Sample lesson

Simplifying Rational Expressions

This is how every lesson in Math GRE Studio works: understand the idea, watch it applied, then try one yourself. Scroll down to start.

The question
$$\text{Simplify } \frac{x^2 - 9}{x^2 - 3x}$$
and state all excluded values.

Only factors cancel — never terms.

When you see a fraction like $\dfrac{x^2 - 9}{x^2 - 3x}$, it's tempting to "cancel" the $x^2$ from the top and bottom. But that's illegal — you can only cancel factors, not individual terms inside a sum.

The first step is always the same: factor the numerator and denominator completely, then see what's actually shared.

Factor

Rewrite the numerator and denominator as products:

$$\frac{x^2 - 9}{x^2 - 3x} = \frac{(x-3)(x+3)}{x(x-3)}$$

$x^2 - 9$ is a difference of squares. $x^2 - 3x$ has a common factor of $x$.

Find restrictions

Before you cancel anything, look at the original denominator $x(x-3)$ and find where it equals zero:

$$x \neq 0 \quad \text{and} \quad x \neq 3$$

These restrictions survive simplification. Write them down now so you don't forget.

Cancel and state result

The factor $(x-3)$ appears in both numerator and denominator — cancel it:

$$\frac{\cancel{(x-3)}(x+3)}{x\cancel{(x-3)}} = \frac{x+3}{x}, \quad x \neq 0,\; x \neq 3$$
Your turn
$$\text{Simplify } \frac{x^2 - 16}{x^2 - 4x}$$

Use the same three moves: factor, find restrictions, then cancel.

Show the solution
Factor: $$\frac{x^2 - 16}{x^2 - 4x} = \frac{(x-4)(x+4)}{x(x-4)}$$
Restrictions: $x \neq 0$ and $x \neq 4$ (from the original denominator)
Cancel and state result: $$\frac{\cancel{(x-4)}(x+4)}{x\cancel{(x-4)}} = \frac{x+4}{x}, \quad x \neq 0,\; x \neq 4$$

That's one lesson.

Concept first, worked example second, practice third. Every topic in Math GRE Studio follows this pattern — from algebra through analysis, in the right prerequisite order. This sample took about 3 minutes. Imagine a full study session where every topic builds on the last.

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