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Algebra

Algebra

Linear equations and changing the subject

Core + Extended 7 min read

On this page
  1. In this note
  2. Formula bank
  3. Solving linear equations
  4. Equations with fractions
  5. Changing the subject
  6. Ways they test it
  7. Mistakes to avoid
  8. Summary
  9. Practice
  10. Can I do this?
  11. Quiz

In this note

Syllabus 2.5 and 2.6: linear equations and changing the subject
  • 2.5Solve linear equations with brackets, x on both sides, and fractions.
  • 2.5Solve simple equations with x in the denominator.
  • 2.6Change the subject of a formula, including squares, roots and the subject appearing twice.

Before you start: Expressions and expanding. Next: Algebraic fractions.

Formula bank

Inverse operationsLearn it+ ↔ −, × ↔ ÷, square ↔ square rootundo what was done to x, in reverse order.Use it when solving or rearranging.
Clearing fractionsLearn itmultiply EVERY term by the LCM of the denominatorsput brackets round numerators with two terms.Use it when an equation with fractions.

Solving linear equations

An equation is a balance: whatever you do to one side, do to the other. To find \(x\), undo what has been done to it, in reverse order.

General methodHow to solve a linear equation

  1. Expand any brackets (or divide both sides by the number outside, if it goes in exactly).
  2. Get the x terms on one side (the side with more x) and the numbers on the other.
  3. Divide by the number in front of x.
  4. Check: put your answer back into the original equation.

Example Solve \(7x-4=3x+10\).

Take \(3x\) from both sides, add 4, divide by 4.

  1. \(4x-4=10\)
  2. \(4x=14\)
  3. \(x=3.5\)

Answer \(x=3.5\)

Try it yourself

1
Solve \(2x-9=5\).
Show answer

Answer 7

2
Solve \(3(2x-5)=21\).
Show answer

Answer 6

3
Solve \(5(x+2)=3x\).
Show answer

Answer −5

Equations with fractions

General methodHow to solve an equation with fractions

  1. Find the LCM of the denominators.
  2. Multiply EVERY term (on both sides) by it. Put the numerators in brackets.
  3. Expand, collect and solve as before.
  4. x in a denominator: multiply both sides by that bracket, e.g. \(\dfrac6{x+1}=2\Rightarrow6=2(x+1)\).

Example Solve \(\dfrac{2x-1}3-\dfrac{x+2}4=1\).

LCM 12: multiply every term by 12, numerators in brackets.

  1. \(4(2x-1)-3(x+2)=12\)
  2. \(8x-4-3x-6=12\)
  3. \(5x=22\)

Answer \(x=4.4\)

Try it yourself

1
Solve \(\dfrac x3+\dfrac x4=7\).
Show answer

Answer 12

2
Solve \(\dfrac x2-3=4\).
Show answer

Answer 14

3
Solve \(\dfrac6{x+1}=2\).
Show answer

Answer 2

Changing the subject

Rearranging a formula is solving an equation where the answer is a letter. Use the same inverse operations.

General methodHow to change the subject

  1. Clear any fractions (multiply both sides by the denominator).
  2. Expand any brackets.
  3. Get every term with the new subject on one side, everything else on the other.
  4. If the subject is in two terms, factorise it out.
  5. Divide by the bracket; then take a square root (or square) if the subject was squared (or under a root).

Example Make \(x\) the subject of \(y=\dfrac{2x+3}{x-4}\).

Multiply up, collect the x terms, factorise x out.

  1. \(y(x-4)=2x+3\)
  2. \(xy-4y=2x+3\)
  3. \(xy-2x=4y+3\)
  4. \(x(y-2)=4y+3\)

Answer \(x=\dfrac{4y+3}{y-2}\)

Try it yourself

1
Make \(x\) the subject of \(y=3x-5\).
Show answer

Answer \(x=\dfrac{y+5}3\)

2
Make \(r\) the subject of \(A=\pi r^{2}\).
Show answer

Answer \(r=\sqrt{\dfrac A\pi}\)

3
Make \(a\) the subject of \(P=2a+ab\).
Hint
Factorise: \(P=a(2+b)\).
Show answer

Answer \(a=\dfrac P{2+b}\)

Ways they test it

Type 1 Solve a linear equation

"Solve \(4-3x=2x-11\)." (2 marks)

Spot it one letter, no squares.

  1. Collect x on one side, numbers on the other.
  2. Divide.

For example \(15=5x\)

    Answer \(x=3\)

    Type 2 Equation with fractions

    "Solve \(\dfrac{x+1}2+\dfrac{x-3}5=3\)." (3 marks)

    Spot it fractions with x on top.

    1. × every term by the LCM (10), bracket the numerators.
    2. Expand and solve.

    For example \(5(x+1)+2(x-3)=30\)

    1. \(7x=31\)

    Answer \(x=\tfrac{31}7\)

    Type 3 Rearrange: square or root

    "Make v the subject of \(E=\tfrac12mv^{2}\)." (3 marks)

    Spot it the new subject is squared or under a root.

    1. Undo the × and ÷ first.
    2. Square root (or square) last.

    For example \(2E=mv^{2}\)

    1. \(v^{2}=\tfrac{2E}m\)

    Answer \(v=\sqrt{\dfrac{2E}m}\)

    Type 4 Rearrange: subject twice

    "Make x the subject of \(K=\dfrac{5-3x}{x+2}\)." (4 marks)

    Spot it the subject appears in two places.

    1. Clear the fraction and expand.
    2. Collect the x terms on one side.
    3. Factorise x out and divide.

    For example \(Kx+2K=5-3x\)

    1. \(Kx+3x=5-2K\)
    2. \(x(K+3)=5-2K\)

    Answer \(x=\dfrac{5-2K}{K+3}\)

    Mistakes to avoid

    • ✗ \(\tfrac{x+1}2\times10=5x+1\)✓ \(5(x+1)=5x+5\)Bracket the numerator.
    • ✗ Only the fractions × 10✓ every term × 10Keep the balance.
    • ✗ \(P=2a+ab\Rightarrow a=\tfrac{P-ab}2\)✓ \(a=\tfrac P{2+b}\)The subject must be on its own: factorise.
    • ✗ \(v=\sqrt{2E}\div m\)✓ \(v=\sqrt{\tfrac{2E}m}\)The root covers the whole fraction.

    Summary

    • Same operation to both sides; undo in reverse order.
    • Fractions: × every term by the LCM; bracket the numerators.
    • Rearranging = solving for a letter.
    • Subject twice: collect, factorise, divide.

    If you see …

    • fractions in an equationmultiply every term by the LCM.
    • the subject in two placesget both terms on one side and factorise.

    Practice

    Practice: easy, medium, hard

    Easy

    1
    Solve \(4x+7=23\).
    Show answer

    Answer 4

    2
    Solve \(5-2x=11\).
    Show answer

    Answer −3

    3
    Make \(t\) the subject of \(v=u+at\).
    Hint
    Subtract u, then divide by a.
    Show working
    1. \(v-u=at\)
    2. \(t=\dfrac{v-u}a\)

    Answer \(t=\dfrac{v-u}a\)

    Medium

    1
    Solve \(4-3x=2x-11\).
    Guide me

    What is it asking? x

    What do we know? x on both sides

    Plan add 3x and 11 to both sides

    Show working
    1. \(15=5x\)
    2. \(x=3\)

    Answer 3

    2
    Solve \(\dfrac{x+1}2+\dfrac{x-3}5=3\).
    Guide me

    What is it asking? x

    What do we know? LCM 10

    Plan × every term by 10

    Show working
    1. \(5(x+1)+2(x-3)=30\)
    2. \(7x-1=30\)
    3. \(x=\tfrac{31}7\)

    Answer \(\tfrac{31}7\)

    3
    Make \(v\) the subject of \(E=\tfrac12mv^{2}\).
    Guide me

    What is it asking? v = ...

    What do we know? \(v\) is squared

    Plan × 2, ÷ m, square root

    Show working
    1. \(2E=mv^{2}\)
    2. \(v^{2}=\dfrac{2E}m\)
    3. \(v=\sqrt{\dfrac{2E}m}\)

    Answer \(v=\sqrt{\dfrac{2E}m}\)

    Hard

    1
    Make \(x\) the subject of \(K=\dfrac{5-3x}{x+2}\).
    Guide me

    What is it asking? x = ...

    What do we know? x appears twice

    Plan clear the fraction, collect x, factorise

    Show working
    1. \(Kx+2K=5-3x\)
    2. \(Kx+3x=5-2K\)
    3. \(x(K+3)=5-2K\)

    Answer \(x=\dfrac{5-2K}{K+3}\)

    PartAnswer / workingMarksGuidance
    \(K(x+2)=5-3x\)M1
    \(Kx+3x=5-2K\)M1
    \(x(K+3)=5-2K\)M1
    \(x=\dfrac{5-2K}{K+3}\)A1
    Total: 4 marks
    2
    Make \(m\) the subject of \(R=\dfrac m{m-k}\).
    Guide me

    What is it asking? m = ...

    What do we know? m appears twice

    Plan \(Rm-Rk=m\)

    Show working
    1. \(R(m-k)=m\)
    2. \(Rm-m=Rk\)
    3. \(m(R-1)=Rk\)

    Answer \(m=\dfrac{Rk}{R-1}\)

    PartAnswer / workingMarksGuidance
    \(Rm-Rk=m\)M1
    \(m(R-1)=Rk\)M1
    \(m=\dfrac{Rk}{R-1}\)A1
    Total: 3 marks
    3
    Make \(x\) the subject of \(y=\dfrac{x+1}{x-2}\).
    Guide me

    What is it asking? x = ...

    What do we know? x appears twice

    Plan \(xy-2y=x+1\)

    Show working
    1. \(y(x-2)=x+1\)
    2. \(xy-x=2y+1\)
    3. \(x(y-1)=2y+1\)

    Answer \(x=\dfrac{2y+1}{y-1}\)

    PartAnswer / workingMarksGuidance
    \(xy-2y=x+1\)M1
    \(x(y-1)=2y+1\)M1
    \(x=\dfrac{2y+1}{y-1}\)A1
    Total: 3 marks

    Can I do this?

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

    Quiz

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