Rearranging Formulas (Cambridge International IGCSE Mathematics)

Simple Rearranging

What are formulas?

• A formula is a rule, definition or relationship between different quantities, written in shorthand using letters(variables)

  • Formulas include an equals sign

• Some examples you should be familiar with are:

  • The equation of a straight line

    • $y=mx+c$

  • The area of a trapezium

    • $\text{Area} = \frac{( a + b ) h}{2}$

  • Pythagoras' theorem

    • $a^{2}+b^{2}=c^{2}$

How do I rearrange formulas?

• The letter (variable) that is on its own on one side is called the subject

  • y is the subject of y = mx + c

• To make a different letter the subject, we need to rearrange the formula

  • This is also called changing the subject

• The method is as follows:

  • First, remove any fractions

    • Multiply both sides by the lowest common denominator

  • Then use inverse (opposite) operations to get the variable on its own

    • This is similar to solving equations

• For example, make $x$ the subject of $\frac{5x+6}{2}=y$

  • First remove fractions

    • Multiply both sides by 2 $5x+6=2y$

  • Then get $x$ on its own

    • Subtract 6 from both sides $5x=2y-6$
    • Divide both sides by 5 $x=\frac{2y-6}{5}$

  • There may be more than one correct way to write an answer

    • The following are acceptable alternative forms $x=\frac{2y}{5}-\frac{6}{5}$ $x = \frac{2 ( y - 3 )}{5}$ $x = 0.4 ( y - 3 )$ $x=0.4y-1.2$

Should I expand brackets?

• If the variable is inside the brackets, you can either:

  • expand the brackets
  • or **divide **both by the coefficient if it is not zero

• For example,**** make $x$ the subject of $3(1+x)=y$

  • You can expand and then rearrange

    • $3+3x=y$
    • $\begin{gathered} 3 x & = y - 3 \end{gathered}$
    • $\begin{gathered} x = \frac{y - 3}{3} \end{gathered}$

  • You can divide and then rearrange

    • $1+x=\frac{y}{3}$
    • $x=\frac{y}{3}-1$

  • Both answers are equivalent

    • $\frac{y-3}{3}$ is the same as $\frac{y}{3}-1$

Exam Hint Stop and think when you see a bracket. If the variable you are trying to make the subject is not inside the bracket, then you do not need to expand.

For example, to make $x$ the subject of $( 1 + k ) x = y$, you can simply divide both sides by $( 1 + k )$ to get $x=\frac{y}{1+k}$.

What if I get fractions in fractions?

• Some rearrangements can lead to fractions in fractions

  • $x=\frac{\frac{3}{t}}{2}$

• Either rewrite with a divide sign, $\div$, then use the method of dividing two fractions

  • $x=\frac{3}{t}\div2$ $x=\frac{3}{t}\div\frac{2}{1}\,x=\frac{3}{t}\times\frac{1}{2}\,x=\frac{3}{2t}$

• Or multiply top and bottom by the the lowest common denominator of the two fractions and cancel

  • $x=\frac{\frac{5}{y}}{\frac{t}{8}}$ becomes $x=\frac{\frac{5}{y}\times8y}{\frac{t}{8}\times8y}=\frac{40}{ty}$

What if I end up dividing by a negative?

• Remember that $\frac{a}{-b}$ (minus below) is the same as $\frac{-a}{b}$ (minus above) and the same as $-\frac{a}{b}$ (minus outside)

  • Though be careful, as $\frac{-a}{-b}$ is $\frac{a}{b}$

• $-2x=y-3$ becomes $x=\frac{y-3}{-2}$ (minus below)

  • This is the same as $x = \frac{- ( y - 3 )}{2}$ (minus above) or $-\frac{y-3}{2}$ (minus outside)

    • brackets are required for minus above
    • brackets are assumed for minus outside

  • You can also expand the brackets $\frac{- ( y - 3 )}{2} = \frac{- y + 3}{2} = \frac{3 - y}{2}$

Exam Hint

• Mark schemes will accept different forms of the same answer, as long as they are correct and fully simplified.

Worked Example Make $x$ the subject of the following.

(a) $4m+5x=3$

Answer:

Get 5x on its own by subtracting 4m from both sides

$5x=3-4m$

Get x on its own by dividing both sides by 5

$x=\frac{3-4m}{5}$

(b) $3t=\frac{2}{x}$

Answer:

Remove fractions by multiplying both sides by the denominator, x

$3tx=2$

Get x on its own by dividing both sides by 3**t

$x=\frac{2}{3t}$

(c) $A = \frac{9 ( 1 - 4 x )}{2 g}$

Answer:

Remove fractions by multiplying both sides by the denominator, 2**g

$2 g A = 9 ( 1 - 4 x )$

x is inside the brackets Expand the brackets to release the x term

$2gA=9-36x$

One way to get x on its own is by subtracting 9 then dividing by -36 Or you can first add 36x to both sides, to create positive 36x on the left

$2gA+36x=9$

Now get x on its own by subtracting 2 gA then dividing by 36

$\begin{gathered} 36 x & = 9 - 2 g A \\ x = \frac{9 - 2 g A}{36} \end{gathered}$

$x=\frac{9-2gA}{36}$

Final answer: Other accepted forms of the answer are:$\frac{2 g A - 9}{- 36} , \frac{- ( 2 g A - 9 )}{36} , - \frac{2 g A - 9}{36} , 1 \frac{1}{4} - \frac{g A}{18}$