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Simultaneous

Simultaneous

Substitution, and a line with a curve

Substitution Core + Extended 6 min read

On this page
  1. In this note
  2. Formula bank
  3. Substitution with two lines
  4. A line and a curve Extended
  5. Ways they test it
  6. Mistakes to avoid
  7. Summary
  8. Practice
  9. Can I do this?
  10. Quiz

In this note

Syllabus 2.5: substitution; a line and a curve
  • 2.5Solve simultaneous equations by substitution when one letter is already the subject.
  • 2.5Solve a linear and a non-linear equation together Extended.
  • 2.10Understand the answers as the points where a line meets a curve.

Before you start: Elimination and solving quadratics.

Formula bank

SubstitutionLearn itif \(y=2x-1\), replace every \(y\) by \((2x-1)\)keep the brackets.Use it when one equation already gives x = … or y = ….

Substitution with two lines

If one equation says \(y=2x-1\), then \(y\) and \(2x-1\) are the same number. Replace \(y\) in the other equation by \((2x-1)\), in brackets, and you have one equation in \(x\).

General methodHow to solve by substitution

  1. Make one letter the subject of the simpler equation (it often is already).
  2. Substitute it, in brackets, into the other equation.
  3. Solve for the remaining letter.
  4. Put that value back into the subject equation to find the other letter.

Example Solve \(y=2x-1\) and \(3x+2y=19\).

Replace \(y\) by \((2x-1)\).

  1. \(3x+2(2x-1)=19\)
  2. \(7x-2=19\Rightarrow x=3\)
  3. \(y=2(3)-1=5\)

Answer \(x=3,\ y=5\)

Try it yourself

1
Solve \(y=3x\) and \(x+y=12\).
Show answer

Answer \(x=3,\ y=9\)

2
Solve \(x=3y+2\) and \(2x-5y=7\).
Hint
\(2(3y+2)-5y=7\).
Show answer

Answer \(x=11,\ y=3\)

A line and a curve Extended

When one equation has \(x^{2}\), \(xy\) or \(y^{2}\), substituting gives a QUADRATIC, so there are usually TWO pairs of answers: the two points where the line crosses the curve.

General methodHow to solve a linear and a non-linear equation

  1. Rearrange the LINEAR equation to y = … (or x = …).
  2. Substitute into the non-linear equation, in brackets, and expand.
  3. Rearrange to a quadratic = 0 and solve (factorise or formula).
  4. For EACH x, find its y from the linear equation. Write the answers as pairs.

Example Solve \(x+y=9\) and \(y=x^{2}+3\).

\(y=9-x\); substitute into the curve.

  1. \(9-x=x^{2}+3\)
  2. \(x^{2}+x-6=0\Rightarrow(x+3)(x-2)=0\)
  3. \(x=-3\Rightarrow y=12\); \(x=2\Rightarrow y=7\)

Answer \((-3,\ 12)\) and \((2,\ 7)\)

Try it yourself

1
Solve \(y=x^{2}\) and \(y=x+6\). Give the two x-values.
Show answer

Answer 3, −2

Ways they test it

Type 1 Substitution (linear)

"Solve \(x=3y+2\) and \(2x-5y=7\)." (3 marks)

Spot it one equation is x = … or y = ….

  1. Substitute in brackets.
  2. Solve.
  3. Find the other letter.

Type 2 Line and parabola

"Solve \(y=x^{2}-2x\) and \(y=x+4\)." (4 marks, Extended)

Spot it \(y=\) quadratic and \(y=\) line.

  1. Set the two expressions equal.
  2. Quadratic = 0.
  3. Two pairs.

For example \(x^{2}-2x=x+4\)

  1. \(x^{2}-3x-4=0\Rightarrow(x-4)(x+1)=0\)

Answer \((4,\ 8)\), \((-1,\ 3)\)

Type 3 xy or a circle

"Solve \(2x+y=7\) and \(x^{2}+xy=10\)." (5 marks, Extended)

Spot it an \(xy\) or \(x^{2}+y^{2}\) term.

  1. y = 7 − 2x.
  2. \(x^{2}+x(7-2x)=10\).
  3. Quadratic = 0; pair each x with its y.

For example \(x^{2}-7x+10=0\)

  1. \(x=2\) or \(5\)

Answer \((2,\ 3)\), \((5,\ -3)\)

Type 4 Where do they meet?

"Find the coordinates of the points where the line ... meets the curve ..."

Spot it 'points of intersection'.

  1. Same method; answer as coordinates.

Mistakes to avoid

  • ✗ \(3x+2\cdot2x-1\)✓ \(3x+2(2x-1)\)Brackets when substituting.
  • ✗ \((x+1)^{2}=x^{2}+1\)✓ \(x^{2}+2x+1\)Expand fully.
  • ✗ x = 2, y = −3 (wrong partner)✓ \((2,\ 3)\) and \((5,\ -3)\)Each x has its own y.

Summary

  • Substitute the subject, in brackets.
  • Line + curve → a quadratic → usually two pairs.
  • Use the LINEAR equation to find each partner value.
  • The answers are where the graphs meet.

If you see …

  • one equation is y = … alreadysubstitute it into the other.
  • \(x^{2}\), \(xy\) or \(y^{2}\) in one equationexpect two pairs of answers.

Practice

Practice: easy, medium, hard

Easy

1
Solve \(y=2x-1\) and \(3x+2y=19\).
Show answer

Answer \(x=3,\ y=5\)

2
Solve \(y=3x\) and \(x+y=12\).
Show answer

Answer \(x=3,\ y=9\)

3
Solve \(x=3y+2\) and \(2x-5y=7\).
Show answer

Answer \(x=11,\ y=3\)

Medium

1
Solve \(y=x^{2}-3x+1\) and \(y=2x-3\). Give the x-values.
Guide me

What is it asking? x-values

What do we know? both are y =

Plan \(x^{2}-3x+1=2x-3\)

Show working
  1. \(x^{2}-5x+4=0\)
  2. \((x-1)(x-4)=0\)

Answer 1, 4

2
Solve \(x+y=9\) and \(y=x^{2}+3\). Give the y-values.
Guide me

What is it asking? y-values

What do we know? \(y=9-x\)

Plan \(9-x=x^{2}+3\)

Show working
  1. \(x^{2}+x-6=0\)
  2. \(x=-3\) or \(2\)
  3. \(y=12\) or \(7\)

Answer 12, 7

3
Solve \(y=x^{2}-2x\) and \(y=x+4\). Give the x-values.
Guide me

What is it asking? x-values

What do we know? set equal

Plan \(x^{2}-3x-4=0\)

Show working
  1. \((x-4)(x+1)=0\)

Answer 4, −1

Hard

1
Solve \(2x+y=7\) and \(x^{2}+xy=10\).
Guide me

What is it asking? two pairs

What do we know? \(y=7-2x\)

Plan \(x^{2}+x(7-2x)=10\)

Show working
  1. \(-x^{2}+7x=10\)
  2. \(x^{2}-7x+10=0\)
  3. \(x=2\Rightarrow y=3\); \(x=5\Rightarrow y=-3\)

Answer \((2,\ 3)\), \((5,\ -3)\)

PartAnswer / workingMarksGuidance
substitutionM1
\(x^{2}-7x+10=0\)M1
\((x-2)(x-5)=0\)M1
x = 2, 5A1
y = 3, −3 pairedA1
Total: 5 marks
2
Solve \(x^{2}+y^{2}=25\) and \(y=x+1\).
Guide me

What is it asking? two pairs

What do we know? substitute y

Plan \(x^{2}+(x+1)^{2}=25\)

Show working
  1. \(2x^{2}+2x-24=0\)
  2. \(x^{2}+x-12=0\)
  3. \(x=3\Rightarrow y=4\); \(x=-4\Rightarrow y=-3\)

Answer \((3,\ 4)\), \((-4,\ -3)\)

PartAnswer / workingMarksGuidance
\(x^{2}+(x+1)^{2}=25\)M1
\(x^{2}+x-12=0\)A1
both pairsA1
Total: 3 marks
3
Solve \(y=3-2x\) and \(xy=1\).
Guide me

What is it asking? two pairs

What do we know? \(x(3-2x)=1\)

Plan \(2x^{2}-3x+1=0\)

Show working
  1. \((2x-1)(x-1)=0\)
  2. \(x=\tfrac12\Rightarrow y=2\); \(x=1\Rightarrow y=1\)

Answer \(\left(\tfrac12,\ 2\right)\), \((1,\ 1)\)

PartAnswer / workingMarksGuidance
\(x(3-2x)=1\)M1
\(2x^{2}-3x+1=0\)M1
\((2x-1)(x-1)=0\)M1
both pairsA2
Total: 5 marks

Can I do this?

Anything unticked: revise that section first. All ticked: take the quiz.

Quiz

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