and y = 3
  1. Multiply the second equation by 3 so the y coefficients match: 6x - 3y = 15.
  2. Add this to the first equation to eliminate y: (4x + 3y) + (6x - 3y) = 25 + 15, giving 10x = 40.
  3. So x = 4.
  4. Substitute into 2x - y = 5: 8 - y = 5, so y = 3. Check in the first equation: 16 + 9 = 25, correct.
How the marks are awarded. 1 mark for a correct method to eliminate one variable, including correct multiplication. 1 mark for x = 4. 1 mark for y = 3.
Where students lose the mark. Adding when the signs require subtracting, or subtracting when they require adding. If the matching terms have opposite signs, add. If they have the same sign, subtract.
Question 2[5 marks]

Solve the simultaneous equations y = x + 1 and x2 + y2 = 25.

Show the worked answer
Answer: x = 3 with y = 4, and x = -4 with y = -3
  1. Substitute y = x + 1 into the second equation: x2 + (x + 1)2 = 25.
  2. Expand: x2 + x2 + 2x + 1 = 25, so 2x2 + 2x - 24 = 0.
  3. Divide through by 2: x2 + x - 12 = 0, which factorises to (x + 4)(x - 3) = 0.
  4. So x = -4 or x = 3. Substitute each back into y = x + 1 to get y = -3 and y = 4 respectively.
How the marks are awarded. 1 mark for substituting correctly. 1 mark for expanding and forming a correct three term quadratic. 1 mark for solving the quadratic. 1 mark for both x values. 1 mark for both y values correctly paired.
Where students lose the mark. Giving all four combinations of x and y values. Each x value pairs with exactly one y value, so state them as ordered pairs.
Question 3[4 marks]

Three pens and two notebooks cost RM26. Five pens and four notebooks cost RM46. Find the cost of one pen and the cost of one notebook.

Show the worked answer
Answer: A pen costs RM6 and a notebook costs RM4.
  1. Let p be the cost of a pen in ringgit and n be the cost of a notebook in ringgit.
  2. Form the equations: 3p + 2n = 26 and 5p + 4n = 46.
  3. Multiply the first equation by 2: 6p + 4n = 52. Subtract the second: 6p - 5p = 52 - 46, so p = 6.
  4. Substitute p = 6 into 3p + 2n = 26: 18 + 2n = 26, so 2n = 8 and n = 4.
How the marks are awarded. 1 mark for defining the variables and forming both correct equations. 1 mark for a correct elimination method. 1 mark for p = 6. 1 mark for n = 4.
Where students lose the mark. Not defining the variables. Writing 3x + 2y = 26 with no statement of what x and y represent regularly loses the first mark.
Question 4[4 marks]

The line y = 3x - 2 meets the curve y = x2 at two points. Find the coordinates of both points.

Show the worked answer
Answer: (1, 1) and (2, 4)
  1. At the points of intersection the y values are equal, so x2 = 3x - 2.
  2. Rearrange to x2 - 3x + 2 = 0.
  3. Factorise: (x - 1)(x - 2) = 0, so x = 1 or x = 2.
  4. Substitute into y = x2: when x = 1, y = 1, and when x = 2, y = 4. The points are (1, 1) and (2, 4).
How the marks are awarded. 1 mark for equating the two expressions. 1 mark for a correct quadratic equal to zero. 1 mark for both x values. 1 mark for both coordinate pairs written correctly.
Where students lose the mark. Giving only the x values. The question asks for coordinates, so both numbers in each pair are needed.

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Simultaneous Equations FAQs

When should I use elimination instead of substitution?

Use elimination when both equations are in the form ax + by = c, because matching one coefficient and adding or subtracting is quicker and less error prone. Use substitution when one equation already expresses one variable in terms of the other, or when one of the equations is not linear.

How do I solve simultaneous equations when one is a quadratic?

Rearrange the linear equation to make one variable the subject, substitute that expression into the non-linear equation, and simplify to a quadratic in one variable. Solve it by factorising or with the quadratic formula, then substitute each root back into the linear equation to find the matching value of the other variable.

How many marks do I lose for not showing working?

Usually most of them. Simultaneous equations questions are marked with method marks for the elimination or substitution step and the rearrangement. A correct final answer with no working typically scores far below full marks, and an answer with an arithmetic slip but clear method often scores nearly all of them.

What if the two equations have no solution?

If eliminating a variable leaves a false statement such as 0 = 7, the equations represent parallel lines and there is no solution. If it leaves a true statement such as 0 = 0, the equations describe the same line and there are infinitely many solutions.

This page covers solving two equations together. For a single linear equation or an inequality, see Linear Equations and Inequalities.

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Written to the published Cambridge IGCSE Mathematics (0580) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.