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Learn7357Paper 2 · Pure and MechanicsPure content (sections A–J)Simultaneous equations and lines meeting curves
Pure content (sections A–J) · Subtopic 5 of 66

Simultaneous Equations and Lines Meeting Curves

Solve one linear and one quadratic equation together by substitution, find the points where a line meets a curve, and use the discriminant to decide whether a line crosses, touches or misses a curve.

Cambridge 9709 Paper 1Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2AQA 7357 Paper 3≈ 40 minRevision 5 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Solve a linear and a quadratic equation simultaneously by substitution.
  • Give the answers as matching pairs, and use them to find lengths and midpoints.
  • Use the discriminant of the resulting quadratic to decide how many times a line meets a curve.
  • Find the condition for a line to be a tangent to a curve, and the point where it touches.
Pause and think

A straight line and a U-shaped parabola. What are all the possible numbers of meeting points?

Section 2 of 14

Core idea

Start hereWhere does the ramp meet the hill?

A skateboard ramp follows the straight line y=x+2y=x+2 and the ground beside it curves as y=x2y=x^2. Where do they meet? A meeting point lies on both, so its coordinates satisfy both equations at once. Solving two equations together is called solving them simultaneously. The answer: (2,4)(2,4) and (−1,1)(-1,1).

The big idea

To solve a linear and a quadratic equation simultaneously, substitute the linear one into the quadratic. This gives a single quadratic equation in one variable. Its roots are the xx-coordinates of the intersection points; put them into the linear equation to get the yy-coordinates. The discriminant tells you how many intersections there are.

1. Two linear equations (revision)

Two unknowns need two equations. Elimination: add or subtract the equations to remove one variable. Substitution: make one variable the subject and replace it in the other. Both work because if two things are equal, you can swap one for the other, or add equal amounts to both sides.

Worked example2x+3y=122x+3y=12 and x−y=1x-y=1

Solve 2x+3y=122x+3y=12 and x−y=1x-y=1.

Goal
The pair (x,y)(x,y) satisfying both.
Strategy
Substitution: the second equation easily gives x=y+1x=y+1.
  1. x=y+1x=y+1
    Add yy to both sides of x−y=1x-y=1.
  2. 2(y+1)+3y=122(y+1)+3y=12
    Replace xx in the first equation.
  3. 5y+2=12⇒y=25y+2=12\Rightarrow y=2
    Expand, collect, solve.
  4. x=2+1=3x=2+1=3
    Back-substitute.
Answerx=3x=3, y=2y=2.
✓ Check2(3)+3(2)=122(3)+3(2)=12 ✓; 3−2=13-2=1 ✓.

2. A linear and a quadratic

y = x² and y = x + 2 cross twice
−3−2−1123−112345670xyy = x²y = x + 2(−1, 1)(2, 4)
The two crossing points are the two solutions of the pair of equations.
Worked exampleThe ramp: y=x+2y=x+2 and y=x2y=x^2

Find where y=x+2y=x+2 meets y=x2y=x^2.

Goal
All intersection points (x,y)(x,y).
Strategy
Both equal yy, so set them equal; solve; back-substitute into the line.
  1. x2=x+2x^2=x+2
    At a meeting point the yy-values are equal.
  2. x2−x−2=0x^2-x-2=0
    Rearrange to =0=0.
  3. (x−2)(x+1)=0⇒x=2 or x=−1(x-2)(x+1)=0\Rightarrow x=2\text{ or }x=-1
    Factorise.
  4. x=2: y=4;x=−1: y=1x=2:\ y=4;\qquad x=-1:\ y=1
    Use the linear equation y=x+2y=x+2 (simplest).
Answer(2,4)(2,4) and (−1,1)(-1,1).
✓ CheckBoth satisfy y=x2y=x^2 too: 22=42^2=4 ✓, (−1)2=1(-1)^2=1 ✓.
Worked exampleA harder pair: x+2y=5x+2y=5 and x2+y2=10x^2+y^2=10

Solve x+2y=5x+2y=5 and x2+y2=10x^2+y^2=10.

Goal
All solutions.
Strategy
Make xx the subject of the linear equation (avoids fractions), substitute into the circle.
  1. x=5−2yx=5-2y
    Rearrange the linear equation.
  2. (5−2y)2+y2=10(5-2y)^2+y^2=10
    Substitute.
  3. 25−20y+4y2+y2=1025-20y+4y^2+y^2=10
    Expand (5−2y)2(5-2y)^2 carefully: three terms.
  4. 5y2−20y+15=0⇒y2−4y+3=05y^2-20y+15=0\Rightarrow y^2-4y+3=0
    Collect; divide by 55.
  5. (y−1)(y−3)=0⇒y=1 or 3(y-1)(y-3)=0\Rightarrow y=1\text{ or }3
    Factorise.
  6. y=1: x=3;y=3: x=−1y=1:\ x=3;\qquad y=3:\ x=-1
    Back-substitute into x=5−2yx=5-2y.
Answer(3,1)(3,1) and (−1,3)(-1,3).
✓ Check9+1=109+1=10 ✓ and 1+9=101+9=10 ✓; 3+2=53+2=5 ✓ and −1+6=5-1+6=5 ✓.
Common misconceptionPairing the answers wrongly Each xx goes with its own yy. Write answers as coordinate pairs, and substitute into the linear equation; substituting into the quadratic can give extra, false yy-values (y2=1y^2=1 gives y=±1y=\pm1).

3. How many intersections?

The line y = x + c moving down past y = x²
−3−2−1123−22460xyy = x²
Substituting gives x2−x−c=0x^2-x-c=0, with discriminant 1+4c1+4c. Two points while c>−14c>-\tfrac14, one (a tangent) at c=−14c=-\tfrac14, none below that.
y = x² + 1 and lines y = mx through the origin
−3−2−1123−112345670xyy = x² + 1y = mx
x2+1=mx⇒x2−mx+1=0x^2+1=mx\Rightarrow x^2-mx+1=0, Δ=m2−4\Delta=m^2-4. Shallow lines (m<2m<2) miss; m=2m=2 touches at (1,2)(1,2); steeper lines cut twice.
Worked exampleTangent condition

Find cc such that y=x+cy=x+c is a tangent to y=x2y=x^2, and find the point of contact.

Goal
cc with exactly one meeting point.
Strategy
Substitute; set Δ=0\Delta=0.
  1. x2=x+c⇒x2−x−c=0x^2=x+c\Rightarrow x^2-x-c=0
    Equate and rearrange.
  2. Δ=1+4c=0⇒c=−14\Delta=1+4c=0\Rightarrow c=-\tfrac14
    a=1a=1, b=−1b=-1, ccoef=−cc_{\text{coef}}=-c.
  3. x2−x+14=(x−12)2=0⇒x=12x^2-x+\tfrac14=\left(x-\tfrac12\right)^2=0\Rightarrow x=\tfrac12
    Repeated root.
  4. y=12−14=14y=\tfrac12-\tfrac14=\tfrac14
    Use the line.
Answerc=−14c=-\frac14; contact at (12,14)\left(\frac12,\frac14\right).
✓ Check(12)2=14\left(\frac12\right)^2=\frac14 ✓: the point is on the parabola.
Linear + quadratic: the method

Always finish with coordinate pairs.

Pause and think

Show that y=2x−5y=2x-5 does not meet y=x2y=x^2.

Summary

  • Substitute the linear equation into the quadratic.
  • Solve; back-substitute into the linear one; pair answers.
  • Δ\Delta counts intersections; Δ=0\Delta=0 is a tangent.
Key words
Simultaneous equations
Equations true at the same time.
Intersection
A point lying on both graphs.
Back-substitute
Put a found value into an earlier equation to get the other unknown.
HistoryTwo thousand years of simultaneous equations. The ancient Chinese book The Nine Chapters on the Mathematical Art, compiled over several centuries up to about the 1st century CE, solves systems of equations about grain and livestock with a method almost identical to the row operations used in computers today.
In the real worldYour phone does this to find you. A GPS receiver knows its distance from several satellites. Each distance puts it on a sphere around a satellite, and its position is where those spheres meet. Finding that point means solving simultaneous quadratic equations, many times a second.
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