Fractions with letters in them

Cancelling is legal for factors and never for terms, and everything else follows from ordinary fraction arithmetic.

A rational expression is one polynomial divided by another, such as $\dfrac{x^{2}-9}{x-3}$. Somebody claims that this one is the same as $x+3$. Test the claim before believing it.

The claim is right everywhere except at $x = 3$, where the fraction is $\frac{0}{0}$ and $x+3$ is a perfectly ordinary 6. The algebra shows why: $\frac{x^{2}-9}{x-3} = \frac{(x-3)(x+3)}{x-3}$, and the matching factor cancels, leaving $x+3$ with the note that $x$ cannot be 3.

The value that has to be excluded is not a technicality. Producing a fraction that says $\frac{0}{0}$, cancelling, and reading off the value the tidy version gives is the whole mechanism of a derivative.

Cancelling: factors yes, terms no

A factor is something being multiplied. A term is something being added. Cancelling means dividing top and bottom by the same quantity, and that only works on factors, because only a factor divides the whole numerator.

Multiplying and dividing

Multiplying rational expressions is the same as multiplying numeric fractions: tops times tops, bottoms times bottoms. The work is all in factoring first so that the cancelling can happen before anything is multiplied out.

That example collapsed to 1 because every factor found a partner, which is tidier than the usual case. Here is one where something survives, every step.

Simplify $\dfrac{x^{2}-1}{x^{2}+4x+3}\cdot\dfrac{x+3}{x}$.

1. Factor every top and every bottom before touching anything else. $x^{2}-1$ is a difference of squares, so it is $(x-1)(x+1)$.
2. $x^{2}+4x+3$ needs two numbers multiplying to 3 and adding to 4, which are 1 and 3, so it is $(x+1)(x+3)$.
3. The other two pieces, $x+3$ and $x$, are already as factored as they get.
4. Write the whole thing as one fraction, tops multiplied and bottoms multiplied: $\dfrac{(x-1)(x+1)(x+3)}{(x+1)(x+3)\,x}$.
5. Cancel the factors that appear top and bottom. $(x+1)$ appears in both, so it goes. $(x+3)$ appears in both, so it goes too.
6. What is left is $\dfrac{x-1}{x}$.
7. Check at $x = 2$: the original is $\frac{4-1}{4+8+3}\cdot\frac{5}{2} = \frac{3}{15}\cdot\frac{5}{2} = \frac{1}{2}$, and the answer is $\frac{2-1}{2} = \frac{1}{2}$.
8. Note what is excluded. The cancelled factors were zero at $x = -1$ and $x = -3$, and the $x$ on the bottom rules out $x = 0$, so the answer holds for every $x$ except those three.

Dividing is multiplying by the reciprocal, that is, by the fraction turned upside down. $\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b}\cdot\dfrac{d}{c}$. Flip the second one only, never the first.

Simplify $\dfrac{x^{2}-4}{x^{2}+6x+9} \div \dfrac{x-2}{x+3}$.

1. Turn the division into a multiplication by flipping the second fraction: $\dfrac{x^{2}-4}{x^{2}+6x+9}\cdot\dfrac{x+3}{x-2}$.
2. Factor what is left. $x^{2}-4 = (x-2)(x+2)$, a difference of squares.
3. $x^{2}+6x+9$ needs two numbers multiplying to 9 and adding to 6, which are 3 and 3, so it is $(x+3)(x+3)$.
4. As one fraction: $\dfrac{(x-2)(x+2)(x+3)}{(x+3)(x+3)(x-2)}$.
5. Cancel $(x-2)$ from top and bottom, and cancel one $(x+3)$ against one of the two on the bottom. One $(x+3)$ stays behind on the bottom, because there were two down there and only one up top.
6. The answer is $\dfrac{x+2}{x+3}$.
7. Check at $x = 1$: the original is $\frac{1-4}{1+6+9} \div \frac{1-2}{1+3} = \frac{-3}{16} \div \frac{-1}{4} = \frac{-3}{16}\cdot\frac{4}{-1} = \frac{3}{4}$, and the answer is $\frac{1+2}{1+3} = \frac{3}{4}$.

Adding and subtracting: the common denominator

Fractions can only be added when their denominators match, because $\frac{a}{d}+\frac{b}{d} = \frac{a+b}{d}$ needs the same $d$ in both. When they do not match, each fraction is multiplied by a disguised 1 until they do.

The lowest common denominator, or LCD, is built from the factorisations: take every distinct factor that appears anywhere, each raised to the highest power it reaches in any one denominator.

Work $\dfrac{1}{x^{2}-1} + \dfrac{2}{x+1}$ all the way through.

1. Factor every denominator. $x^{2}-1$ is a difference of squares, so it is $(x-1)(x+1)$. The second denominator is already factored: $(x+1)$.
2. Collect the distinct factors: $(x-1)$ and $(x+1)$. Each appears to the first power at most, so the LCD is $(x-1)(x+1)$.
3. The first fraction already has that denominator, so it is left alone: $\dfrac{1}{(x-1)(x+1)}$.
4. The second is missing the factor $(x-1)$, so multiply it by $\dfrac{x-1}{x-1}$, which is 1: it becomes $\dfrac{2(x-1)}{(x-1)(x+1)}$.
5. Now the denominators match, so add the numerators: $1 + 2(x-1) = 1 + 2x - 2 = 2x - 1$.
6. The sum is $\dfrac{2x-1}{(x-1)(x+1)}$, which can also be written $\dfrac{2x-1}{x^{2}-1}$.
7. Can it simplify further? The top would have to share a factor with the bottom, and $2x-1$ is zero at $x = \frac{1}{2}$ while the bottom is zero at $x = 1$ and at $x = -1$, so no.
8. Check at $x = 3$: the original is $\frac{1}{8}+\frac{2}{4} = 0.625$, and the answer is $\frac{5}{8} = 0.625$.