## Literal Equations

Contents

A literal equation is one that has several letters or variables. Examples include the area of a circle \((A=\pi {r}^{2})\) and the formula for speed \((v=\cfrac{D}{t})\). In this section we solve literal equations in terms of one variable. To do this, we use the principles we have learnt about solving equations and apply them to rearranging literal equations. Solving literal equations is also known as changing the subject of the formula.

Keep the following in mind when solving literal equations:

We isolate the unknown variable by asking “what is it joined to?” and “how is it joined?” We then perform the opposite operation to both sides as a whole.

If the unknown variable is in two or more terms, then we take it out as a common factor.

If we have to take the square root of both sides, remember that there will be a positive and a negative answer.

If the unknown variable is in the denominator, we multiply both sides by the lowest common denominator (LCD) and then continue to solve.

The following video shows an example of solving literal equations.

## Example

### Question

The area of a triangle is \(A=\cfrac{1}{2}bh\). What is the height of the triangle in terms of the base and area?

### Isolate the required variable

We are asked to isolate the height, so we must rearrange the equation with \(h\) on one side of the equals sign and the rest of the variables on the other.

\begin{align*} A & = \cfrac{1}{2}bh \\ 2A & = bh \\ \cfrac{2A}{b} & = h \end{align*}

### Write the final answer

The height of a triangle is given by: \(h = \dfrac{2A}{b}\)

## Example

### Question

Given the formula: \[h = R \times \cfrac{H}{R + {r}^{2}}\] make \(R\) the subject of the formula.

### Isolate the required variable

\begin{align*} h(R + {r}^{2}) & = R \times H \\ hR + h{r}^{2} & = HR \\ h{r}^{2} & = HR – hR \\ h{r}^{2} & = R(H – h) \\ \therefore R & = \cfrac{h{r}^{2}}{H – h} \end{align*}