Linear Formula (SQA National 5 Maths): Revision Note

Changing the subject of a linear formula

What is a formula?

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

  • A formula includes 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

    • $A=\frac{\left(a+b\right)h}{2}$

  • Pythagoras' theorem

    • $a^2+b^2=c^2$

What is the subject of a formula?

• Usually a formula will have a single letter on its own on one side of the equals sign

  • That letter is known as the subject of the formula

    • For $C=2\pi r$, the subject is $C$

    • For $E=\frac{1}{2}m{v}^2$, the subject is $E$

  • You can use the formula to calculate the value of the subject, if you know all the other values

• By rearranging a formula, you can change the subject

  • This means rewriting the formula so a different letter is on its own on one side of the equals sign

How do I rearrange a formula to change 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 you want 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 
      $x=\frac{2y}{5}-\frac{6}{5}$
      $x=\frac{2}{5}y-\frac{6}{5}$
      $x=\frac{2\left(y-3\right)}{5}$
      $x=0.4\left(y-3\right)$
      $x=0.4y-1.2$

Should I expand brackets?

• Expand brackets if it releases the variable you want from inside the brackets

  • If not, you can leave them in

• To make $x$ the subject of $3(1+x)=y$

  • $x$ is inside the brackets, so expand

    • $3+3x=y$

  • Rearrange

    • $\begin{array}{rcl}3x& =& y-3\end{array}$
        $\begin{array}{rcl}x& =& \frac{y-3}{3}\end{array}$

• To make $x$ the subject of $\left(1+k\right)x=y$

  • $x$ is not inside the brackets, so you do not need to expand

  • Instead, divide both sides by the bracket $\left(1+k\right)$

    • $x=\frac{y}{1+k}$

What if I get fractions in fractions?

• Some rearrangements can lead to fractions in fractions 

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

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

  • $x=\frac{3}{t}÷2$
    $$\begin{gathered} x=\frac{3}{t}÷\frac{2}{1} \\ x=\frac{3}{t}\times \frac{1}{2} \\ x=\frac{3}{2t} \end{gathered}$$

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

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

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 $x=\frac{-\left(y-3\right)}{2}$ (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
    $\frac{-\left(y-3\right)}{2}=\frac{-y+3}{2}=\frac{3-y}{2}$