1.26: Solving Fractional Equations

A fractional equation is an equation involving fractions which has the unknown in the denominator of one or more of its terms.

Example 24.1

The following are examples of fractional equations:

a) \(\frac{3}{x}=\frac{9}{20}\)

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

c) \(\frac{3}{x-3}=\frac{4}{x-5}\)

d) \(\frac{3}{4}-\frac{1}{8 x}=0\)

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

The Cross-Product property can be used to solve fractional equations.

Cross-Product Property

If \(\frac{A}{B}=\frac{C}{D}\) then \(A \cdot D=B \cdot C\).

Using this property we can transform fractional equations into non-fractional ones. We must take care when applying this property and use it only when there is a single fraction on each side of the equation. So, fractional equations can be divided into two categories.

I. Single Fractions on Each Side of the Equation

Equations a), b) and c) in Example 24.1 fall into this category. We solve these equations here.

a) Solve \(\frac{3}{x}=\frac{9}{20}\)

\[\begin{array}{ll} \text{Cross-Product} & 3 \cdot 20=9 \cdot x \\ \text{Linear Equation} & 60=9 x \\ \text{Divide by 9 both sides} & \frac{60}{9}=x \end{array}\nonumber\]

The solution is \(x=\frac{60}{9}=\frac{20}{3}\).

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

\[\begin{array}{ll} \text{Cross-Product} & 5 \cdot(x-2)=3 \cdot(x+2) \\ \text{Remove parentheses} & 5 x-10=3 x+6 \\ \text{Linear Equation: isolate the variable} & 5 x-3 x=10+6 \\ & 2 x=16 \\ \text{Divide by 2 both sides} & \frac{2 x}{2}=\frac{16}{2}\end{array}\nonumber\]

the solution is \(x=8\).

c) \(\frac{3}{x-3}=\frac{4}{x-5}\)

\[\begin{array}{ll} \text{Cross-Product} & 3 \cdot(x-5)=4 \cdot(x-3) \\ \text{Remove parentheses} & 3 x-15=4 x-12 \\ \text{Linear Equation: isolate the variable} & 3 x-4 x=15-12 \\ & -x=3 \\ \text{Divide by 2 both sides} & \frac{-x}{-1}=\frac{3}{-1}\end{array}\nonumber\]

The solution is \(x=-3\)

Note: If you have a fractional equation and one of the terms is not a fraction, you can always account for that by putting 1 in the denominator. For example:

Solve

\[\frac{3}{x}=15\nonumber\]

We re-write the equation so that all terms are fractions.

\[\frac{3}{x}=\frac{15}{1}\nonumber\]

\[\begin{array}{ll} \text{Cross-Product} & 3 \cdot 1=15 \cdot x \\ \text{Linear Equation: isolate the variable} & 3=15 x \\ \text{Divide by 15 both sides} & \frac{3}{15}=\frac{15 x}{15} \end{array}\nonumber\]

The solution is \(x=\frac{3}{15}=\frac{3 \cdot 1}{3 \cdot 5}=\frac{1}{5}\).

II. Multiple Fractions on Either Side of the Equation

Equations d) and e) in Example 24.1 fall into this category. We solve these equations here.

We use the technique for combining rational expressions we learned in Chapter 23 to reduce our problem to a problem with a single fraction on each side of the equation.

d) Solve \(\frac{3}{4}-\frac{1}{8 x}=0\)

First we realize that there are two fractions on the LHS of the equation and thus we cannot use the Cross-Product property immediately. To combine the LHS into a single fraction we do the following:

\[\begin{array}{ll} \text{Find the LCM of the denominators} & 8 x \\ \text{Rewrite each fraction using the LCM} & \frac{3 \cdot 2 x}{8 x}-\frac{1}{8 x}=0 \\ \text{Combine into one fraction} & \frac{6 x-1}{8 x}=0 \\ \text{Re-write the equation so that all terms are fractions} & \frac{6 x-1}{8 x}=\frac{0}{1} \\ \text{Cross-Product} & (6 x-1) \cdot 1=8 x \cdot 0 \\ \text{Remove parentheses} & 6 x-1=0 \\\text{Linear Equation: isolate the variable} & 6 x=1 \\ \text{Divide by 6 both sides} & \frac{6 x}{6}=\frac{1}{6} \end{array}\nonumber\]

The solution is \(x=\frac{1}{6}\).

e) Solve \(\frac{x}{6}+\frac{2}{3 x}=\frac{2}{3}\)

\[\begin{array}{ll} \text{Find the LCM of the denominators of LHS} & 6x \\ \text{Rewrite each fraction on LHS using their LCM} & \frac{x \cdot x}{6 x}+\frac{2 \cdot 2}{6 x}=\frac{2}{3} \\ \frac{x^{2}+4}{6 x}=\frac{2}{3} \text{Combine into one fraction} & \left(x^{2}+4\right) \cdot 3=6 x \cdot 2 \\ \text{Cross-Product} & 3 x^{2}+12=12 x \\ \text{Remove parentheses} & 3 x^{2}-12 x+12=0 \\ \text{Quadratic Equation: Standard form} & 3 x^{2}-12 x+12=0 \\\text{Quadratic Equation: Factor} & 3 \cdot x^{2}-3 \cdot 4 x+3 \cdot 4=0 \\ & 3\left(x^{2}-4 x+4\right)=0 \\ & 3(x-2)(x-2)=0 \\ \text{Divide by 3 both sides} & \frac{3(x-2)(x-2)}{3}=\frac{0}{3} \\ & (x-2)(x-2)=0 \\ \text{Quadratic Equation: Zero-Product Property} & (x-2)=0 \text { or }(x-2)=0 \end{array}\nonumber\]

Since both factors are the same, then \(x-2=0\) gives \(x=2\). The solution is \(x=2\)

Note: There is another method to solve equations that have multiple fractions on either side. It uses the LCM of all denominators in the equation. We demonstrate it here to solve the following equation: \(\frac{3}{2}-\frac{9}{2 x}=\frac{3}{5}\)

\[\begin{array} \text{Find the LCM of all denominators in the equation} & 10x \\ \text{Multiply every fraction (both LHS and RHS) by the LCM} & 10 x \cdot \frac{3}{2}-10 x \cdot \frac{9}{2 x}=10 x \cdot \frac{3}{5} \\ & \frac{10 x \cdot 3}{2}-\frac{10 x \cdot 9}{2 x}=\frac{10 x \cdot 3}{5} \\ \text{Simplify every fraction} & \frac{5 x \cdot 3}{1}-\frac{5 \cdot 9}{1}=\frac{2 x \cdot 3}{1} \\ \text{See how all denominatiors are now 1, thus can be disregarded} & 5 x \cdot 3-5 \cdot 9=2 x \cdot 3 \\ \text{Solve like you would any other equation} & 15 x-45=6 x \\ \text{Linear equation: islolate the variable} & 15 x-6 x=45 \\ & 9 x=45 \\ & x=\frac{45}{9} \\ & x=5 \end{array} \nonumber\]

The solution is \(x=5\)

Exit Problem

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