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Rearranging Formulae

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AQA Level 2 Further Maths practice questions, each with a hint and a full worked solution.

Have a pen and paper handy, to work out your answers.

Question 13 marks

Make r the subject of

p=\dfrac{2r-a}{r+3}

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Multiply both sides by (r+3) first, then get every term containing r on the same side.

Worked solution
  1. Multiply both sides by (r+3): \;p(r+3)=2r-a
  2. Expand: \;pr+3p=2r-a
  3. Collect the r terms on one side: \;pr-2r=-a-3p
  4. Factorise out r: \;r(p-2)=-a-3p
  5. Divide by (p-2): \;r=\dfrac{-a-3p}{p-2}
  6. Multiply top and bottom by -1: \;r=\dfrac{a+3p}{2-p}

Question 23 marks

Make n the subject of

A=3\sqrt{\dfrac{k}{n}}-1

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Undo the operations in reverse order: deal with the -1 and the 3 before you square.

Worked solution
  1. Add 1 to both sides: \;A+1=3\sqrt{\dfrac{k}{n}}
  2. Divide by 3: \;\dfrac{A+1}{3}=\sqrt{\dfrac{k}{n}}
  3. Square both sides: \;\dfrac{(A+1)^2}{9}=\dfrac{k}{n}
  4. Multiply by 9n: \;n(A+1)^2=9k
  5. Divide by (A+1)^2: \;n=\dfrac{9k}{(A+1)^2}

Question 3Challenge6 marks

x and y are connected by the formula

\dfrac{x^2+4}{3}=\dfrac{x^2y}{y+2}

where x>0 and y>1

(a)

Make x the subject of the formula.

4 marks

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

(b)

Work out the value of x when y=5

Give your answer in the form \dfrac{\sqrt{a}}{b} where a and b are positive integers and a is as small as possible.

2 marks

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Cross-multiply to clear both fractions, expand the brackets, then collect every x^2 term on one side and factorise.

Worked solution

Part (a)

  1. Cross-multiply: \;(x^2+4)(y+2)=3x^2y
  2. Expand: \;x^2y+2x^2+4y+8=3x^2y
  3. Collect the x^2 terms on one side: \;2x^2+x^2y-3x^2y=-4y-8
  4. Simplify: \;2x^2-2x^2y=-4y-8
  5. Factorise out x^2: \;x^2(2-2y)=-4y-8
  6. Divide: \;x^2=\dfrac{-4y-8}{2-2y}=\dfrac{2(y+2)}{y-1}
  7. Square root (x>0): \;x=\sqrt{\dfrac{2(y+2)}{y-1}}

Part (b)

  1. Substitute y=5 into your formula from part (a): \;x^2=\dfrac{2\times 7}{4}=\dfrac{14}{4}
  2. x=\sqrt{\dfrac{14}{4}}=\dfrac{\sqrt{14}}{\sqrt{4}}
  3. x=\dfrac{\sqrt{14}}{2}

More on this topic: Rearranging Formulae worksheet with full solutions

All AQA Level 2 Further Maths practice questions

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