1.3: Rational Equations
- Page ID
- 38264
A rational expression is the quotient of two polynomials.
A rational equation is an \( \underline{\textrm{equation}} \) containing one or more rational expressions.
Rational expressions typically contain a variable in the denominator. For this reason, care must be taken to ensure that the denominator is not \(0\) by making note of restrictions and checking that no solution makes a denominator zero in the original equation. Solving rational equations involves clearing fractions by multiplying both sides of the equation by the least common denominator (LCD).
How to: Solve a Rational Equation.
- Factor all denominators to determine the LCD.
Note the restrictions to \(x\). These are the values of \(x\) that make the denominator zero. They are values that \(x\) can't be because a fraction with a zero denominator is undefined. - Multiply both sides of the equal sign by the LCD. Every term in the equation is multiplied by the LCD. This results in the fractions getting "cleared" from the equation.
- Solve the resulting equation
- Check for extraneous solutions. Check each solution to confirm the value does not make the denominator in the original equation equal to zero. If a solution is found to make a denominator in the original equation zero, it must be rejected as a solution.
Example \(\PageIndex{1}\):
Solve: \(\dfrac { 1 } { x } + \dfrac { 2 } { x ^ { 2 } } = \dfrac { x + 9 } { 2 x ^ { 2 } }\).
Solution
Both sides of the equation are multiplied by the LCD, which is \(2x^{2}\). Note the restriction on \(x\) is \( x \ne 0 \).
\(\begin{aligned} \color{Cerulean}{2 x ^ { 2} }\color{black}{ \cdot} \left( \frac { 1 } { x } + \frac { 2 } { x ^ { 2 } } \right) & =\color{Cerulean}{ 2 x ^ { 2} }\color{black}{ \cdot} \left( \frac { x + 9 } { 2 x ^ { 2 } } \right)\quad\quad\color{Cerulean}{Multiply\:both\:sides\:by\:the\:LCD.} \\ \color{Cerulean}{2 x ^ { 2} }\color{black}{ \cdot} \frac { 1 } { x } + \color{Cerulean}{2 x ^ { 2} }\color{black}{ \cdot} \frac { 2 } { x ^ { 2 } } & = \color{Cerulean}{2 x ^ { 2 }}\color{black}{ \cdot} \frac { x + 9 } { 2 x ^ { 2 } }\quad\quad\quad\:\:\color{Cerulean}{Distribute.} \\ 2 x + 4 & = x + 9\quad\quad\quad\quad\quad\quad\color{Cerulean}{Simplify\:and\:then\:solve.} \\ x & = 5 \end{aligned}\)
The answer must always be checked that it is not a restricted value. In other words, answers must be checked so that when substituted back into the original equation, no division by zero occurs.
The solution set is \( \{5 \}\).
After multiplying both sides of the previous example by the LCD, we were left with a linear equation to solve. This is not always the case; sometimes we will be left with quadratic equation.
Example \(\PageIndex{2}\):
Solve: \(\dfrac { 3 ( x + 2 ) } { x - 4 } - \dfrac { x + 4 } { x - 2 } = \dfrac { x - 2 } { x - 4 }\).
Solution
In this example, there are two restrictions, \(x≠4\) and \(x≠2\). Begin by multiplying both sides by the LCD, \((x−2)(x−4)\).
\(\begin{aligned} \color{Cerulean}{(x-2)(x-4)}\color{black}{\cdot} \left( \frac{3(x+2)}{x-4}-\frac{x+4}{x-2} \right) &= \color{Cerulean}{(x-2)(x-4)}\color{black}{\cdot} \left( \frac{x-2}{x-4} \right)\\ \color{Cerulean}{(x-2)\cancel{(x-4)}}\color{black}{\cdot} \frac{3(x+2)}{\cancel{x-4}} - \color{Cerulean}{\cancel{(x-2)}(x-4)}\color{black}{\cdot} \frac{x+4}{\cancel{x-2}} &= \color{Cerulean}{(x-2)\cancel{(x-4)}}\color{black}{\cdot} \frac{x-2}{\cancel{x-4}} \\ 3 ( x + 2 ) ( x - 2 ) - ( x + 4 ) ( x - 4 ) & = ( x - 2 ) ( x - 2 ) \\ 3 \left( x ^ { 2 } - 4 \right) - \left( x ^ { 2 } - 16 \right) & = x ^ { 2 } - 2 x - 2 x + 4 \\ 3 x ^ { 2 } - 12 - x ^ { 2 } + 16 & = x ^ { 2 } - 4 x + 4\\ 2x^{2} + 4 & = x^{2}-4x+4 \end{aligned}\)
To solve, rewrite the quadratic equation in standard form, factor, and then set each factor equal to 0.
\(\begin{array} { l } { 2 x ^ { 2 } + 4 = x ^ { 2 } - 4 x + 4 } \\ { x ^ { 2 } + 4 x = 0 } \\ { x ( x + 4 ) = 0 } \end{array}\)
\(\begin{aligned} x = 0 \text { or } x + 4 & = 0 \\ x & = - 4 \end{aligned}\)
These values are checked, confirming they are in the domain of the original equation.
The solution set is \( \{ 0, −4 \} \).
Up to this point, all of the possible solutions have solved the original equation. However, this may not always be the case. Multiplying both sides of an equation by variable factors may lead to extraneous solutions, which are solutions that do not solve the original equation. A complete list of steps for solving a rational equation is outlined in the following example.
Example \(\PageIndex{3}\):
Solve: \(\dfrac { 2 x } { 3 x + 1 } = \dfrac { 1 } { x - 5 } - \dfrac { 4 ( x - 1 ) } { 3 x ^ { 2 } - 14 x - 5 }\).
Solution
Step 1: Factor all denominators and determine the LCD.
\(\begin{array} { l } { \dfrac { 2 x } { 3 x + 1 } = \dfrac { 1 } { x - 5 } - \dfrac { 4 ( x - 1 ) } { 3 x ^ { 2 } - 14 x - 5 } } \\ { \dfrac { 2 x } { ( 3 x + 1 ) } = \dfrac { 1 } { ( x - 5 ) } - \dfrac { 4 ( x - 1 ) } { ( 3 x + 1 ) ( x - 5 ) } } \end{array}\)
The LCD is \((3x+1)(x−5)\).
Identify the restrictions. In this case, \(x≠−\frac{1}{3}\) and \(x≠5\).
Step 2: Multiply both sides of the equation by the LCD. Distribute carefully and then simplify.
\(\begin{aligned} \color{Cerulean}{( 3 x + 1 ) ( x - 5 )}\color{black}{ \cdot} \frac { 2 x } { ( 3 x + 1 ) } &=\color{Cerulean}{ ( 3 x + 1 ) ( x - 5 )}\color{black}{ \cdot} \left( \frac { 1 } { ( x - 5 ) } - \frac { 4 ( x - 1 ) } { ( 3 x + 1 ) ( x - 5 ) }\right)\\ \color{Cerulean}{\cancel{(3x+1)}(x-5)}\color{black}{ \cdot}\frac{2x}{\cancel{(3x+1)}} &= \color{Cerulean}{(3x+1)\cancel{(x-5)}}\color{black}{\cdot} \frac{1}{\cancel{(x-5)}} -\color{Cerulean}{\cancel{ (3x+1)}\cancel{(x-5)}}\color{black}{ \cdot} \frac{4(x-1)}{\cancel{(3x+1)}\cancel{(x-5)}} \\ 2x(x-5) &=(3x+1)-4(x-1) \end{aligned}\)
Step 3: Solve the resulting equation. Here the result is a quadratic equation. Rewrite it in standard form, factor, and then set each factor equal to \(0\).
\(\begin{aligned} 2 x ( x - 5 ) & = ( 3 x + 1 ) - 4 ( x - 1 ) \\ 2 x ^ { 2 } - 10 x & = 3 x + 1 - 4 x + 4 \\ 2 x ^ { 2 } - 10 x & = - x + 5 \\ 2 x ^ { 2 } - 9 x - 5 & = 0 \\ ( 2 x + 1 ) ( x - 5 ) & = 0 \end{aligned}\)
\(\begin{aligned} 2 x + 1 & = 0 \quad\quad \text { or } &x - 5 &= 0 \\ 2 x & = - 1 & x &= 5 \\ x &= - \frac { 1 } { 2 } \end{aligned}\)
Step 4: Check for extraneous solutions. Here \(5\) is a restriction and so \(5\) produces a zero denominator in the original equation and so is an extraneous solution. Therefore it is not included in the solution set.
The solution set is \( \Big\{− \dfrac{1}{2} \Big\} \).
If this process produces a solution that happens to be a restriction, always disregard it as a solution.
Try It \(\PageIndex{3}\)
- Answer
- \(−\dfrac{3}{2}\)
Sometimes all potential solutions are extraneous, in which case we say that there is no solution to the original equation. In the next two examples, we demonstrate two ways in which rational equation can have no solutions.
Example \(\PageIndex{4}\):
Solve: \(1 + \dfrac { 5 x + 22 } { x ^ { 2 } + 3 x - 4 } = \dfrac { x + 4 } { x - 1 }\)
Solution
To identify the LCD, first factor the denominators.
\(\begin{array} { c } { 1 + \dfrac { 5 x + 22 } { x ^ { 2 } + 3 x - 4 } = \dfrac { x + 4 } { x - 1 } } \\ { 1 + \dfrac { 5 x + 22 } { ( x + 4 ) ( x - 1 ) } = \dfrac { x + 4 } { ( x - 1 ) } } \end{array}\)
Multiply both sides by the LCD, \((x+4)(x−1)\), distributing carefully.
\(\begin{aligned} \color{Cerulean}{ ( x + 4 ) ( x - 1 )}\color{black}{ \cdot} \left( 1 + \frac { 5 x + 22 } { ( x + 4 ) ( x - 1 ) } \right) &= \color{Cerulean}{( x + 4 ) ( x - 1 )}\color{black}{ \cdot} \frac { x + 4 } { ( x - 1 ) } \\ \color{Cerulean}{( x + 4 ) ( x - 1 )}\color{black}{ \cdot} 1 +\color{Cerulean}{ ( x + 4 ) ( x - 1 )}\color{black}{ \cdot} \frac { ( 5 x + 22 ) } { ( x + 4 ) ( x - 1 ) }& = \color{Cerulean}{( x + 4 ) ( x - 1 ) }\color{black}{\cdot} \frac { ( x + 4 ) } { ( x - 1 ) } \\ (x+4)(x-1) + (5x+22) &=(x+4)(x+4) \\ x ^ { 2 } - x + 4 x - 4 + 5 x + 22& = x ^ { 2 } + 4 x + 4x + 16 \\ x ^ { 2 } + 8 x + 18 &= x ^ { 2 } + 8 x + 16 \\ 18 &= 16 \:\: \color{red} { False } \end{aligned}\)
The equation is a contradiction and thus has no solution. The solution set is \( \{ \:\: \} \) or equivalently, \(Ø\)
Example \(\PageIndex{5}\):
Solve: \(\dfrac { 3 x } { 2 x - 3 } - \dfrac { 3 ( 4 x + 3 ) } { 4 x ^ { 2 } - 9 } = \dfrac { x } { 2 x + 3 }\).
Solution
First, factor the denominators.
\(\dfrac { 3 x } { ( 2 x - 3 ) } - \dfrac { 3 ( 4 x + 3 ) } { ( 2 x + 3 ) ( 2 x - 3 ) } = \dfrac { x } { ( 2 x + 3 ) }\)
Take note that the restrictions on the domain are \(x≠±\dfrac{3}{2}\). To clear the fractions, multiply by the LCD, \((2x+3)(2x−3)\).
\(\begin{aligned} \frac { 3 x \cdot \color{Cerulean}{( 2 x + 3 ) ( 2 x - 3 )} } { \color{black}{( 2 x - 3 )} } - \frac { 3 ( 4 x + 3 ) \cdot \color{Cerulean}{( 2 x + 3 ) ( 2 x - 3 )} } {\color{black}{ (} 2 x + 3 ) ( 2 x - 3 ) } &= \frac { x \cdot \color{Cerulean}{ ( 2 x + 3 ) ( 2 x - 3 )} } { \color{black}{(} 2 x + 3 ) } \\ 3 x ( 2 x + 3 ) - 3 ( 4 x + 3 ) &= x ( 2 x - 3 ) \\ 6 x ^ { 2 } + 9 x - 12 x - 9 &= 2 x ^ { 2 } - 3 x \\ 6 x ^ { 2 } - 3 x - 9 &= 2 x ^ { 2 } - 3 x \\ 4 x ^ { 2 } - 9 &= 0 \\ ( 2 x + 3 ) ( 2 x - 3 ) &= 0 \end{aligned}\)
\(\begin{aligned} 2 x + 3 & = 0 \quad\quad \text { or }& 2 x - 3& = 0 \\ 2 x & = - 3 &2 x& = 3 \\ x &= - \frac { 3 } { 2 } & x& = \frac { 3 } { 2 } \end{aligned}\)
Both of these values are restrictions of the original equation; hence both are extraneous, and the solution set is \(Ø\)
Rational Expressions and Rational Equations
It is important to point out that this technique for clearing algebraic fractions only works for equations. Do not try to clear algebraic fractions when simplifying expressions. As a reminder, an example of each is provided below.
| Expression | Equation | ||
|---|---|---|---|
|
\(\dfrac { 1 } { x } + \dfrac { x } { 2 x + 1 }\) |
\(\dfrac { 1 } { x } + \dfrac { x } { 2 x + 1 } =0\) |
Expressions are to be simplified and equations are to be solved. If we multiply the expression by the LCD, \(x (2x + 1)\), we obtain another expression that is not equivalent.
| Incorrect | Correct if an Expression | Correct if an Equation |
|---|---|---|
|
\(\begin{array} { l } { \dfrac { 1 } { x } + \dfrac { x } { 2 x + 1 } } \\ { \neq \color{red}{x ( 2 x + 1 )}\color{black}{ \cdot} \left( \dfrac { 1 } { x } + \dfrac { x } { 2 x + 1 } \right) } \\ { = 2 x + 1 + x ^ { 2 } } \color{red}{✗} \end{array}\) |
\(\begin{array} { l } { \dfrac { 1 } { x } + \dfrac { x } { 2 x + 1 } } \\ = \dfrac { 1 } { x } {\color{Cerulean}{ \dfrac{2x+1}{2x+1} }} + \dfrac { x } { 2 x + 1 } {\color{Cerulean}{ \dfrac{x}{x} }} \\ =\dfrac{2x+1 +x^2}{x(2x+1)}\\ =\dfrac{ (x+1)^2 }{x(2x+1)} \\ \end{array}\) |
\(\begin{aligned} \frac { 1 } { x } + \frac { x } { 2 x + 1 } & = 0 \\ \color{Cerulean}{x(2x+1)} \color{black}{\cdot} ( \frac { 1 } { x } + \frac { x } { 2 x + 1 } ) & = \color{Cerulean}{x ( 2 x + 1 )} \color{black}{\cdot}0 \\ 2 x + 1 + x ^ { 2 } & = 0 \\ x ^ { 2 } + 2 x + 1 & = 0 \color{Cerulean}{✓}\end{aligned}\) |
Rational equations are sometimes expressed using negative exponents.
Example \(\PageIndex{6}\):
Solve: \(6+x^{−1}=x^{−2}\).
Solution:
Begin by removing the negative exponents.
\(\begin{aligned} 6 + x ^ { - 1 } & = x ^ { - 2 } \\ 6 + \frac { 1 } { x } & = \frac { 1 } { x ^ { 2 } } \end{aligned}\)
Here we can see the restriction, \(x≠0\). Next, multiply both sides by the LCD, \(x^{2}\).
\(\begin{aligned} \color{Cerulean}{x ^ { 2 }}\color{black}{ \cdot} \left( 6 + \frac { 1 } { x } \right) & =\color{Cerulean}{ x ^ { 2} }\color{black}{ \cdot} \left( \frac { 1 } { x ^ { 2 } } \right) \\ \color{Cerulean}{x ^ { 2} }\color{black}{ \cdot} 6 + \color{Cerulean}{x ^ { 2} } \color{black}{\cdot} \frac { 1 } { x } & =\color{Cerulean}{ x ^ { 2} }\color{black}{ \cdot} \frac { 1 } { x ^ { 2 } } \\ 6 x ^ { 2 } + x & = 1 \\ 6 x ^ { 2 } + x - 1 & = 0 \\ ( 3 x - 1 ) ( 2 x + 1 ) & = 0 \end{aligned}\)
\(\begin{aligned} 3 x - 1 &= 0 \quad\quad \text { or } & 2 x + 1 &= 0 \\ 3 x &= 1 & 2 x & = - 1 \\ x &= \frac { 1 } { 3 } &x &= - \frac { 1 } { 2 } \end{aligned}\)
Neither solution is a restricted value and will not produce division by zero in the original equation, so the solution set is \( \Large\{−\frac{1}{2}, \frac{1}{3} \Large\} \)
Proportions
A proportion is a statement of equality of two ratios, \(\frac{a}{b}=\frac{c}{d}\). This proportion is often read “\(a\) is to \(b\) as \(c\) is to \(d\).” Given any nonzero real numbers \(a, b, c\), and \(d\) that satisfy a proportion, multiply both sides by the product of the denominators to obtain the following:
\(\begin{aligned} \frac { a } { b } & = \frac { c } { d } \\ \color{Cerulean}{b d}\color{black}{ \cdot} \frac { a } { b } & = \color{Cerulean}{b d}\color{black}{ \cdot} \frac { c } { d } \\ a d & = b c \end{aligned}\)
This shows that cross products are equal, and is commonly referred to as cross multiplication.
If \(\dfrac{a}{b}=\dfrac{c}{d}\) then \(ad = bc\)
Cross multiply to solve proportions where terms are unknown.
Example \(\PageIndex{7}\):
Solve: \(\dfrac { 5 n - 1 } { 5 } = \dfrac { 3 n } { 2 }\).
Solution
When cross multiplying, be sure to group \(5n−1\).
\(( 5 n - 1 ) \cdot 2 = 5 \cdot 3 n\)
Apply the distributive property in the next step.
\(\begin{aligned} ( 5 n - 1 ) \cdot 2 & = 5 \cdot 3 n \\ 10 n - 2 & = 15 n \quad \color{Cerulean} { Distribute. } \\ - 2 & = 5 n \quad\:\:\color{Cerulean}{Solve.} \\ \frac { - 2 } { 5 } & = n \end{aligned}\)
Neither fraction can ever have a denominator of zero, so there are no extraneous solutions and the solution set is \( \{−\frac{2}{5} \} \)
Cross multiplication can be used as an alternate method for solving rational equations, but in order to employ this technique, each side of the equation must be simplified to a single algebraic fraction and then cross multiplication can be done.
Example \(\PageIndex{8}\):
Solve: \(\dfrac { 1 } { 2 } - \dfrac { 4 } { x } = - \dfrac { x } { 8 }\).
Solution
Obtain a single algebraic fraction on the left side by subtracting the equivalent fractions with a common denominator.
\(\begin{aligned} \frac { 1 } { 2 } \cdot \color{Cerulean}{\frac { x } { x }}\color{black}{ -} \frac { 4 } { x } \cdot \color{Cerulean}{\frac { 2 } { 2 }} & \color{black}{=} - \frac { x } { 8 } \\ \frac { x } { 2 x } - \frac { 8 } { 2 x } & = - \frac { x } { 8 } \\ \frac { x - 8 } { 2 x } & = - \frac { x } { 8 } \end{aligned}\)
Note that \(x≠0\), cross multiply, and then solve for \(x\).
\(\begin{aligned} \frac { x - 8 } { 2 x } & = \frac { - x } { 8 } \\ 8 ( x - 8 ) & = - x \cdot 2 x \\ 8 x - 64 & = - 2 x ^ { 2 } \\ 2 x ^ { 2 } + 8 x - 64 & = 0 \\ 2 \left( x ^ { 2 } + 4 x - 32 \right) & = 0 \\ 2 ( x - 4 ) ( x + 8 ) & = 0 \end{aligned}\)
Next, set each variable factor equal to zero.
\(\begin{aligned} x - 4 & = 0 \quad\quad\text { or } &x + 8 &= 0 \\ x & = 4 \quad &x &= - 8 \end{aligned}\)
The restrictions to \( x \) are \( x \ne 0 \), so both solutions are legitimate. The solution set is \( \{−8, 4 \} \)
Try It \(\PageIndex{3}\)
- Answer
- \(−\dfrac{3}{2}\)\(2, 3\)
A note of caution should accompany proposed use of cross multiplication because it sometimes results in producing a more complicated equation to solve.
Example \(\PageIndex{9}\):
Solve \( \dfrac{2x+3}{x^2-1}=\dfrac{x+1}{2x-2} \).
Solution
\( \underline{\textrm{Method 1}} \). Using cross multiplication produces: \( 4x^2+2x-6 = x^3+x^2-x-1\). Zeroing one side of the equal sign and combining like terms produces \(0 = x^3 -3x^2 -3x+5\) which is not an equation for which a solution can easily be found.
\( \underline{\textrm{Method 2}} \). Starting with determining the LCD, which is \(2(x^2-1)\) or \( 2(x-1)(x+1) \), and multiplying both sides of the equal sign by that LCD produces \( 4x+6 = x^2+2x+1\). Simplification produces the equation \( 5 = x^2 - 2x\) which is easily solved. The solution set is \( \{ 1 \pm \sqrt{6} \} \).
In order to avoid complications like this, denominators should be examined before performing cross multiplication, and any common factors in the denominators should be canceled first. In the above example, the two denominators are \( (x-1)(x+1) \) and \(2(x-1)\). If the common factor in both denominators, \( (x-1) \), is removed (by multiplying both sides of the equal sign by \( (x-1) \), then when cross multiplication is done, the result would be a quadratic \( 2(2x+3) = (x+1)(x+1) \) which is more easily solved than a cubic.