GCSE Simultaneous (linear/quadratic) Practice Questions

Free GCSE Simultaneous (linear/quadratic) practice questions with full step-by-step worked solutions. Covers substitution, quadratic and horizontal line, factorising by taking out x, common factor. Practise exam-style problems and check your method.

substitutionquadratic and horizontal linefactorising by taking out xcommon factorcirclevertical line
GCSE Higher70 questionsStep-by-step solutions
Question 1
1 markeasy
Solve the simultaneous equations y=x2y = x^2 and y=9y = 9.
Show worked solution

Worked solution

  1. Substitute the line into the curve

    x2=9x^{2} = 9

    Both equations give yy, so where the graphs meet the two expressions for yy are equal. That leaves one equation in xx alone.

  2. Rearrange and solve for x

    x2−9=0 ⇒ (x+3)(x−3)=0 ⇒ x=−3 or x=3x^{2} - 9 = 0 \ \Rightarrow \ (x + 3)(x - 3) = 0 \ \Rightarrow \ x = -3 \ \text{or} \ x = 3

    Rearrange so the quadratic equals zero, factorise, then set each factor to zero. Never divide through by xx — that would throw away the root x=0x = 0.

  3. State the two solutions

    x=−3, y=9orx=3, y=9x = -3,\ y = 9 \quad \text{or} \quad x = 3,\ y = 9

    The line meets the curve at (-3, 9) and (3, 9).

Answer
x=−3, y=9orx=3, y=9x = -3,\ y = 9 \quad \text{or} \quad x = 3,\ y = 9
Question 2
2 markseasy
Solve the simultaneous equations y=x2y = x^2 and y=x+12y = x + 12.
Show worked solution

Worked solution

  1. Substitute the line into the curve

    x2=x+12x^{2} = x + 12

    Both equations give yy, so where the graphs meet the two expressions for yy are equal. That leaves one equation in xx alone.

  2. Rearrange and solve for x

    x2−x−12=0 ⇒ (x+3)(x−4)=0 ⇒ x=−3 or x=4x^{2} - x - 12 = 0 \ \Rightarrow \ (x + 3)(x - 4) = 0 \ \Rightarrow \ x = -3 \ \text{or} \ x = 4

    Rearrange so the quadratic equals zero, factorise, then set each factor to zero. Never divide through by xx — that would throw away the root x=0x = 0.

  3. State the two solutions

    x=−3, y=9orx=4, y=16x = -3,\ y = 9 \quad \text{or} \quad x = 4,\ y = 16

    The line meets the curve at (-3, 9) and (4, 16).

Answer
x=−3, y=9orx=4, y=16x = -3,\ y = 9 \quad \text{or} \quad x = 4,\ y = 16
Question 3
2 marksintermediate
The line y=2x−1y = 2x - 1 is a tangent to the curve y=x2y = x^2. Work out the coordinates of the point of contact.
Show worked solution

Worked solution

  1. Substitute the line into the curve

    x2=2x−1x^{2} = 2 x - 1

    Both equations give yy, so where the graphs meet the two expressions for yy are equal. That leaves one equation in xx alone.

  2. Rearrange so the quadratic equals zero

    x2−2x+1=0x^{2} - 2 x + 1 = 0

    Collect every term on one side. A quadratic can only be factorised once it is equal to zero.

  3. Factorise

    (x−1)2=0(x - 1)^2 = 0

    Look for the pair of numbers that multiply to give the constant term and add to give the coefficient of xx.

  4. Solve for x

    x=1x = 1

    A repeated root: there is only one value of xx, so only one point of contact.

  5. Find the y that goes with the first x-value

    y=2(1)−1=1y = 2 \left(1\right) - 1 = 1

    Substitute this xx into the LINEAR equation — it is far less work than the quadratic, and it gives the yy that belongs to this xx.

  6. State the point of contact

    (1, 1)\left(1,\ 1\right)

    The line touches the curve at exactly one point, (1, 1), so it is a tangent.

Answer
(1, 1)\left(1,\ 1\right)
Question 4
3 markshard
The curve y=x2y = x^2 and the line y=6−xy = 6 - x are drawn on the same axes. They cross at two points whose xx-coordinates are −3-3 and 22. Work out the coordinates of both crossing points, and confirm them algebraically.
Show worked solution

Worked solution

  1. Read where the graphs cross

    x=−3andx=2x = -3 \quad \text{and} \quad x = 2

    The solutions of the simultaneous equations are exactly the points that lie on both graphs, so read off the two crossing points.

  2. Read the y-coordinate at each crossing

    (−3, 9)and(2, 4)\left(-3,\ 9\right) \quad \text{and} \quad \left(2,\ 4\right)

    Go up (or down) from each crossing to the yy-axis. Each xx keeps the yy it was read with.

  3. Label the two equations

    (1) y=x2(2) y=6−x(1)\ y = x^{2} \qquad (2)\ y = 6 - x

    One equation is quadratic and one is linear, so you cannot eliminate a variable by adding or subtracting. Substitution is the only route.

  4. Substitute the line into the curve

    x2=6−xx^{2} = 6 - x

    Both equations give yy, so where the graphs meet the two expressions for yy are equal. That leaves one equation in xx alone.

  5. Rearrange so the quadratic equals zero

    x2+x−6=0x^{2} + x - 6 = 0

    Collect every term on one side. A quadratic can only be factorised once it is equal to zero.

  6. Factorise

    (x+3)(x−2)=0(x + 3)(x - 2) = 0

    Look for the pair of numbers that multiply to give the constant term and add to give the coefficient of xx.

  7. Solve for x

    x=−3orx=2x = -3 \quad \text{or} \quad x = 2

    These are the xx-coordinates of the two points where the line meets the curve.

  8. Find the y that goes with the first x-value

    y=6−(−3)=9y = 6 - \left(-3\right) = 9

    Substitute this xx into the LINEAR equation — it is far less work than the quadratic, and it gives the yy that belongs to this xx.

  9. Find the y that goes with the second x-value

    y=6−(2)=4y = 6 - \left(2\right) = 4

    Substitute this xx into the LINEAR equation — it is far less work than the quadratic, and it gives the yy that belongs to this xx.

  10. State the solutions

    x=−3, y=9orx=2, y=4x = -3,\ y = 9 \quad \text{or} \quad x = 2,\ y = 4

    The algebra confirms exactly what the graph shows: the two crossing points are (-3, 9) and (2, 4).

Answer
(−3, 9)and(2, 4)\left(-3,\ 9\right) \quad \text{and} \quad \left(2,\ 4\right)
Question 5
6 markschallenging
The line y=2x+ky = 2x + k crosses the circle x2+y2=5x^2 + y^2 = 5 at two distinct points. Find the range of values of kk.
Show worked solution

Worked solution

  1. Substitute the line into the circle equation

    x2+(2x+k)2=5x^2 + \left(2x + k\right)^2 = 5

    Substitution leaves a quadratic in xx whose coefficients involve kk.

  2. Expand the bracket

    x2+4x2+4kx+k2=5x^2 + 4x^{2} + 4kx + k^{2} = 5

    Use (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 with b=kb = k.

  3. Rearrange so the quadratic equals zero

    5x2+4kx+(k2−5)=05x^{2} + 4 kx + \left(k^{2} - 5\right) = 0

    Collect like terms and take the constant across.

  4. Write down a, b and c

    a=5,b=4k,c=k2−5a = 5,\quad b = 4 k,\quad c = k^{2} - 5

    Both bb and cc contain the unknown kk.

  5. "Two distinct points" means two distinct real roots

    b2−4ac>0b^2 - 4ac > 0

    Each real root of the quadratic gives one point of intersection, so two distinct points require a strictly positive discriminant.

  6. Write down the discriminant

    b2−4ac=(4k)2−4(5)(k2−5)b^2 - 4ac = \left(4 k\right)^2 - 4\left(5\right)\left(k^{2} - 5\right)

    Substitute the coefficients.

  7. Simplify

    =100−4k2= 100 - 4 k^{2}

    The k2k^2 terms partly cancel, leaving a quadratic in kk.

  8. Form the inequality

    100−4k2>0100 - 4 k^{2} > 0

    This is the condition for two distinct intersection points.

  9. Rearrange

    k2<25k^2 < 25

    Divide by the negative coefficient of k2k^2 and REVERSE the inequality sign — the classic slip here.

  10. Take the square root

    −5<k<5-5 < k < 5

    k2<25k^2 < 25 means ∣k∣<5|k| < 5, which is the double inequality −5<k<5-5 < k < 5.

  11. Check the boundary values

    k=±5: b2−4ac=0k = \pm 5: \ b^2 - 4ac = 0

    At the boundaries the discriminant is zero: the line is a tangent, touching the circle at one point rather than cutting it at two. So the boundaries are excluded.

  12. Check a value inside the range

    k=0: b2−4ac=100>0 ✓k = 0: \ b^2-4ac = 100 > 0 \ \checkmark

    The line y=2xy = 2x passes through the centre of the circle, so of course it cuts it twice.

  13. Check a value outside the range

    k=6: b2−4ac=−44<0 ✓k = 6: \ b^2-4ac = -44 < 0 \ \checkmark

    A negative discriminant means the line misses the circle entirely — outside the required range, as expected.

  14. Interpret geometrically

    tangents at k=−5 and k=5\text{tangents at } k = -5 \text{ and } k = 5

    As kk increases the line slides upwards. Between the two tangent positions it cuts the circle twice; outside them it misses.

  15. State the range

    −5<k<5-5 < k < 5

    The line cuts the circle at two distinct points exactly when −5<k<5-5 < k < 5.

Answer
−5<k<5-5 < k < 5

Unlock 65 more Simultaneous (linear/quadratic) questions

Create a free account to work through every GCSE Simultaneous (linear/quadratic) question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Simultaneous (linear/quadratic) practice

Hard QuestionsWorked Solutions

Related Algebra topics

Algebraic fractionsEquation of a circleCoordinatesCubic and reciprocal graphsExponential and trig graphsExpanding binomials