Toggle Main Menu Toggle Search



The process known as substitution is the way in which symbols in formulas are replaced by actual numerical values, other symbols or even other formulas.

Worked Examples

Example 1

Find the value of $I=x P + n$ when $x=0.05,\,P=100,\,n=3$.


\[I = x P + n = (0.05)(100) + (3) = 5+3 = 8\]

Example 2

Find the value of $\dfrac{i}{(1+i)^n-1}$ when $n=10$ and $i=0.11$.


\begin{align} \frac{i}{(1+i)^n-1}&=\frac{0.11}{(1+0.11)^{10}-1}\\\\ &=\frac{0.11}{1.11^{10}-1}\\\\ &=0.060 \text{ (to 3 d.p.)} \end{align}

Example 3

This is an example of substituting a formula for a variable in an expression.

The kinetic energy of a body of mass $m$ and velocity $v$ is given by $K=\frac{1}{2}mv^2$.

Suppose that the body starts at time $t=0$ with velocity $u$ and constant acceleration $a$ then at time $t$ the velocity is $v=u+at$.

Express the kinetic energy of the body as a function of time $t$.


In $K=\frac{1}{2}mv^2$ substitute $v=u+at$ to obtain: \[K=\frac{1}{2}m(u+at)^2\] This gives the kinetic energy in terms of the time travelled $t$, the initial velocity $u$ and the acceleration $a$.

Video Example

Prof. Robin Johnson evaluates $(1+x)^5-(1-x)^7$ at $x=0.1$.


This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.

Test Yourself

Test yourself: Numbas test on substitution

External Resources

Whiteboard maths

More Support

You can get one-to-one support from Maths-Aid.