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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 $2^{2 x} - 2^{x + 1} + 1 = 0$

This one looks a bit weird, but with a little help from the rules of calculating with powers it gets easier.

\begin{array}{l} 2^{2 x} - 2^{x + 1} + 1 & = 0 \end{array}

Transform the equation so you can use substitution:

\begin{array}{l} {(2^{x})}^{2} - 2 \cdot 2^{x} + 1 & = 0 \end{array}

You can see that $2^{x} = u$ gives us an ordinary square equation:

\begin{array}{l} u^{2} - 2 \cdot u + 1 & = 0 \\ u^{2} - 2 u + 1 & = 0 \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}{l} u^{2} - 2 u + 1 & = 0 \\ (u - 1) (u - 1) & = 0 \\ {(u - 1)}^{2} & = 0 \\ u - 1 & = 0 \\ u & = 1 \end{array}

Put the values for $u$ in to $2^{x} = u$ and solve for $x$ : $\begin{array}{l} 2^{x} & = 1 \\ \log 2^{x} & = \log 1 \\ x \cdot \log 2 & = 0 & (\log 1 = 0) \\ x & = \frac{0}{\log 2} = 0 \end{array}$

Example 2

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

To solve an equation like this, it helps to recognize that $x^{4} = {(x^{2})}^{2}$ . You can substitute and solve it as a normal quadratic equation:

\begin{array}{l} x^{4} - 5 x^{2} + 6 & = 0 \\ {(x^{2})}^{2} - 5 x^{2} + 6 & = 0 \end{array}

Let $x^{2} = u$ and substitute:

\begin{array}{l} u^{2} - 5 u + 6 & = 0 \end{array}

You can solve this equation either with the quadratic formula, or with inspection:

\begin{array}{l} u & = \frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot 1 \cdot 6}}{2 \cdot 1} \\ = \frac{5 \pm \sqrt{25 - 24}}{2} \\ = \frac{5 \pm \sqrt{1}}{2} \\ = \frac{5 \pm 1}{2} \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}{l} u_{1} & = \frac{5 + 1}{2} & u_{2} & = \frac{5 - 1}{2} \\ = \frac{6}{2} & = \frac{4}{2} \\ = 3 & = 2 \end{array}

At the end you put these values back in the substitution $2^{x} = u$ to find the values for $x$ : $\begin{array}{l} x^{2} & = 3 & x^{2} & = 2 \\ x & = \pm \sqrt{3} & x & = \pm \sqrt{2} \end{array}$

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

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