Module 4: Rational Expressions, Functions, and Equations
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 [latex]\dfrac{x-2}{3}=\dfrac{1}{2}[/latex], which is equivalent to [latex]\dfrac{1}{3}x-\dfrac{2}{3}=\dfrac{1}{2}[/latex], 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]
Show Solution
Let’s first determine the restricted values of the variable in the equation. Only one rational expression in the equation has a variable in the denominator. Since [latex]2x+1=0[/latex] only if [latex]x=-\dfrac{1}{2}[/latex], the only restricted value of [latex]x[/latex] is [latex]-\dfrac{1}{2}[/latex].
We next determine the LCD by finding the LCM of all denominators. Since [latex]2x+1[/latex] is a prime polynomial and [latex]3[/latex] is a prime number, [latex]LCM(2x+1, 3)=3\cdot (2x+1)[/latex]. We will next multiply both sides of the equation by the LCD.
[latex]\require{color} \begin{align} \frac{4}{2x+1}{\color{red}{} \cdot 3\cdot (2x+1)} &= \frac{2x}{3}{\color{red}{} \cdot 3\cdot (2x+1)} && {\color{blue}\textsf{multiply by the LCD}}\\[5pt] \frac{4}{{\color{red}\cancel{\color{black}{(2x+1)}}}\cdot 1} \cdot \frac{3\cdot {\color{red}\cancel{\color{black}{(2x+1)}}}}{1} &= \frac{2x}{{\color{red}\cancel{\color{black}{3}}}\cdot 1} \cdot \frac{{\color{red}\cancel{\color{black}{3}}}\cdot (2x+1)}{1} && {\color{blue}\textsf{divide out common factors}}\\[5pt] 12 &= 2x\cdot (2x+1) && {\color{blue}\textsf{simplify}}\\[5pt] \end{align}[/latex]
The resulting equation is quadratic. We solve it by factoring and using the Zero-Product Property.
[latex]\begin{align} {\color{red}{\dfrac{{\color{black}{{\color{red}\cancel{\color{black}{2}}} \cdot 6}}}{\cancel{2}}}} &= {\color{red}{\dfrac{{\color{black}{{\color{red}\cancel{\color{black}{2}}} \cdot x \cdot (2x+1)}}}{\cancel{2}}}} && {\color{blue}\textsf{divide by $2$ and divide out common factors}}\\[5pt] \underset{\large{\color{red}{-6}}}{6} &= 2x^2+x+\underset{\large{\color{red}{-6}}}{0} && {\color{blue}\textsf{distribute $x$ and subtract $6$}}\\[5pt] 0 &= 2x^2+x-6 && {\color{blue}\textsf{simplify}}\\[5pt] 0 &= (2x-3)(x+2) && {\color{blue}\textsf{factor}}\\[5pt] 2x-\underset{\large{\color{red}{+3}}}{3} &= \underset{\large{\color{red}{+3}}}{0}\ \ \textsf{or}\ \ x+\underset{\large{\color{red}{-2}}}{2} = \underset{\large{\color{red}{-2}}}{0} && \color{blue}{\textsf{Zero-Product Property}}\\[5pt] {\color{red}{\dfrac{{\color{black}{{\color{red}\cancel{\color{black}{2}}} \cdot x}}}{\cancel{2}}}} &= {\color{red}{\dfrac{{\color{black}{3}}}{2}}} \ \ \textsf{or}\ \ x = -2 \\[5pt] x &= \dfrac{3}{2} \ \ \textsf{or}\ \ x = -2 \end{align}[/latex]
We must check the obtained solutions against the list of restricted values of the variable!
We determined earlier that the only restricted value of [latex]x[/latex] is [latex]-\dfrac{1}{2}[/latex]. Hence both, [latex]x = \dfrac{3}{2}[/latex] and [latex]x = -2[/latex] are valid solutions.
It’s always a good idea to verify the solutions.
Substituting [latex]x={\color{Green}{\dfrac{3}{2}}}[/latex] in the left hand side (LHS), [latex]\dfrac{4}{2{\color{Green}{x}}+1}[/latex], results in
Since [latex]-\dfrac{4}{3}=-\dfrac{4}{3}[/latex], LHS [latex]=[/latex] RHS. [latex]\large\checkmark[/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 [latex]b\ne 0[/latex] and [latex]d\ne 0[/latex], the equation [latex]\dfrac{a}{b} = \dfrac{c}{d}[/latex] is equivalent to the equation [latex]a\cdot d = b\cdot c[/latex].
The means and extremes terminology comes from the theory of ratios and proportions.
In the example above, the equation [latex]\dfrac{\;\fcolorbox{Green}{SpringGreen}{$4$}\;}{\;\fcolorbox{Blue}{SkyBlue}{$2x+1$}\;} = \dfrac{\;\fcolorbox{Blue}{SkyBlue}{$2x$}\;}{\;\fcolorbox{Green}{SpringGreen}{$3$}\;}[/latex] was equivalent to the equation [latex]\fcolorbox{Green}{SpringGreen}{$4$} \cdot \fcolorbox{Green}{SpringGreen}{$3$} = \fcolorbox{Blue}{SkyBlue}{$(2x+1)$} \cdot \fcolorbox{Blue}{SkyBlue}{$2x$}[/latex]. We will say that the former equation has been cross multiplied to obtain the latter equation.
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]
Show Solution
Let’s start with the restricted values of the variable in the equation. There are two rational expression in the equation with a variable in the denominator. Since [latex]x-2=0[/latex] only if [latex]x=2[/latex], and [latex]x+3=0[/latex] only if [latex]x=-3[/latex], the restricted values of [latex]x[/latex] are [latex]-3[/latex] and [latex]2[/latex].
The LCD (the LCM of all denominators) is [latex](x-2)(x+3)[/latex] since both [latex]x-2[/latex] and [latex]x+3[/latex] are prime polynomials. We’re ready to multiply both sides of the equation by the LCD.
Recall that [latex]x=-8[/latex] is so far a proposed solution and must be checked against the list of restricted values of the variable. Since the restricted values of [latex]x[/latex] are [latex]-3[/latex] and [latex]2[/latex], [latex]x=-8[/latex] is a valid solution.
We leave it to you to substitute [latex]x=-8[/latex] into the original equation to verify that it results in a true statement.
Let’s start with the restricted values of the variable in the equation. There are three rational expression in the equation with a variable in the denominator. Since [latex]x-1=0[/latex] only if [latex]x=1[/latex], and [latex]x+1=0[/latex] only if [latex]x=-1[/latex], [latex]x=-1[/latex] and [latex]x=1[/latex] are restricted. Since [latex]x^2-1 = (x+1)(x-1)[/latex], [latex]x^2-1 = 0[/latex] only if [latex]x=-1[/latex] or [latex]x=1[/latex] by the Zero-Product Property. Hence all restricted values of [latex]x[/latex] are [latex]-1[/latex] and [latex]1[/latex].
Since [latex]x^2-1 = (x+1)(x-1)[/latex], The LCD is [latex](x-1)(x+1)[/latex]. We’re ready to multiply both sides of the equation by the LCD.
Of course, [latex]x=2[/latex] and [latex]x=-1[/latex] are so far proposed solutions and each must be checked against the list of restricted values of the variable. Since the restricted values of [latex]x[/latex] are [latex]-1[/latex] and [latex]1[/latex], [latex]x=2[/latex] is a valid solution while [latex]x=-1[/latex] is not. In particular, [latex]x=-1[/latex] is an extraneous solution.
We leave it to you to substitute [latex]x=2[/latex] into the original equation to verify that it results in a true statement. Also note what happens when you substitute the extraneous solution.
Rational Equations and Function Notation
Since rational expressions can be used to define rational functions, if [latex]f(x)[/latex] and [latex]g(x)[/latex] are rational functions, an equation of the form [latex]f(x)=g(x)[/latex] will be a rational equation.
ExAMPLE
Let [latex]f(x)=\dfrac{16}{x+4}[/latex] and [latex]g(x)=\dfrac{x^2}{x+4}[/latex]. Determine all values of [latex]x[/latex] for which [latex]f(x)=g(x)[/latex].
Show Solution
Notice that once we substitute the rational expressions defining [latex]f(x)[/latex] and [latex]g(x)[/latex] into the equation [latex]f(x)=g(x)[/latex], we obtain the rational equation
[latex]\dfrac{16}{x+4} = \dfrac{x^2}{x+4}[/latex]
For the restricted values of [latex]x[/latex], since [latex]x+4=0[/latex] only if [latex]x=-4[/latex], the only restricted values of [latex]x[/latex] is [latex]-4[/latex].
Although it’s tempting to cross multiply, since the denominators of the rational expressions on both sides are the same, it will be more convenient to multiply both sides by that common denominator and simplify.
Since [latex]-4[/latex] is a restricted value of [latex]x[/latex], [latex]x = 4[/latex] is the only valid solution and the only value of [latex]x[/latex] for which [latex]f(x)=g(x)[/latex].
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 [latex]x\ne -4[/latex], the equation [latex]\dfrac{16}{x+4} = \dfrac{x^2}{x+4}[/latex] is equivalent to [latex]16 = x^2[/latex]. 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]
Show Solution
The restricted values of [latex]x[/latex] are [latex]-3[/latex] and [latex]2[/latex].
Thus the original equation is so far equivalent to
[latex]\dfrac{2}{x-2} = \dfrac{1}{x+3}[/latex]
The LCD of both sides is [latex](x-2)(x+3)[/latex]. Now, [latex]\dfrac{2{\color{red}{} \cdot (x+3)}}{(x-2){\color{red}{} \cdot (x+3)}} = \dfrac{2x+6}{(x-2)(x+3)}[/latex], and [latex]\dfrac{1{\color{red}{} \cdot (x-2)}}{(x+3){\color{red}{} \cdot (x-2)}} = \dfrac{x-2}{(x-2)(x+3)}[/latex], so the original equation is equivalent to
We solve accordingly to obtain [latex]x=-8[/latex]. Since it’s not a restricted value, it’s a valid solution of the original equation.
Another fact is that a fraction is equal to [latex]0[/latex] only if its numerator is equal to [latex]0[/latex]. 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 [latex]0[/latex] 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.
Refer to the Example further up the page how to identify the restricted values of [latex]x[/latex]: [latex]-1[/latex] and [latex]1[/latex].
To isolate [latex]0[/latex] on the RHS, we start by subtracting [latex]\dfrac{1}{x+1}[/latex] from both sides of the equation and proceed to rewrite the LHS as a simplified single rational expression.
We now solve [latex]x-2=0[/latex] resulting in [latex]x=2[/latex]. It’s not a restricted value of [latex]x[/latex], so its a valid (only) solution of the original equation.
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 [latex]x^2-x=x(x-1)[/latex], the restricted values of [latex]x[/latex] are [latex]0[/latex] and [latex]1[/latex]. After isolating [latex]0[/latex] on the RHS and rewriting the LHS as a single rational expression, we obtain the equation [latex]\dfrac{x^2}{x\cdot(x-1)}=0[/latex]. We can divide out [latex]x[/latex] on the LHS to obtain the simplified form of the equation, [latex]\dfrac{x}{x-1}=0[/latex]. Equating the numerator, [latex]x[/latex], and [latex]0[/latex], we get [latex]x=0[/latex]. It’s a restricted value of [latex]x[/latex] 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].
Candela Citations
CC licensed content, Original
Revision and Adaptation. Provided by: Lumen Learning. License: CC BY: Attribution
Unit 15: Rational Expressions, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education. License: CC BY: Attribution
Unit 15: Rational Expressions, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education. License: CC BY: Attribution
