IGCSE Additional Mathematics: Simultaneous Equations Practice Questions
When one equation is linear and the other is not, substitute the linear equation into the other to produce a quadratic in one variable. If the line is a tangent to the curve, that quadratic has a discriminant of zero.
Topic 6 of Cambridge IGCSE Additional Mathematics 0606 almost always pairs a line with a curve. The tangency condition is the part worth learning cold, since it links this topic directly to the discriminant. The questions below cover both cases.
What you need to know for Simultaneous Equations
- Substitution methodRearrange the linear equation for one variable and substitute into the other. Always substitute into the non-linear equation.
- Pairing solutionsEach value of x pairs with exactly one value of y. Substitute back into the linear equation, which is simpler.
- TangencyA line is a tangent to a curve when the resulting quadratic has exactly one solution, so its discriminant is zero.
- No intersectionIf the discriminant of the resulting quadratic is negative, the line and curve do not meet.
- CheckingSubstitute each coordinate pair back into both original equations.
IGCSE Additional Mathematics Simultaneous Equations questions and answers
4 exam-style questions written to the 0606 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.
Solve the simultaneous equations y = x + 2 and y = x2 minus 4.
Show the worked answer
- Both expressions equal y, so set them equal: x + 2 = x2 minus 4.
- Rearrange to a quadratic equal to zero: x2 minus x minus 6 = 0.
- Factorise: (x minus 3)(x + 2) = 0, so x = 3 or x = minus 2.
- Substitute into y = x + 2: when x = 3, y = 5, and when x = minus 2, y = 0.
Solve the simultaneous equations x + y = 5 and x2 + y2 = 13.
Show the worked answer
- Rearrange the linear equation: y = 5 minus x.
- Substitute into the second equation: x2 + (5 minus x)2 = 13.
- Expand: x2 + 25 minus 10x + x2 = 13, so 2x2 minus 10x + 12 = 0.
- Divide by 2: x2 minus 5x + 6 = 0, which factorises to (x minus 2)(x minus 3) = 0.
- So x = 2 or x = 3, giving y = 3 and y = 2 respectively. The solutions are (2, 3) and (3, 2).
The line y = 2x + k is a tangent to the curve y = x2. Find the value of k.
Show the worked answer
- Set the expressions equal: x2 = 2x + k.
- Rearrange: x2 minus 2x minus k = 0.
- A tangent touches the curve at exactly one point, so the discriminant is zero: (minus 2)2 minus 4 x 1 x (minus k) = 0.
- 4 + 4k = 0, so k = minus 1.
Solve the simultaneous equations 2x + y = 7 and xy = 6.
Show the worked answer
- Rearrange the linear equation: y = 7 minus 2x.
- Substitute into xy = 6: x(7 minus 2x) = 6.
- Expand and rearrange: 7x minus 2x2 = 6, so 2x2 minus 7x + 6 = 0.
- Factorise: (2x minus 3)(x minus 2) = 0, so x = 1.5 or x = 2.
- Substitute back: when x = 1.5, y = 4, and when x = 2, y = 3.
Common mistakes in this topic
- Substituting into the linear equation instead of the non-linear one.
- Expanding a squared bracket without the middle term.
- Failing to pair each x with its own y.
- Forgetting that tangency means a discriminant of zero.
- Sign errors when the constant is negative in the discriminant.
Exam tips
- Always substitute the linear equation into the non-linear one, never the reverse.
- Present answers as coordinate pairs so the pairing is unambiguous.
- Tangent means one solution means discriminant zero. Learn that chain.
- Make the leading coefficient positive before factorising.
- Substitute your pairs back into both original equations as a check.
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Simultaneous Equations FAQs
How do I solve one linear and one quadratic equation?
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, then substitute each root back into the linear equation to find the matching value.
How do I find the condition for a line to be a tangent to a curve?
Set the two expressions equal and rearrange to a quadratic equal to zero. A tangent touches at exactly one point, so that quadratic has one repeated root, which means its discriminant equals zero. Solve that condition for the unknown.
How do I know if a line and curve do not intersect?
Form the quadratic in the usual way and calculate its discriminant. If the discriminant is negative there are no real solutions, so the line and the curve never meet. A positive discriminant gives two points of intersection.
Why must answers be given as coordinate pairs?
Each value of x corresponds to exactly one value of y, so listing the x values and y values separately does not show which goes with which. Writing coordinates makes the pairing explicit, and marks are awarded for the correct pairing.
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Written to the published Cambridge IGCSE Additional Mathematics (0606) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.