Overview

In all levels of competition math, you often encounter expressions that look messy at first glance, but actually have an elegant structure underneath. Substitution is a common and powerful strategy to find these hidden structures, turning messy expressions into familiar ones. Look for repeated terms/patterns like $x+1/x$, $x^2$, or symmetric sums.

Definition and Usage

Substitution is a technique that replaces a (possibly hidden) repeated expression with a single variable in order to simplify a problem. Consider using substitution when, for instance:

• The same expression appears multiple times
• Only even powers of a term are present (e.g. $y=x^2$ for $x^4+x^2+1$)
• The expression is symmetric
• The answer choices look simple, but the expression looks unnecessarily complicated

Common Types of Substitution

1. Symmetric Substitution (Two Variables)

An expression is symmetric in two variables when switching $x$ and $y$ does not change its value. It is often useful to make the substitution

A useful identity is

Also,

Some substitutions only become evident once the problem is factored or rewritten in a clever way. In fact, it is always possible to write any symmetric expression in terms of $s$ and $p.$

2. Even-Power Substitution

Consider letting $t=x^2$ when only even powers of a term appear.

Note: This is not limited to powers of 2. If you see the equation $x^6+2x^3+1=0,$ for example, you can turn it into a quadratic by letting $t=x^3.$

3. Reciprocal Substitution

Consider making the substitution $t=x+1/x$ when the expression is unchanged when $x$ is replaced with $1/x,$ or when you see terms involving $x + \frac{1}{x}, x^2+\frac{1}{x^2},$ etc.

Each term $x^n+\frac{1}{x^n}$ can be expressed fully in terms of $t.$ Notice that $t^2 = \left(x+\frac{1}{x}\right)\left(x+\frac{1}{x}\right)=x^2+2+\frac{1}{x^2},$ so

Similarly, $\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)=x^3+x+\frac{1}{x}+\frac{1}{x^3},$ so

For higher powers, this process continues. Note: This is a special case of symmetric substitution!

4. Repeated Structure (General)

Look for the same expression appearing multiple times, then replace it with a single variable to reduce complexity.

Example: If $2^{2x}+2^{x+1}+1=25,$ find $x.$

Solution: Let $t=2^x.$ Then $t^2+2t+1=25,$ so $(t+1)^2=25$ and $t=-6,4.$ However, no power of $2$ is negative, so $2^x=4$ and so $x=2.$

Common Mistakes

• Leftover Terms: Ensure your substitution accounts for all of the old variable's occurrences. For instance, the substitution $t=x^2$ is not useful for the equation $x^4+x^2+x+1=0,$ due to the $x$ term.
• Forgetting to substitute back for the original variable: It may help to circle or write a note on your scratch paper to remember this.
• Extraneous solutions: After solving in the new variable, substitute back and check that all solutions fit the original domain. Some substitutions may introduce extraneous solutions.

  • Example: Solve for $x$ if $\sqrt{x}^2+2\sqrt{x}+1=25.$
  • Incorrect Solution: Let $t=\sqrt{x}.$ Then we have $t^2+2t+1=25,$ so $(t+1)^2=5$ and $t = -6, 4.$ Then $\sqrt{x}=-6,4,$ so $x=36,16.$
  • In making the substitution $t = \sqrt{x},$ we implicitly assumed that $t \ge 0,$ the range of the square root function. Thus $t=-6$ is not a solution (we can also check this by inputting the resulting solution for $x$ back into the original equation), but $t=4 \implies x=16$ is.
  • Ensure you exclude $x \ne 0$ as a solution if the equation involves $1/x,$ as well.
• The substitution doesn't simplify the equation: In general, a good substitution should immediately make the problem simpler.

Worked Examples

Example 1. Reciprocal Form

If $x^2+\frac{1}{x^2}=7$ and $x>0,$ find $x^3+\frac{1}{x^3}.$

Solution: Let $t=x+\frac{1}{x}.$ Notice that

Therefore, $t^2=9,$ so $t=3, -3.$ We discard $-3$ as $x>0.$ Next we apply the identity

obtaining a final answer of $3^3-3\cdot 3=18.$

Example 2. Harder Reciprocal Form

If $x+\frac{1}{x}=3,$ find $x^5+\frac{1}{x^5}.$

Solution: Let $t=x+\frac{1}{x}.$ Then $t=3,$ so we know that

Also,

so

Next,

Finally, we have

so

Example 3: Even Powers

If $x^4-5x^2+4=0$, solve for $x$.

Solution: Let $u=x^2$. Then $u^2-5u+4=0,$ so we factor as $(u-1)(u-4)=0,$ yielding $u=1$ or $4$. Hence $x=\pm 1,\pm 2$.

Example 4: Symmetric System

If $x^2+y^2=8$ and $\frac{1}{x}+\frac{1}{y}=1,$ find the positive value of $x+y.$

Solution: These expressions are both symmetric, so we let $s=x+y$ and $p=xy.$ Then, notice that

Thus $s^2-2p=8,$ and $s=p.$ Inputting, we have

Checking, $s=4$ and $p=4$ yields $x,y=2,$ which is indeed a solution.

Practice Problems