Formulas where Subject Appears Twice (Cambridge International IGCSE Mathematics)

Subject Appears Twice

How do I rearrange formulae where the subject appears twice?

• If the subject appears twice, you will need to factorise at some point

  • E.g. When making $x$ the subject of $x+xy=3-2y$

  • Factorise out $x$on the left to get $x ( 1 + y ) = 3 - 2 y$

    • Notice that the subject now only appears once!

  • Then divide both sides by $( 1 + y )$ to get $x=\frac{3-2y}{1+y}$

• If the subject appears twice, and any of these are inside a set of brackets, you will need to expand these brackets first

  • E.g. When making $x$ the subject of $c ( x + 2 ) - x = f$
  • Expand the bracket first to $cx+2c-x=f$
  • Keep the $x$ terms on one side $cx-x=f-2c$
  • $x$ can then be made the subject using factorising as above

• If the subject appears on two sides of a formula, you will need to bring those terms to the same side before you can factorise

  • E.g. When making $x$the subject of $3x=y-px$
  • Add $px$ to both sides first to form $3x+px=y$
  • $x$ can then be made the subject using factorising as above

How do I factorise powers of a subject?

• If the subject appears twice, and both have the same power, you will need to collect these terms together before applying their inverse

  • E.g. When making $x$the subject of $x^{2}=-px^{2}+r$

  • Add $px^{2}$ to both sides first to form $x^{2}+px^{2}=r$

  • $x^{2}$ can then be factorised out $x^{2} ( 1 + p ) = r$ to give $x^{2}=\frac{r}{1+p}$

    • Now take plus-or-minus square roots
    • $x=\pm\sqrt{\frac{r}{1+p}}$

• Be careful when square rooting, or cube rooting etc

  • E.g. To make $x$ the subject of $x^{3}=\frac{t^{3}+1}{t^{3}+8}$

    • The whole right hand side must be cube rooted
    • $x = \sqrt[3]{ \frac{t^{3} + 1}{t^{3} + 8} }$
    • This cannot be simplified further
    • The right hand side is not equal to $\frac{t+1}{t+2}$, (this is a common error)

Worked Example Rearrange the formula $p=\frac{2-ax}{x-b}$ to make $x$ the subject.

Answer:

Get rid of the fraction by multiplying both sides by the expression on the denominator

$p ( x - b ) = 2 - a x$

Expand the brackets on the left hand side to 'release' the $x$

$px-pb=2-ax$

Bring the terms containing $x$ to one side of the equals sign and any other terms to the other side

$\begin{gathered} p x - p b & = 2 - a x \end{gathered}$

$\begin{gathered} ( + a x ) ( + a x ) \end{gathered}$

$\begin{gathered} p x - p b + a x & = 2 \end{gathered}$

$\begin{gathered} ( + p b ) ( + p b ) \end{gathered}$

$\begin{array}{rcl} p x + a x & = 2 + p b \end{array}$

Factorise the left-hand side to bring $x$ outside of the brackets, so that it appears only once

$x ( p + a ) = 2 + p b$

Make $x$ the subject by dividing by the whole expression $( p + a )$

$x=\frac{2+pb}{p+a}$