Exercises: Rational Expression Arithmetic (Add, Subtract, Multiply, Divide)

$x \neq 4$

$x \neq 2$ and $x \neq -2$

$x \neq -5$

No restriction — all real numbers are allowed

$\dfrac{3}{7} + \dfrac{2}{7} = \dfrac{3 + 2}{7} = \dfrac{5}{7}$

$\dfrac{3}{7} + \dfrac{2}{5} = \dfrac{3 + 2}{7 + 5} = \dfrac{5}{12}$

$\dfrac{3}{7} \times \dfrac{2}{7} = \dfrac{3 \times 2}{7 \times 7} = \dfrac{6}{49}$

$\dfrac{3}{7} \div \dfrac{2}{7} = \dfrac{3}{7} \times \dfrac{7}{2} = \dfrac{3}{2}$

LCD $= (x-2)(x+3)$; rewrite as $\dfrac{x+3}{(x-2)(x+3)} + \dfrac{x-2}{(x-2)(x+3)}$

LCD $= (x-2) + (x+3) = 2x+1$; rewrite as $\dfrac{1}{2x+1} + \dfrac{1}{2x+1}$

LCD $= (x-2)(x+3)$; rewrite as $\dfrac{1}{(x-2)(x+3)} + \dfrac{1}{(x-2)(x+3)}$

LCD $= (x-2)^2(x+3)^2$; rewrite as $\dfrac{(x+3)^2}{(x-2)^2(x+3)^2} + \dfrac{(x-2)^2}{(x-2)^2(x+3)^2}$

The $x$ in $x + 3$ and the $x$ in $x + 5$ are terms being added, not factors multiplying the entire numerator and denominator. Only common factors (not terms) can be canceled.

The $x$ can cancel, but only if the student also cancels the constants: $\frac{x+3}{x+5} = \frac{3}{5}$ is correct when $x = 0$.

The student should have subtracted the $x$'s instead: $\frac{x+3}{x+5} = \frac{3}{5}$ is wrong; $\frac{0+3}{0+5} = \frac{3}{5}$ is correct.

There is no error — $\frac{x+3}{x+5}$ is always equal to $\frac{3}{5}$.