Adding and Subtracting Rational Expressions: Practice

The denominators are already the same, so the two fractions count pieces of the same size. Add the numerators and keep the denominator.

$\frac{3}{x} + \frac{5}{x} = \frac{3+5}{x} = \frac{8}{x}$

The denominator names the size of the piece, so it never gets added.

Same denominator, so add the numerators and leave the denominator alone.

$\frac{x}{x-4} + \frac{3}{x-4} = \frac{x+3}{x-4}$

The numerator $x+3$ shares no factor with $x-4$, so this is already in lowest terms, with $x \neq 4$.

Subtract the numerators, bracketing the one being subtracted so the minus sign reaches both of its terms.

$(4x+3) - (x+3) = 4x + 3 - x - 3 = 3x$

So the difference is the fraction below.

$\frac{4x+3}{x+6} - \frac{x+3}{x+6} = \frac{3x}{x+6}$

Dropping the brackets would give $4x + 3 - x + 3 = 3x + 6$, which is wrong.

To turn the denominator $x-1$ into $(x-1)(x+4)$ you multiply it by $x+4$. To keep the value unchanged you must multiply the numerator by the same factor.

$\frac{3}{x-1} \cdot \frac{x+4}{x+4} = \frac{3(x+4)}{(x-1)(x+4)} = \frac{3x+12}{(x-1)(x+4)}$

Multiplying only the denominator would change the value of the expression.

The denominators share no factor, so the LCD is $x(x+1)$. The first fraction is missing $x+1$ and the second is missing $x$.

$\frac{2(x+1)}{x(x+1)} + \frac{3x}{x(x+1)} = \frac{2x + 2 + 3x}{x(x+1)}$

Collect the like terms in the numerator.

$\frac{5x+2}{x(x+1)}$

The LCD is $x(x-3)$. Build both fractions up to it, then subtract, bracketing the numerator being taken away.

$\frac{5x}{x(x-3)} - \frac{2(x-3)}{x(x-3)} = \frac{5x - (2x - 6)}{x(x-3)}$

Distribute the minus sign across both terms.

$\frac{5x - 2x + 6}{x(x-3)} = \frac{3x+6}{x(x-3)}$

Forgetting to distribute would give $5x - 2x - 6 = 3x - 6$, which is the trap.

A fraction is undefined exactly where its denominator is zero, so set the denominator to zero.

$x - 7 = 0 \;\Rightarrow\; x = 7$

The numerator plays no part in this: $\frac{4}{x-7}$ is perfectly fine at $x = 4$ and at $x = 0$.

Every denominator in the sum contributes its own restriction, so set each one to zero.

$x = 0, \qquad x - 2 = 0 \;\Rightarrow\; x = 2$

Both values are excluded, because at either one the sum has a term that does not exist.

The denominators match, so subtract the numerators, bracketing the one being subtracted.

$(x+1) - (x-1) = x + 1 - x + 1 = 2$

Notice that $-(-1) = +1$. Now finish the division.

$\frac{2}{2} = 1$

Dropping the brackets would give $x + 1 - x - 1 = 0$, which is the most common wrong answer here.

The factors $x-2$ and $x+2$ are different, and neither divides the other, so each appears once in the LCD.

$\text{LCD} = (x-2)(x+2) = x^2 - 4$

When two denominators have no factor in common, their product is the least common denominator.

The denominators $2$ and $x$ share no factor, so the LCD is $2x$. Build both fractions up to it.

$\frac{1}{2} = \frac{x}{2x}, \qquad \frac{1}{x} = \frac{2}{2x}$

Add the numerators over the common denominator.

$\frac{x}{2x} + \frac{2}{2x} = \frac{x+2}{2x}$

The LCD is $x(x-1)$. Build each fraction up to it and bracket the numerator being subtracted.

$\frac{6x}{x(x-1)} - \frac{6(x-1)}{x(x-1)} = \frac{6x - (6x - 6)}{x(x-1)}$

Distribute the minus sign across both terms of $6x - 6$.

$\frac{6x - 6x + 6}{x(x-1)} = \frac{6}{x(x-1)}$

Failing to distribute would give $6x - 6x - 6 = -6$ and the wrong sign.