4.5 Solving Rational Equations

Learning Outcomes

  • Solve rational equations.
  • Solve rational equations resulting from setting functions equal to each other.
  • Solve rational equations with extraneous solutions.

Equations that contain rational expressions with a variable in at least one denominator are called rational equations. For example, [latex]\dfrac{4}{2x+1}=\dfrac{2x}{3}[/latex] is a rational equation. Rational equations can be useful for representing real-life situations and for finding answers to real problems. In particular, they are quite good for describing a variety of proportional relationships. Equations involving rational expressions with only numeric denominators, like $$\dfrac{x-2}{3}=\dfrac{1}{2}$$, which is equivalent to $$\dfrac{1}{3}x-\dfrac{2}{3}=\dfrac{1}{2}$$, a linear equation, aren’t considered rational equations, although the methods of solving these are very similar to the ones used for solving rational equations.

One of the most straightforward ways to solve a rational equation is to transform it into a polynomial (possibly linear) equation by multiplying both sides of the equation by the LCD of all denominators and simplifying. This step in general is not completely mathematically legitimate since only multiplication of both sides of an equation by a nonzero expression results in an equivalent equation. Since the LCD of a rational equation will always have a variable, it’s possible that for certain values of that variable, the LCD might be equal to zero. Fortunately, as long as we identify the restricted values of the variable, we will be okay. Moreover, we will never lose any valid solutions. Once we identify the restricted values of the variable, multiply both sides of the equation by the LCD of all denominators and simplify, we solve the resulting equation accordingly. No restricted value can be a solution to the equation. This is the reason we have developed the habit of creating the list of restricted values this entire chapter.

ExAMPLE

Solve.

[latex]\dfrac{4}{2x+1}=\dfrac{2x}{3}[/latex]

Since the equation resulting from multiplying both sides of the original equation by the LCD might result in an inequivalent equation, we will refer to the solutions of the resulting polynomial equation as proposed solutions. They must be checked against the list of restricted values of the variable in the original equation or verified by substituting into the original equation. Each proposed solution must be checked and everything can happen. For example, even if there are say three proposed solutions, none, one, two, or even all three can be restricted values. So, it might for example turn out that none of the proposed solutions works since the original equation didn’t have any solutions. If a proposed solution is not a valid solution of the original equation, we will refer to it as an extraneous solution.

Another thing worth mentioning for solving rational equations having just a single rational expression on each side is what is often referred to as “cross multiplication” and mathematically as the Means-Extremes Property:

MEANS-EXTREMES PROPERTY

If $$b\ne 0$$ and $$d\ne 0$$, the equation $$\dfrac{a}{b} = \dfrac{c}{d}$$ is equivalent to the equation $$a\cdot d = b\cdot c$$.

The means and extremes terminology comes from the theory of ratios and proportions.

[latex]\dfrac{\;\fcolorbox{Green}{SpringGreen}{$a$}\;}{\;\fcolorbox{Blue}{SkyBlue}{$b$}\;} = \dfrac{\;\fcolorbox{Blue}{SkyBlue}{$c$}\;}{\;\fcolorbox{Green}{SpringGreen}{$d$}\;} \quad \textsf{equaivalent to}\quad \underbrace{\fcolorbox{Green}{SpringGreen}{$a$} \cdot \fcolorbox{Green}{SpringGreen}{$d$}\ \ }_{{\color{Green}\textsf{extremes}}} = \underbrace{\fcolorbox{Blue}{SkyBlue}{$b$} \cdot \fcolorbox{Blue}{SkyBlue}{$c$}\ \ }_{{\color{Blue}\textsf{means}}}[/latex]

In the example above, the equation $$\dfrac{\;\fcolorbox{Green}{SpringGreen}{$4$}\;}{\;\fcolorbox{Blue}{SkyBlue}{$2x+1$}\;} = \dfrac{\;\fcolorbox{Blue}{SkyBlue}{$2x$}\;}{\;\fcolorbox{Green}{SpringGreen}{$3$}\;}$$ was equivalent to the equation $$\fcolorbox{Green}{SpringGreen}{$4$} \cdot \fcolorbox{Green}{SpringGreen}{$3$} = \fcolorbox{Blue}{SkyBlue}{$(2x+1)$} \cdot \fcolorbox{Blue}{SkyBlue}{$2x$}$$. We will say that the former equation has been cross multiplied to obtain the latter equation.

Caution CAUTION! Caution

DO NOT cross multiply when either side of the equation has more than one term!

The following examples will include more than just two rational expressions.

ExAMPLE

Solve.

[latex]\dfrac{x}{x-2}-1 = \dfrac{1}{x+3}[/latex]

ExAMPLE

Solve.

[latex]\dfrac{x}{x-1}-\dfrac{x+3}{x^2-1} = \dfrac{1}{x+1}[/latex]

Rational Equations and Function Notation

Since rational expressions can be used to define rational functions, if $$f(x)$$ and $$g(x)$$ are rational functions, an equation of the form $$f(x)=g(x)$$ will be a rational equation.

ExAMPLE

Let $$f(x)=\dfrac{16}{x+4}$$ and $$g(x)=\dfrac{x^2}{x+4}$$. Determine all values of $$x$$ for which $$f(x)=g(x)$$.

Other Methods of Solving Rational Equations

In the example above, we had equality of rational expressions with like denominators. Notice that in general, two rational expressions with like denominators are equal only if their numerators are equal. We still have to identify the restricted values of the variable, so we could have acknowledged that as long as $$x\ne -4$$, the equation [latex]\dfrac{16}{x+4} = \dfrac{x^2}{x+4}[/latex] is equivalent to $$16 = x^2$$. This general observation leads to another method of solving rational equations. We can present each side of a rational equation as a single rational expression by combing the rational expressions and rewriting each side in an equivalent form as a rational expression with a common denominator of both sides. Then, after identifying the restricted values of the variable, we equate the numerators of both sides and solve that equation to obtain the proposed solutions. Let’s revisit an earlier example.

ExAMPLE

Solve.

[latex]\dfrac{x}{x-2}-1 = \dfrac{1}{x+3}[/latex]

Another fact is that a fraction is equal to $$0$$ only if its numerator is equal to $$0$$. It generalizes to rational expressions as long as we identify the restricted values of the variable. This fact leads to yet another method of solving rational equations. After identifying the restricted values of the variable, we can isolate $$0$$ on one side of the equation and present the other side as a single rational expression. Then we solve the equation [latex]\mathsf{“The\ Numerator”} = 0[/latex] to obtain the proposed solutions. Let’s revisit another earlier example.

Example

Solve.

[latex]\dfrac{x}{x-1}-\dfrac{x+3}{x^2-1} = \dfrac{1}{x+1}[/latex]

Caution CAUTION! Caution

DO NOT get a generally wrong impression that the method above eliminates extraneous solutions and hence there is no necessity to identify the restricted values of the variable. Consider the equation [latex]\dfrac{2x^2-1}{x^2-x}=1 + \dfrac{1}{x}[/latex]. Since $$x^2-x=x(x-1)$$, the restricted values of $$x$$ are $$0$$ and $$1$$. After isolating $$0$$ on the RHS and rewriting the LHS as a single rational expression, we obtain the equation $$\dfrac{x^2}{x\cdot(x-1)}=0$$. We can divide out $$x$$ on the LHS to obtain the simplified form of the equation, $$\dfrac{x}{x-1}=0$$. Equating the numerator, $$x$$, and $$0$$, we get $$x=0$$. It’s a restricted value of $$x$$ though, so it’s not a valid solution of the original equation. Since it’s the only proposed solution, the original equation doesn’t actually have a solution.

The following videos show additional examples of solving rational equations.

Summary

You can solve rational equations by finding a common denominator. You can multiply both sides of the equation by the least common denominator of all rational expressions so that all terms become polynomials instead of rational expressions. Or, you can rewrite the equation so that all terms have the common denominator, and solve for the variable using just the numerators.

An important step in solving rational equations is to reject any extraneous solutions from the final answer. Extraneous solutions are solutions that do not satisfy the original form of the equation because they produce untrue statements or are excluded values that cause a denominator to equal [latex]0[/latex].