# How to Solve Equations Using Substitution

Sometimes expressions are especially complex, and a clever trick to solve them is to simplify them by using substitution. When you substitute, you exchange an expression with a variable. The variables often used are $u$, $v$, $w$, and $z$.

Theory

### Substitution

Substitution is when you substitute a variable for part of an expression:

1.
You substitute a variable, such as $u$, for a complex factor in the expression, then calculate with the variable in place.
2.
Once your calculation is complete, you substitute the original factor back into the result for the variable $u$, and solve.

Example 1

Solve the equation ${2}^{2x}-{2}^{x+1}+1=0$

This one looks a bit weird, but with a little help from the rules of calculating with powers it gets easier. $\begin{array}{llll}\hfill {2}^{2x}-{2}^{x+1}+1& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Transform the equation so you can use substitution: $\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}{\left({2}^{x}\right)}^{2}-2\cdot {2}^{x}+1& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You can see that ${2}^{x}=u$ gives us an ordinary square equation: $\begin{array}{llll}\hfill {u}^{2}-2\cdot u+1& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {u}^{2}-2u+1& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You can see that this is a quadratic equation which is solved either by using the quadratic formula, or by factorizing. You can therefore find the $u-$values: $\begin{array}{llll}\hfill {u}^{2}-2u+1& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \phantom{\rule{-0.17em}{0ex}}\left(u-1\right)\phantom{\rule{-0.17em}{0ex}}\left(u-1\right)& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \phantom{\rule{-0.17em}{0ex}}{\left(u-1\right)}^{2}& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill u-1& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill u& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Put the values for $u$ in to ${2}^{x}=u$ and solve for $x$: $\begin{array}{llllll}\hfill {2}^{x}& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \mathrm{log}{2}^{x}& =\mathrm{log}1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x\cdot \mathrm{log}2& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{-0.17em}{0ex}}\left(\mathrm{log}1=0\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =\frac{0}{\mathrm{log}2}=0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 2

Solve the equation ${x}^{4}-5{x}^{2}+6=0$

To solve an equation like this, it helps to recognize that ${x}^{4}=\phantom{\rule{-0.17em}{0ex}}{\left({x}^{2}\right)}^{2}$. You can substitute and solve it as a normal quadratic equation: $\begin{array}{llll}\hfill {x}^{4}-5{x}^{2}+6& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \phantom{\rule{-0.17em}{0ex}}{\left({x}^{2}\right)}^{2}-5{x}^{2}+6& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Let ${x}^{2}=u$ and substitute: $\begin{array}{llll}\hfill {u}^{2}-5u+6& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You can solve this equation either with the quadratic formula, or with inspection: $\begin{array}{llll}\hfill u& =\frac{5±\sqrt{\phantom{\rule{-0.17em}{0ex}}{\left(-5\right)}^{2}-4\cdot 1\cdot 6}}{2\cdot 1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{5±\sqrt{25-24}}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{5±\sqrt{1}}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{5±1}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Now you split the two equations, one with a positive root in the numerator and one with a negative root in the numerator: $\begin{array}{llllllll}\hfill {u}_{1}& =\frac{5+1}{2}\phantom{\rule{2em}{0ex}}& \hfill {u}_{2}& =\frac{5-1}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{6}{2}\phantom{\rule{2em}{0ex}}& \hfill & =\frac{4}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =3\phantom{\rule{2em}{0ex}}& \hfill & =2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

At the end you put these values back in the substitution ${2}^{x}=u$ to find the values for $x$: $\begin{array}{llllllll}\hfill {x}^{2}& =3\phantom{\rule{2em}{0ex}}& \hfill {x}^{2}& =2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =±\sqrt{3}\phantom{\rule{2em}{0ex}}& \hfill x& =±\sqrt{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The answer you get is then ${x}_{1}=\sqrt{3}$, ${x}_{2}=-\sqrt{3}$, ${x}_{3}=\sqrt{2}$ and ${x}_{4}=-\sqrt{2}$.