\begin{align*} \require{cancel} \frac{x^2 + 3x}{x + 3} & = \frac{x(x + 3)}{x + 3} \\ & = \frac{x\cancel{(x + 3)}}{\cancel{x + 3}} \qquad [x + 3 \text{ is a common factor}] \\ & = {x \over 1} \\ & = x \end{align*}

Common mistake - Canceling before factorising:

\begin{align*} \require{cancel} \frac{x^2 + 3x}{x + 3} & \ne \frac{x^2 + \cancel{3}x}{x + \cancel{3}} \qquad [\text{Cannot cancel } 3 \text{ since it is not a common factor}] \\ & = {x^2 + x \over x} \\ & = {\cancel{x}(x + 1) \over \cancel{x}} \\ & = {x + 1 \over 1} \\ & = x + 1 \end{align*}

The lowest common denominator of $2$, $3x$ and $4x$ is $12x$:

\begin{align*} \frac{1}{2} + \frac{1}{3x} - \frac{1}{4x} & = \frac{1 \times 6x}{2 \times 6x} + \frac{1 \times 4}{3x \times 4} - \frac{1 \times 3}{4x \times 3} \\ & = {6x \over 12x} + {4 \over 12x} - {3 \over 12x} \\ & = {6x + 1 \over 12x} \qquad \qquad [\text{No common factor so cannot simplify anymore!}] \end{align*}

Alternative method using $2 \times 3x \times 4x = 24x^2$ as the common denominator (this method works, but it is usually longer.):

\begin{align*} \frac{1}{2} + \frac{1}{3x} - \frac{1}{4x} & = \frac{1 \times 3x \times 4x}{2 \times 3x \times 4x} + \frac{1 \times 2 \times 4x}{3x \times 2 \times 4x} - \frac{1 \times 2 \times 3x}{4x \times 2 \times 3x} \\ & = \frac{12x^2}{24x^2} + \frac{8x}{24x^2} - \frac{6x}{24x^2} \\ & = \frac{12x^2 + 8x - 6x}{24x^2} \\ & = \frac{12x^2 + 2x}{24x^2} \\ & = \frac{2x(6x + 1)}{24x^2} \\ & = \frac{6x + 1}{12x} \end{align*}

The common denominator is $(x - 1)(2x + 1)$.

\begin{align*} \frac{2}{x - 1} - \frac{3}{2x + 1} & = {2(2x + 1) \over (x - 1)(2x + 1)} - {3(x - 1) \over (x - 1)(2x + 1)} \\ & = {2(2x + 1) - 3(x - 1) \over (x - 1)(2x + 1)} \\ & = {4x + 2 - 3x + 3 \over (x - 1)(2x + 1)} \\ & = {x + 5 \over (x - 1)(2x + 1)} \end{align*}

When multiplying algebraic fractions, multiply the numerators together and multiply the denominators together:

\begin{align*} \frac{3x}{4y} \times \frac{8y}{9x^2} &= \frac{3x \times 8y}{4y \times 9x^2} \\ &= \frac{24xy}{36x^2y} \\ &= \frac{2}{3x} \end{align*}

Change division to multiplication by the reciprocal:

\begin{align*} \require{cancel} \frac{x + 2}{x - 1} \div \frac{x + 2}{2x + 1} & = \frac{x + 2}{x - 1} \times \frac{2x + 1}{x + 2} \\ & = \frac{(x + 2)(2x + 1)}{(x - 1)(x + 2)} \\ & = \frac{\cancel{(x + 2)}(2x+1)}{(x - 1)\cancel{(x + 2)}} \\ & = \frac{2x + 1}{x - 1} \end{align*}

Since there is no equals sign, we are not solving for $x$. We are simplifying the expression.

\begin{align*} \frac{x}{2} \times \frac{2x + 1}{3} &= \frac{x(2x + 1)}{2 \times 3} \\ &= \frac{2x^2 + x}{6} \end{align*}

Solve by cross-multiplication:

\begin{align*} \frac{x}{2} &= \frac{2x + 1}{3} \\ 3x &= 2(2x + 1) \\ 3x &= 4x + 2 \\ 3x - 4x &= 2 \\ -x &= 2 \\ x &= -2 \end{align*}

Solve by making denominator the same on both sides:

\begin{align*} \frac{x}{2} &= \frac{2x + 1}{3} \\ \frac{x \times 3}{2 \times 3} & = \frac{ (2x + 1) \times 2 }{3 \times 2} \\ \frac{3x}{6} & = \frac{4x + 2}{6} \\ \implies 3x & = 4x + 2 \\ 3x - 4x & = 2 \\ -x & = 2 \\ x & = -2 \end{align*}