Equations with rational expressions are usually most easily solved with cross-multiplication. Multiplication is only allowed over the equals sign, and even then only when both sides of the equation are in either product or quotient form. Thus, additions and subtractions can only occur in the numerator, in the denominator or inside brackets.

The zeros of a rational equation's denominator are not valid solutions for the equation, as you cannot divide a value by zero. Therefore, before answering, it is necessary to check whether the value obtained is valid.

Example 1

Solve the proportional equation [[$ \displaystyle\frac {x} {3} = \displaystyle\frac {x + 1} {5} $]].​

Rational expressions are eliminated by multiplying by cross.

[[$ \begin{align*} \displaystyle\frac {x} {3} &= \displaystyle\frac {x + 1} {5} \;\;\;\;\; {\color {green} {\text {Cross-multiplication can be performed because both sides are in the quotient form.}}} \\ 5x &= 3(x + 1) \\ 5x &= 3x + 3 \\ 5x - 3x &= 3 \\ 2x &= 3 \;\;\; | :2 \\ x &= \displaystyle\frac {3} {2} \\ \end{align*} $]]​

Answer: [[$ x = \displaystyle\frac {3} {2} $]]​

Example 2

Solve the rationlal equation [[$ \displaystyle\frac {3} {x - 2} = \displaystyle\frac {4} {x - 1} $]].​

The zeros of the denominators are [[$ x = 2 $]]​ and [[$ x = 1 $]]​, so the equation is defined when [[$ x ≠ 2 $]]​ and [[$ x ≠ 1 $]]​.

Multiply the expressions crosswise.

[[$ \begin{align*} \displaystyle\frac {3} {x - 2} &= \displaystyle\frac {4} {x - 1} \;\;\;\;\; {\color {green} {\text {Cross-multiplication can be performed because both sides are in the quotient form.}}} \\ 3(x - 1) &= 4(x - 2) \\ 3x - 3 &= 4x - 8 \\ 5x - 3x &= 3 \\ 3x - 4x &= -8 + 3 \\ -x &= -5 \;\;\; | :(-1) \\ x &= 5 \\ \end{align*} $]]​

This is a valid solution, as neither denominator receives a value of zero.

Answer: [[$ x = 5 $]]​

Example 3

Solve the equation [[$ \displaystyle\frac {3} {x} - \displaystyle\frac {1} {x - 2} = 0 $]].​

The zeros of the denominators are [[$ x = 0 $]] and [[$ x = 2 $]], so the equation is defined when and [[$ x ≠ 0 $]] and [[$ x ≠ 2 $]].

[[$ \begin{align*} \displaystyle\frac {3} {x} - \displaystyle\frac {1} {x - 2} &= 0 \;\;\;\;\; {\color {red} {\text {Cross-multiplication is only allowed over the equal sign.}}} \\ \displaystyle\frac {3} {x} &= \displaystyle\frac {1} {x - 2} \\ x \cdot 1 &= 3(x - 2) \\ x &= 3x - 6 \\ x - 3x &= -6 \\ -2x &= -6 \;\;\; | :(-2) \\ x &= 3 \\ \end{align*} $]]​

This is a valid solution.

Answer: [[$ x = 3 $]]​

Example 4

Solve the equation [[$ \displaystyle\frac {18} {2x} + 2 = 5 $]].​

The zero of the denominator is [[$ x = 0 $]], so the equation is defined when [[$ x ≠ 0 $]].

[[$ \begin{align*} \displaystyle\frac {18} {2x} + 2 &= 5\;\;\;\;\; {\color {red} {\text {Cross-multiplication cannot be done because the left side of the equation is in sum form.}}} \\ \displaystyle\frac {18} {2x} &= 5 - 2 \\ \displaystyle\frac {18} {2x} &= 3 \;\;\;\;\; {\color {green} {\text {Now cross-multiplication is allowed. NUmber 3 is also changed to a quotient format, which makes cross-multiplication better understood.}}} \\\\ \displaystyle\frac {18} {2x} &= \displaystyle\frac {3} {1} \\ 2x \cdot 3 &= 18 \cdot 1 \;\;\; | : 6 \\ x &= \displaystyle\frac {18} {6}\\ x &= 3 \\ \end{align*} $]]​

This is a valid solution.

Answer: [[$ x = 3 $]]​