8.7: Rational Equations
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- 49392
Rational Equations
Rational Equations
When one rational expression is set equal to another rational expression, a rational equation results.
Some examples of rational equations are the following (except for number 5):
\(\dfrac{3x}{4} = \dfrac{15}{2}\)
\(\dfrac{x+1}{x-2} = \dfrac{x-7}{x-3}\)
\(\dfrac{5a}{2} = 10\)
\(\dfrac{3}{x} + \dfrac{x-3}{x+1} = \dfrac{6}{5x}\)
\(\dfrac{x-6}{x+1}\) is a rational expression, not a rational equation.
The Logic Behind The Process
It seems most reasonable that an equation without any fractions would be easier to solve than an equation with fractions. Our goal, then, is to convert any rational equation to an equation that contains no fractions. This is easily done.
To develop this method, let’s consider the rational equation
\(\dfrac{1}{6} + \dfrac{x}{4} = \dfrac{17}{12}\)
The LCD is 12. We know that we can multiply both sides of an equation by the same nonzero quantity, so we’ll multiply both sides by the LCD, 12.
\(12(\dfrac{1}{6} + \dfrac{x}{4}) = 12 \cdot \dfrac{17}{12}\)
Now distribute 12 to each term on the left side using the distributive property.
\(12 \cdot \dfrac{1}{6} + 12 \cdot \dfrac{x}{4} = 12 \cdot \dfrac{17}{12}\)
Now divide to eliminate all denominators.
\(\begin{array}{flushleft}
2 \cdot 1 + 3 \cdot x &= 17\\
2 + 3x &= 17
\end{array}\)
Now there are no more fractions, and we can solve this equation using our previous techniques to obtain 5 as the solution.
The Process
We have cleared the equation of fractions by multiplying both sides by the LCD. This development generates the following rule.
To clear an equation of fractions, multiply both sides of the equation by the LCD.
When multiplying both sides of the equation by the LCD, we use the distributive property to distribute the LCD to each term. This means we can simplify the above rule.
To clear an equation of fractions, multiply every term on both sides of the equation by the LCD.
The complete method for solving a rational equation is
1. Determine all the values that must be excluded from consideration by finding the values that will produce zero in the denominator (and thus, division by zero). These excluded values are not in the domain of the equation and are called nondomain values.
2. Clear the equation of fractions by multiplying every term by the LCD.
3. Solve this nonfractional equation for the variable. Check to see if any of these potential solutions are excluded values.
4. Check the solution by substitution.
Extraneous Solutions
Potential solutions that have been excluded because they make an expression undefined (or produce a false statement for an equation) are called extraneous solutions. Extraneous solutions are discarded. If there are no other potential solutions, the equation has no solution.
Sample Set A
Solve the following rational equations.
\(\begin{array}{flushleft}
\dfrac{3x}{4} &= \dfrac{15}{2} & \text{ Since the denominators are constants, there are no excluded values.}\\
&& \text{ No values must be excluded. The LCD is 4. Multiply each term by 4}\\
4 \cdot \dfrac{3x}{4} &= 4 \cdot \dfrac{15}{2}\\
\cancel{4} \cdot \dfrac{3x}{\cancel{4}} &= _{\cancel{4}}^{2} \cdot \dfrac{15}{\cancel{2}}\\
3x &= 2 \cdot 15\\
3x &= 30\\
x &= 10 & 10 \text{ is not an excluded value. Check it as a solution}.
\end{array}\)
Check:
\(\begin{array}{flushleft}
\dfrac{3x}{4} &= \dfrac{15}{2}\\
\dfrac{3(10)}{4} &= \dfrac{15}{2} & \text{ Is this correct? }\\
\dfrac{30}{4} &= \dfrac{15}{2} & \text{ Is this correct? }\\
\dfrac{15}{2} &= \dfrac{15}{2} & \text{ Yes, this is correct }
\end{array}\)
\(\begin{aligned}
\dfrac{4}{x-1} &= \dfrac{2}{x+6} & 1 \text{ and } -6 \text{ are nondomain values. Exclude them from the solution}\\
&& \text{The LCD is } (x-1)(x+6) \text{ Multiply every term by the LCD }\\
(x-1)(x+6) \cdot \dfrac{4}{x-1} &= (x-1)(x+6) \cdot \dfrac{2}{x+6}\\
\cancel{(x-1)}(x+6) \cdot \dfrac{4}{\cancel{x-1}} &= (x-1)\cancel{(x+6)} \cdot \dfrac{2}{\cancel{x+6}}\\
4(x+6) &= 2(x-1) & \text{ Solve this nonfractional equation }\\
4x + 24 &= 2x - 2\\
2x &= -26\\
x &= -13 & -13 \text{ is not an excluded value. Check it as a solution}
\end{aligned}\)
Check:
\(\begin{array}{flushleft}
\dfrac{4}{x-1} &= \dfrac{2}{x+6}\\
\dfrac{4}{-13-1} &= \dfrac{2}{-13 + 6} & \text{ Is this correct?}\\
\dfrac{4}{-14} &= \dfrac{2}{-7} & \text{ Is this correct?}\\
\dfrac{2}{-7} &= \dfrac{2}{-13 + 6} & \text{ Yes, this is correct }\\
\end{array}\)
\(-13\) is the solution.
\(\begin{array}{flushleft}
\dfrac{4a}{a-4} &= 2 + \dfrac{16}{a-4}. & 4 \text{ is a nondomain value. Exclude it from consideration}\\
&& \text{ The LCD is } a-4 \text{. Multiply every term by } a-4\\
(a-4) \cdot \dfrac{4a}{a-4} &= 2(a-4) + (a-4) \cdot \dfrac{16}{a-4}\\
\cancel{(a-4)} \cdot \dfrac{4a}{\cancel{a-4}} &= 2(a-4) + \cancel{(a-4)} \cdot \dfrac{16}{\cancel{a-4}}\\
4a &= 2(a-4) + 16 & \text{ Solve this nonfractional equation }\\
4a &= 2a - 8 + 16\\
4a &= 2a + 8\\
2a &= 8\\
a &= 4
\end{array}\)
This value, \(a = 4\), has been excluded from consideration. It is not to be considered as a solution. It is extraneous. As there are no other potential solutions to consider, we conclude that this equation has no solution.
Practice Set A
Solve the following rational equations.
\(\dfrac{2x}{5} = \dfrac{x-14}{6}\)
- Answer
-
\(x=−10\)
\(\dfrac{3a}{a-1} = \dfrac{3a + 8}\)
- Answer
-
\(a=−2\)
\(\dfrac{3}{y-3} + 2 = \dfrac{y}{y-3}\)
- Answer
-
\(y = 3\) is extraneous, so no solution.
Sample Set B
Solve the following rational equations.
\(\begin{array}{flushleft}
\dfrac{3}{x} + \dfrac{4x}{x-1} &= \dfrac{4x^2 + x + 5}{x^2 - x} & \text{ Factor all denominators to find any excluded values and the LCD }\\
&& \text{ Nondomain values are } 0 \text{ and } 1. \text{ Exclude them from consideration. }\\
\dfrac{3}{x} + \dfrac{4x}{x-1} &= \dfrac{4x^2 + x + 5}{x(x-1)} & \text{ The LCD is } x(x-1) \text{. Multiply each term by } x(x-1) \text{ and simplify }
\end{array}\)
\(\cancel{x}(x-1) \cdot \dfrac{3}{\cancel{x}} + x(\cancel{x-1}) \cdot \dfrac{4x}{\cancel{x-1}} = \cancel{x(x-1)} \cdot \dfrac{4x^2 + x + 5}{\cancel{x(x-1)}}\).
\(\begin{array}{flushleft}
3(x-1) + 4x \cdot x &= 4x^2 + x + 5 & \text{ Solve this nonfractional equation to obtain the potential solutions }\\
3x - 3 + 4x^2 &= 4x^2 + x + 5\\
3x - 3 &= x + 5\\
2x &= 8\\
x &= 4 & 4 \text{ is not an excluded value. Check it as a solution }
\end{array}\)
Check:
\(\begin{array}{flushleft}
\dfrac{3}{x} + \dfrac{4x}{x-1} &= \dfrac{4x^2 + x + 5}{x^2 - x}\\
\dfrac{3}{4} + \dfrac{4 \cdot 4}{4-1} &= \dfrac{4 \cdot 4^2 + 4 + 5}{16 - 4} & \text{ Is this correct? }\\
\dfrac{3}{4} + \dfrac{16}{3} &= \dfrac{64 + 4 + 5}{12} & \text{ Is this correct? }\\
\dfrac{9}{12} + \dfrac{64}{12} &= \dfrac{73}{12} & \text{ Is this correct? }\\
\dfrac{73}{12} &= \dfrac{73}{12} & \text{ Yes, this is correct }
\end{array}\)
\(4\) is the solution.
The zero-factor property can be used to solve certain types of rational equations. We studied the zero-factor property in Section 5.1, and you may remember that it states that if \(a\) and \(b\) are real numbers and that \(a \cdot b=0\), then either or both \(a=0\) or \(b=0\).The zero-factor property is useful in solving the following rational equation.
\(\begin{array}{flushleft}
\dfrac{3}{a^2} - \dfrac{2}{a} &= 1 & \text{ Zero is an excluded value. }\\
&& \text{ The LCD is } a^2 \text{ Multiply each term by } a^2 \text{ and simplify }\\
\cancel{a^2} \cdot \dfrac{3}{\cancel{a^2}} - \cancel{a^2} \cdot \dfrac{2}{\cancel{a}} &= 1 \cdot a^2\\
3-2a &= a^2 & \text{ Solve this nonfractional quadratic equation. Set it equal to zero }\\
0 &= a^2 + 2a - 3\\
0 &= (a+3)(a-1)\\
a&= - 3, a = 1 & \text{ Check these as solutions }
\end{array}\)
Check:
\(\begin{array}{flushleft}
\text{If } a = -3: & \dfrac{3}{(-3)^2} - \dfrac{2}{-3} &= 1 & \text{ Is this correct? }\\
& \dfrac{3}{9} + \dfrac{2}{3} &= 1 & \text{ Is this correct? }\\
& \dfrac{1}{3} + \dfrac{2}{3} &= 1 & \text{ Is this correct? }\\
& 1 &= 1 & \text{ Yes, this is correct }\\
& a &= -3 & \text{ Checks and is a solution }\\
\text{If } a = 1: & \dfrac{3}{(1)^2} - \dfrac{2}{1} &= 1 & \text{ Is this correct? }\\
& \dfrac{3}{1} - \dfrac{2}{1} &= 1 & \text{ Is this correct? }\\
& 1 &= 1 & \text{ Yes, this is correct. }\\
& a &= 1 & \text{ Checks and is a solution }
\end{array}\)
\(-3\) and \(1\) are the solutions.
Practice Set B
Solve the equation \(\dfrac{a+3}{a-2} = \dfrac{a+1}{a-1}\)
- Answer
-
\(a = \dfrac{1}{3}\)
Solve the equation \(\dfrac{1}{x-1} - \dfrac{1}{x+1} = \dfrac{2x}{x^2 - 1}\)
- Answer
-
This equation has no solution. \(x=1\) is extraneous.
Section 7.6 Exercises
For the following problems, solve the rational equations.
\(\dfrac{32}{x} = \dfrac{16}{3}\)
- Answer
-
\(x = 6\)
\(\dfrac{54}{y} = \dfrac{27}{4}\)
\(\dfrac{8}{y} = \dfrac{2}{3}\)
- Answer
-
\(y=12\)
\(\dfrac{x}{28} = \dfrac{3}{7}\)
\(\dfrac{x + 1}{4} = \dfrac{x-3}{2}\)
- Answer
-
\(x = 7\)
\(\dfrac{a + 3}{6} = \dfrac{a - 1}{4}\)
\(\dfrac{y-3}{6} = \dfrac{y + 1}{4}\)
- Answer
-
\(y=−9\)
\(\dfrac{x-7}{8} = \dfrac{x+5}{6}\)
\(\dfrac{a + 6}{9} - \dfrac{a-1}{6} = 0\)
- Answer
-
\(a=15\)
\(\dfrac{y + 11}{4} = \dfrac{y + 8}{10}\)
\(\dfrac{b + 1}{2} + 6 = \dfrac{b- 4}{3}\)
- Answer
-
\(b=−47\)
\(\dfrac{m+3}{2} + 1 = \dfrac{m-4}{5}\)
\(\dfrac{a - 6}{2} + 4 = -1\)
- Answer
-
\(a=−4\)
\(\dfrac{b + 11}{3} + 8 = 6\)
\(\dfrac{y - 1}{y + 2} = \dfrac{y + 3}{y - 2}\)
- Answer
-
\(y = -\dfrac{1}{2}\)
\(\dfrac{x + 2}{x - 6} = \dfrac{x - 1}{x + 2}\)
\(\dfrac{3m + 1}{2m} = \dfrac{4}{3}\)
- Answer
-
\(m=−3\)
\(\dfrac{2k + 7}{3k} = \dfrac{5}{4}\)
\(\dfrac{4}{x + 2} = 1\)
- Answer
-
\(x=2\)
\(\dfrac{-6}{x - 3} = 1\)
\(\dfrac{a}{3} + \dfrac{10 + a}{4} = 6\)
- Answer
-
\(a=6\)
\(\dfrac{k + 17}{5} - \dfrac{k}{2} = 2k\)
\(\dfrac{2b + 1}{3b - 5} = \dfrac{1}{4}\)
- Answer
-
\(b = -\dfrac{9}{5}\)
\(\dfrac{-3a + 4}{2a - 7} = \dfrac{-7}{9}\)
\(\dfrac{x}{x + 3} - \dfrac{x}{x-2} = \dfrac{10}{x^2 + x - 6}\)
- Answer
-
\(x=−2\)
\(\dfrac{3y}{y-1} + \dfrac{2y}{y-6} = \dfrac{5y^2 - 15y + 20}{y^2 - 7y + 6}\)
\(\dfrac{4a}{a+2} - \dfrac{3a}{a-1} = \dfrac{a^2 - 8a - 4}{a^2 + a - 2}\)
- Answer
-
\(a=2\)
\(\dfrac{3a - 7}{a-3} = \dfrac{4a - 10}{a - 3}\)
\(\dfrac{2x - 5}{x - 6} = \dfrac{x+1}{x-6}\)
- Answer
-
No solution; 6 is an excluded value.
\(\dfrac{3}{x + 4} + \dfrac{5}{x + 4} = \dfrac{3}{x - 1}\)
\(\dfrac{2}{y + 2} + \dfrac{8}{y + 2} = \dfrac{9}{y + 3}\)
- Answer
-
\(y=−12\)
\(\dfrac{4}{a^2 + 2a} = \dfrac{3}{a^2 + a - 2}\)
\(\dfrac{2}{b(b+2)} = \dfrac{3}{b^2 + 6b + 8}\)
- Answer
-
\(b=8\)
\(\dfrac{x}{x-1} + \dfrac{3x}{x-4} = \dfrac{4x^2 - 8x + 1}{x^2 - 5x + 4}\)
\(\dfrac{4x}{x+2} - \dfrac{x}{x+1} = \dfrac{3x^2 + 4x + 4}{x^2 + 3x + 2}\)
- Answer
-
no solution
\(\dfrac{2}{a-5} - \dfrac{4a - 2}{a^2 - 6a + 5} = \dfrac{-3}{a-1}\)
\(\dfrac{-1}{x+4} - \dfrac{2}{x+1} = \dfrac{4x + 19}{x^2 + 5x + 4}\)
- Answer
-
No solution; \(−4\) is an excluded value.
\(\dfrac{2}{x^2} + \dfrac{1}{x} = 1\)
\(\dfrac{6}{y^2} - \dfrac{5}{y} = 1\)
- Answer
-
\(y=−6, 1\)
\(\dfrac{12}{a^2} - \dfrac{4}{a} = 1\)
\(\dfrac{20}{x^2} - \dfrac{1}{x} = 1\)
- Answer
-
\(x=4, −5\)
\(\dfrac{12}{y} + \dfrac{12}{y^2} = -3\)
\(\dfrac{16}{b^2} + \dfrac{12}{b} = 4\)
- Answer
-
\(y=4,−1\)
\(\dfrac{1}{x^2} = 1\)
\(\dfrac{16}{y^2} = 1\)
- Answer
-
\(y=4,−4\)
\(\dfrac{25}{a^2} = 1\)
\(\dfrac{36}{y^2} = 1\)
- Answer
-
\(y=6,−6\)
\(\dfrac{2}{x^2} + \dfrac{3}{x} = 2\)
\(\dfrac{2}{a^2} - \dfrac{5}{a} = 3\)
- Answer
-
\(a = \dfrac{1}{3}, -2\)
\(\dfrac{2}{x^2} + \dfrac{7}{x} = -6\)
\(\dfrac{4}{a^2} + \dfrac{9}{a} = 9\)
- Answer
-
\(a = -\dfrac{1}{3}, \dfrac{4}{3}\)
\(\dfrac{2}{x} = \dfrac{3}{x+2} + 1\)
\(\dfrac{1}{x} = \dfrac{2}{x+4} - \dfrac{3}{2}\)
- Answer
-
\(x = -\dfrac{4}{3}, -2\)
\(\dfrac{4}{m} - \dfrac{5}{m-3} = 7\)
\(\dfrac{6}{a + 1} - \dfrac{2}{a-2} = 5\)
- Answer
-
\(a = \dfrac{4}{5}, 1\)
For the following problems, solve each literal equation for the designated letter.
\(V = \dfrac{GMm}{D}\) for \(D\)
\(PV = nrt\) for \(n\).
- Answer
-
\(n = \dfrac{PV}{rt}\)
\(E = mc^2\) for \(m\)
\(P = 2(1 + w)\) for \(w\).
- Answer
-
\(W = \dfrac{P - 2}{2}\)
\(A = \dfrac{1}{2}h(b + B)\) for \(B\).
\(A = P(1 + rt)\) for \(r\).
- Answer
-
\(r = \dfrac{A - P}{Pt}\)
\(z = \dfrac{x-\hat{x}}{s}\) for \(\hat{x}\)
\(F=\dfrac{S_{x}^{2}}{S_{y}^{2}} \text { for } S_{y}^{2}\)
- Answer
-
\(S_{y}^{2}=\dfrac{S_{x}^{2}}{F}\)
\(\dfrac{1}{R} = \dfrac{1}{E} + \dfrac{1}{F}\) for \(F\).
\(K = \dfrac{1}{2}h(s_1 + s_2)\) for \(s_2\).
- Answer
-
\(S_{2}=\dfrac{2 K}{h}-S_{1} \text { or } \dfrac{2 K-h S_{1}}{h}\)
\(Q = \dfrac{2mn}{s + t}\) for \(s\).
\(V = \dfrac{1}{6}\pi(3a^2 + h^2)\) for \(h^2\).
- Answer
-
\(h_{2}=\dfrac{6 V-3 \pi a^{2}}{\pi}\)
\(I = \dfrac{E}{R + r}\) for \(R\).
Exercises For Review
Write \((4x^3y^{-4})^{-2}\) so that only positive exponents appear.
- Answer
-
\(\dfrac{y^8}{16x^6}\)
Factor \(x^4 - 16\)
Supply the missing word. An slope of a line is a measure of the _____ of the line.
- Answer
-
steepness
Find the product \(\dfrac{x^{2}-3 x+2}{x^{2}-x-12} \cdot \dfrac{x^{2}+6 x+9}{x^{2}+x-2} \cdot \dfrac{x^{2}-6 x+8}{x^{2}+x-6}\)
Find the sum. \(\dfrac{2x}{x+1} + \dfrac{1}{x-3}\)
- Answer
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\(\dfrac{2x^2 - 5x + 1}{(x+1)(x-3)}\)