Rational Expressions

A restriction is a value of the variable that must be excluded because it makes a denominator zero somewhere in the original expression — even if that factor later cancels out. Division by zero is undefined for a simple reason: $\frac{6}{2}=3$ because $2\times 3=6$, but $\frac{6}{0}$ would need a number that gives $6$ when multiplied by $0$, and none exists. To find restrictions, factor each denominator and set every factor equal to zero. For $\frac{x+1}{x^2-5x}$, the denominator is $x(x-5)$, so $x\ne 0$ and $x\ne 5$. You can write the allowed values in set notation: $\{x\in \mathbb{R}\mid x\ne 0,5\}$. Some denominators are never zero: $x^2+4\ge 4$ for every real $x$, so $\frac{1}{x^2+4}$ has no restrictions. Always state restrictions using the factored, un-cancelled denominator(s).

Factor the numerator and denominator completely, then cancel any common factors. Cancelling works because $\frac{a\cdot k}{b\cdot k}=\frac{a}{b}\cdot \frac{k}{k}=\frac{a}{b}\cdot 1$, and that is only true when $k$ multiplies the whole numerator and the whole denominator. So you may cancel factors, never terms: in $\frac{x+3}{x}$ the $x$ on top is added to $3$, not multiplied, so nothing cancels (test $x=1$: $\frac{4}{1}=4$, not $3$). Watch for opposite factors: $5-x=-(x-5)$, so $\frac{x-5}{5-x}=-1$ for $x\ne 5$. Take out common factors first, too: $\frac{3x+6}{x^2+2x}=\frac{3(x+2)}{x(x+2)}=\frac{3}{x}\text{,}$ with $x\ne 0,-2$.

MCR3U - Simplifying Rational Expressions with Restrictions Part 3 - Grade 11 Functions

To multiply, factor everything first, then cancel across the whole product before multiplying out. Cancelling first keeps the numbers small: a factor in either numerator may cancel with a factor in either denominator, since the product is really one big fraction. To divide, flip the second fraction (multiply by its reciprocal) and proceed exactly as with multiplication — remembering that the divisor's numerator also creates a restriction (it can't be the value that makes the divisor equal to zero). In $\frac{x+3}{x-1}÷\frac{x^2+6x+9}{x^2-1}$, the restrictions are $x\ne 1$ and $x\ne -1$ from the denominators, and also $x\ne -3$, because at $x=-3$ the divisor equals $0$ and flipping it would put $0$ in a denominator. Collect restrictions from the original denominators and from the numerator of every divisor before flipping.

Just like numeric fractions, you need a common denominator before combining. Factor every denominator first to find the lowest common denominator (LCD): use each different factor, as many times as it appears in any one denominator. Rewrite each fraction over the LCD by multiplying its top and bottom by the missing factors, then add or subtract the numerators only. For $\frac{2}{x^2-9}+\frac{1}{x+3}$, the denominators are $(x-3)(x+3)$ and $(x+3)$, so the LCD is $(x-3)(x+3)$, not the product of both denominators. The second fraction becomes $\frac{x-3}{(x-3)(x+3)}$, and the sum is $\frac{2+x-3}{(x-3)(x+3)}=\frac{x-1}{(x-3)(x+3)}\text{,}$ with $x\ne \pm 3$. When subtracting, put brackets around the second numerator so the negative reaches every term: $\frac{5}{x+1}-\frac{2}{x-3}=\frac{5(x-3)-2(x+1)}{(x+1)(x-3)}=\frac{3x-17}{(x+1)(x-3)}\text{.}$

1. Factor every numerator and denominator completely (common factor first, then trinomials and differences of squares).
2. State the restrictions from every original denominator, and from the numerator of any divisor.
3. Multiply, divide (flip the divisor) or rewrite over the LCD, depending on the operation.
4. Cancel common factors only, never terms.
5. Write the answer with its restrictions, and test one value of x in both forms as a check.

Adding & Subtracting Rational Expressions Easily