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# Literal Equations

## 4.6 Literal equations (EMA3G)

A literal equation is one that has several letters or variables. Examples include the area of a circle $$\left(A=\pi {r}^{2}\right)$$ and the formula for speed $$\left(v=\frac{D}{t}\right)$$. 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.

Video: 2FFQ

## Worked example 15: Solving literal equations

The area of a triangle is $$A=\frac{1}{2}bh$$. 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 & = \frac{1}{2}bh \\ 2A & = bh \\ \frac{2A}{b} & = h \end{align*}

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

## Worked example 16: Solving literal equations

Given the formula: $h = R \times \frac{H}{R + {r}^{2}}$ make $$R$$ the subject of the formula.

### Isolate the required variable

\begin{align*} h\left(R + {r}^{2}\right) & = R \times H \\ hR + h{r}^{2} & = HR \\ h{r}^{2} & = HR - hR \\ h{r}^{2} & = R\left(H - h\right) \\ \therefore R & = \frac{h{r}^{2}}{H - h} \end{align*}

# Success in Maths and Science unlocks opportunities

Exercise 4.5

Solve for $$x$$ in the following formula: $$2x + 4y = 2$$.

\begin{align*} 2x + 4y &= 2 \\ 2x &= 2 - 4y \\ \frac{1}{2}(2x) &= (2 - 4y)\frac{1}{2} \\ x &= 1 - 2y \end{align*}

Make $$a$$ the subject of the formula: $$s = ut + \frac{1}{2}at^{2}$$.

\begin{align*} s & = ut + \frac{1}{2}a^{2} \\ s - ut & = \frac{1}{2}at^{2} \\ 2s - 2ut & = at^{2} \\ \frac{2(s - ut)}{t^{2}} & = a \end{align*}

Note restriction: $$t \ne 0$$

Solve for $$n$$: $$pV = nRT$$.
\begin{align*} pV & = nRT \\ \frac{pV}{RT} & = n \end{align*}

Note restrictions: $$R \ne 0$$, $$T \ne 0$$

Make $$x$$ the subject of the formula: $$\dfrac{1}{b} + \dfrac{2b}{x} = 2$$.

\begin{align*} \frac{1}{b} + \frac{2b}{x} & = 2 \\ \frac{x + b(2b)}{bx} & = 2 \\ x + 2b^{2} & = 2bx \\ x - 2bx & = -2b^{2} \\ x(1 - 2b) & = -2b^{2} \\ x & = \frac{-2b^{2}}{1 - 2b} \end{align*}

Note restriction: $$1 \ne 2b$$

Solve for $$r$$ : $$V = \pi r^{2}h$$.

\begin{align*} V & = \pi r^{2}h \\ \frac{V}{\pi h} & = r^{2} \\ \pm \sqrt{\frac{V}{\pi h}} & = r \end{align*}

Note restriction: $$h \ne 0$$

Solve for $$h$$: $$E = \dfrac{hc}{\lambda}$$.

\begin{align*} E & = \frac{hc}{\lambda} \\ E \lambda & = hc \\ \frac{E \lambda}{c} & = h \end{align*}

Note restriction: $$c \ne 0$$

Solve for $$h$$: $$A = 2\pi rh + 2\pi r$$.

\begin{align*} A & = 2\pi rh + 2\pi r \\ A - 2\pi r & = 2\pi rh \\ \frac{A - 2 \pi r}{2 \pi r} & = h \end{align*}

Note restriction: $$r \ne 0$$

Make $$\lambda$$ the subject of the formula: $$t = \dfrac{D}{f \lambda}$$.

\begin{align*} t & = \frac{D}{f \lambda} \\ t (\lambda) & = \frac{D}{f} \\ \lambda & = \frac{D}{tf} \end{align*}

Note restrictions: $$t \ne 0$$, $$f \ne 0$$

Solve for $$m$$: $$E = mgh + \frac{1}{2}mv^{2}$$.

\begin{align*} E & = mgh + \frac{1}{2}mv^{2} \\ E & = m\left(gh + \frac{1}{2}v^{2}\right) \\ \frac{E}{gh + \frac{1}{2}v^{2}} & = m \end{align*}

Note restriction: $$gh + \frac{1}{2}v^{2} \ne 0$$

Solve for $$x$$: $$x^{2} + x(a + b) + ab = 0$$.

\begin{align*} x^{2} + x(a + b) + ab & = 0 \\ x^{2} + xa + xb + ab & = 0 \\ (x + a)(x + b) & = 0 \\ x = -a & \text{ or } x = -b \end{align*}

Solve for $$b$$: $$c = \sqrt{a^{2} + b^{2}}$$.

\begin{align*} c & = \sqrt{a^{2} + b^{2}} \\ c^{2} & = a^{2} + b^{2} \\ c^{2} - a^{2} & = b^{2} \\ b & = \pm \sqrt{c^{2} - a^{2}} \end{align*}

Make $$U$$ the subject of the formula: $$\dfrac{1}{V} = \dfrac{1}{U} + \dfrac{1}{W}$$.

\begin{align*} \frac{1}{V} & = \frac{1}{U} + \frac{1}{W} \\ \frac{UW}{UVW} & = \frac{VW}{UVW} + \frac{UV}{UVW} \\ UW & = VW + UV \\ UW - UV & = VW \\ U & = \frac{VW}{W - V} \end{align*}

Note restriction: $$W \ne V$$

Solve for $$r$$: $$A = \pi R^{2} - \pi r^{2}$$.

\begin{align*} A & = \pi R^{2} - \pi r^{2} \\ A - \pi R^{2} & = \pi r^{2} \\ \frac{A - \pi R^{2}}{\pi} & = r^{2} \\ r & = \pm \sqrt{\frac{A - \pi R^{2}}{\pi}} \end{align*}

$$F = \frac{9}{5}C + 32°$$ is the formula for converting temperature in degrees Celsius to degrees Fahrenheit. Derive a formula for converting degrees Fahrenheit to degrees Celcius.

\begin{align*} F & = \frac{9}{5}C + 32° \\ F - 32° & = \frac{9}{5}C \\ 5(F - 32°) & = 9C \\ \frac{5(F - 32°)}{9} & = C \end{align*}

To convert degrees Fahrenheit to degrees Celsius we use: $$C = \frac{5}{9}\left(F - 32°\right)$$

$$V = \frac{4}{3}\pi r^{3}$$ is the formula for determining the volume of a soccer ball. Express the radius in terms of the volume.

\begin{align*} V & = \frac{4}{3}\pi r^{3} \\ \frac{3}{4} V & = \pi r^{3} \\ \frac{\frac{3}{4}V}{\pi} & = r^{3} \\ \sqrt[3]{\frac{\frac{3}{4}V}{\pi}} & = r \end{align*}

Therefore expressing the radius in terms of the volume gives: $$r = \sqrt[3]{\frac{\frac{3}{4}V}{\pi}}$$

Solve for $$x$$ in: $$x^2 - ax - 3x = 4 + a$$

\begin{align*} x^2 - ax - 3x &= 4 + a \\ x^2 - ax - 3x + a + 4 &= 0\\ x^2 - x(a + 3) + (a + 4) &= 0 \\ (x + 1)(x - (a+4)) &= 0 \\ \therefore x = a + 4 &\text{ or } x = -1 \end{align*}

Solve for $$x$$ in: $$ax^2 - 4a + bx^{2} - 4b = 0$$

\begin{align*} ax^2 - 4a + bx^{2} - 4b &= 0 \\ a(x^2 - 4) + b(x^2 - 4) &= 0 \\ (a + b)(x^2 - 4) &= 0 \\ (a + b)(x - 2)(x + 2) &= 0 \\ \therefore x = 2 &\text{ or } x = -2 \end{align*}

Solve for $$x$$ in $$v^2 = u^2 + 2ax$$ if $$v = 2$$, $$u = \text{0,3}$$, $$a = \text{0,5}$$

\begin{align*} v^2 &= u^2 + 2ax \\ 2ax &= v^2 - u^2 \\ x &= \frac{v^2 - u^2}{2a} \\ x &= \frac{2^2 - \text{0,3}^2}{2(\text{0,5})} \\ x &= \text{3,91} \end{align*}

Solve for $$u$$ in $$f' = f \dfrac{v}{v-u}$$ if $$v = 13$$, $$f = 40$$, $$f' = 50$$

\begin{align*} f' &= f \frac{v}{v-u} \\ f'(v - u) & = fv \\ v - u &= \frac{fv}{f'} \\ -u &= \frac{fv}{f'} - v \\ u &= v - \frac{fv}{f'} \\ u &= 13 - \frac{40(13)}{50} \\ u &= \text{2,6} \end{align*}

Solve for $$h$$ in $$I = \dfrac{bh^2}{12}$$ if $$b = 18$$, $$I = 384$$

\begin{align*} I &= \frac{bh^2}{12} \\ h^2 & = \frac{12I}{b} \\ h & = \pm \sqrt{\frac{12I}{b}} \\ h & = \pm \sqrt{\frac{12(384)}{18}} \\ h & = \pm 16 \end{align*}

Solve for $$r_2$$ in $$\dfrac{1}{R} = \dfrac{1}{r_1} + \dfrac{1}{r_2}$$ if $$R = \frac{3}{2}$$, $$r_1 = 2$$

\begin{align*} \frac{1}{R} & = \frac{1}{r_1} + \frac{1}{r_2} \\ \frac{1}{r_{2}} & = \frac{1}{R} - \frac{1}{r_{1}} \\ \frac{1}{r_{2}} & = \frac{r_{1} - R}{Rr_{1}} \\ Rr_{1} & = r_{2}(r_{1} - R) \\ \frac{Rr_{1}}{r_{1} - R} & = r_{2} \\ r_{2} & = \frac{\frac{3}{2} \times 2}{2 - \frac{3}{2}}\\ & = \frac{3}{\frac{1}{2}} \\ & = 6 \end{align*}