2.4: Rational Equations
- Page ID
- 94044
- Solve rational equations
- Use rational functions
- Solve a rational equation for a specific variable
Before you get started, take this readiness quiz.
1. Solve \(\dfrac{1}{6} x+\dfrac{1}{2}=\dfrac{1}{3}\).
2. Solve
3. Solve the formula \(5x+2y=10\) for \(y\)
After defining the terms ‘expression’ and ‘equation’ earlier, we have used them throughout this book. We have simplified many kinds of expressions and solved many kinds of equations. We have simplified many rational expressions so far in this chapter. Now we will solve rational equations.
A rational equation is an equation that contains a rational expression.
Let's recall the difference between rational expressions and rational equations. The equation contains an equal sign.
| Rational Expression | Rational Equation |
| \(\dfrac{1}{8} x+\dfrac{1}{2}\) |
\(\dfrac{1}{8} x+\dfrac{1}{2}=\dfrac{1}{4}\) |
| \(\dfrac{y+6}{y^{2}-36}\) | \(\dfrac{y+6}{y^{2}-36}=y+1\) |
| \(\dfrac{1}{n-3}+\dfrac{1}{n+4} \) | \(\dfrac{1}{n-3}+\dfrac{1}{n+4}=\dfrac{15}{n^{2}+n-12} \) |
As we explained in the introduction to this unit, in the previous unit we have used equal signs during a simplification of an expression. These equal signs implicitly mean that equation holds for all values of the variables except those that leave an expression undefined. In the case of the equations dealt with here, these equalities say something about the variable, and for 'most' values of the variable the equation is false. We will seek the values which make the equality true, i.e., we seek to solve these equations.
Solve Rational Equations
We have already solved linear equations that contained fractions. We found the LCD of all the fractions in the equation and then multiplied both sides of the equation by the LCD to “clear” the fractions.
We will use the same strategy to solve rational equations. As is often the case when facing a new kind of problem, we aim to somehow reformulate it in a way that is more familiar. So, we will aim at rewriting the equation as a polynomial equation. We will multiply both sides of the equation by the LCD. Then, we will have an equation that does not contain rational expressions and thus is much easier for us to solve since we already know how to solve linear and quadratic equations. But because the original equation may have a variable in a denominator, we must be careful that we don’t end up with a solution that would make a denominator equal to zero.
So before we begin solving a rational equation, we examine it first to find the values that would make any denominators zero. That way, when we solve a rational equation we will know if there are any algebraic solutions we must discard. Alternatively, we can check to make sure your possible solutions make sense in our equation.
A solution to an an equation that is equivalent (except for a few values) to a rational equation for which the rational expressions are undefined is called an extraneous solution to a rational equation.
An extraneous solution to a rational equation is a solution to an equation which is that is equivalent to the original except for a certain finite number of values that would cause any of the expressions in the original equation to be undefined.
We note any possible extraneous solutions, \(c\), by writing \(x\neq c\) next to the equation.
Solve \(\dfrac{1}{x}+\dfrac{1}{3}=\dfrac{5}{6}\).
Solution
| \(\dfrac{1}{x}+\dfrac{1}{3}=\dfrac{5}{6}\) | |
| Note any value of the variable that would make any denominator zero. |
If \(x=0\), then \(\dfrac{1}{x}\) is undefined. So we'll write \(x \neq 0\) next to the equation. \(\dfrac{1}{x}+\dfrac{1}{3}=\dfrac{5}{6}, x \neq 0 \nonumber \) |
| Find the least common denominator of all denominators in the equation. | Find the LCD of \(\dfrac{1}{x}\), \(\dfrac{1}{3}\), and \(\dfrac{5}{6}\). The LCD is \(6x\). |
| Clear the fractions by multiplying both sides of the equation by the LCD. |
Multiply both sides of the equation by the LCD, \(6x\). \({\color{red}6 x} \cdot\left(\dfrac{1}{x}+\dfrac{1}{3}\right)={\color{red}6 x} \cdot\left(\dfrac{5}{6}\right) \nonumber \) Use the Distributive Property. \({\color{red}6 x} \cdot \dfrac{1}{x}+{\color{red}6 x} \cdot \dfrac{1}{3}={\color{red}6 x} \cdot\left(\dfrac{5}{6}\right) \nonumber \) Simplify - and notice, no more fractions! \(6+2 x=5 x \nonumber \) |
| Solve the resulting equation. |
Simplify. \(\begin{aligned} &6=3 x\\ &2=x \end{aligned} \nonumber \) |
| Check. If any values found in Step 1 are extraneous solutions, discard them. Check any remaining solutions in the original equation. (This is not strictly necessary, but is helpful in case an error was made). |
Our solution is not an extraneous solution. \(\dfrac{1}{x}+\dfrac{1}{3}=\dfrac{5}{6} \nonumber \) We substitute \(x=2\) into the original equation. \(\begin{aligned} \frac{1}{2}+\frac{1}{3}&\overset{?}{=}\frac{5}{6} \\ \frac{3}{6}+\frac{2}{6}&\overset{?}{=}\frac{5}{6} \\ \frac{5}{6}&=\frac{5}{6} \surd \end{aligned} \nonumber \) |
| Conclude. | The solution is \(x=2\). |
Solve \(\dfrac{1}{y}+\dfrac{2}{3}=\dfrac{1}{5}. \nonumber \)
- Answer
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\(y=-\dfrac{15}{7}\)
Solve \(\dfrac{2}{3}+\dfrac{1}{5}=\dfrac{1}{x}. \nonumber \)
- Answer
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\(x=\dfrac{15}{13}\)
The steps of this method are shown.
- Step 1. Note any value of the variable that would make any denominator zero.
- Step 2. Find the least common denominator of all denominators in the equation.
- Step 3. Clear the fractions by multiplying both sides of the equation by the LCD.
- Step 4. Solve the resulting equation.
- Step 5. Check:
- If any values found in Step 1 are algebraic solutions, discard them.
- Check any remaining solutions in the original equation.
We always start by noting the values that would cause any denominators to be zero.
Solve \(1-\dfrac{5}{y}=-\dfrac{6}{y^{2}}. \nonumber \)
Solution
| \(1-\dfrac{5}{y}=-\dfrac{6}{y^{2}} \nonumber\) | |
| Note any value of the variable that would make any denominator zero. | \(1-\dfrac{5}{y}=-\dfrac{6}{y^{2}}, y \neq 0 \nonumber \) |
|
Find the least common denominator of all denominators in the equation. |
The LCD is \(y^2\). |
| Clear the fractions by multiplying both sides of the equation by the LCD. | \(y^{2}\left(1-\dfrac{5}{y}\right)=y^{2}\left(-\dfrac{6}{y^{2}}\right) \nonumber\) |
| Distribute. | \(y^{2} \cdot 1-y^{2}\left(\dfrac{5}{y}\right)=y^{2}\left(-\dfrac{6}{y^{2}}\right) \nonumber\) |
| Multiply. | \(y^{2}-5 y=-6 \nonumber \) |
| Solve the resulting equation. First write the quadratic equation in standard form. | \(y^{2}-5 y+6=0 \nonumber \) |
| Factor. | \((y-2)(y-3)=0 \nonumber \) |
| Use the Zero Product Property. |
\(y-2=0 \text { or } y-3=0 \nonumber \) |
| Solve. | \(y=2 \text { or } y=3 \nonumber \) |
|
Check. There are no extraneous solutions. Check \(y=2\) and \(y=3\) in the original equation. |
\(1-\dfrac{5}{y}=-\dfrac{6}{y^{2}} \quad \quad \quad 1-\dfrac{5}{y}=-\dfrac{6}{y^{2}} \nonumber \) \(1-\dfrac{5}{2} \overset{?}{=} -\dfrac{6}{2^{2}} \quad \quad \quad 1-\dfrac{5}{3} \overset{?}{=} -\dfrac{6}{3^{2}} \nonumber \) \(1-\dfrac{5}{2} \overset{?}{=}-\dfrac{6}{4} \quad \quad \quad 1-\dfrac{5}{3} \overset{?}{=} -\dfrac{6}{9} \nonumber \) \(\dfrac{2}{2}-\dfrac{5}{2} \overset{?}{=} -\dfrac{6}{4} \quad \quad \quad \dfrac{3}{3}-\dfrac{5}{3} \overset{?}{=} -\dfrac{6}{9} \nonumber \) \(-\dfrac{3}{2} \overset{?}{=} -\dfrac{6}{4} \quad \quad \quad -\dfrac{2}{3} \overset{?}{=} -\dfrac{6}{9} \nonumber \) \(-\dfrac{3}{2}=-\dfrac{3}{2} \surd \quad \quad \quad -\dfrac{2}{3}=-\dfrac{2}{3} \surd \nonumber \) |
| Conclude. | The solutions are \(y=2\) and \(y=3\). |
Solve \(1-\dfrac{2}{x}=\dfrac{15}{x^{2}}. \nonumber \)
- Answer
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\(x=-3, x=5\)
Solve \(1-\dfrac{4}{y}=\dfrac{12}{y^{2}}. \nonumber \)
- Answer
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\(y=-2, y=6\)
In the next example, the last denominators is a difference of squares. Remember to factor it first to find the LCD.
Solve \(\dfrac{2}{x+2}+\dfrac{4}{x-2}=\dfrac{x-1}{x^{2}-4}. \nonumber \)
Solution
| \(\dfrac{2}{x+2}+\dfrac{4}{x-2}=\dfrac{x-1}{x^{2}-4} \nonumber \) | |
| Note any value of the variable that would make any denominator zero. | \(\dfrac{2}{x+2}+\dfrac{4}{x-2}=\dfrac{x-1}{(x+2)(x-2)}, x \neq-2, x \neq 2 \nonumber \) |
| Find the least common denominator of all denominators in the equation. | The LCD is \((x+2)(x-2)\). |
| Clear the fractions by multiplying both sides of the equation by the LCD. | \((x+2)(x-2)\left(\dfrac{2}{x+2}+\dfrac{4}{x-2}\right)=(x+2)(x-2)\left(\dfrac{x-1}{x^{2}-4}\right) \nonumber \) |
| Distribute. |
\((x+2)(x-2) \dfrac{2}{x+2}+(x+2)(x-2) \dfrac{4}{x-2}=(x+2)(x-2)\left(\dfrac{x-1}{x^{2}-4}\right) \nonumber \) |
| Remove common factors. | \(\cancel {(x+2)}(x-2) \dfrac{2}{\cancel {x+2}}+(x+2){\cancel {(x-2)}} \dfrac{4}{\cancel {x-2}}=\cancel {(x+2)(x-2)}\left(\dfrac{x-1}{\cancel {x^{2}-4}}\right) \nonumber \) |
| Simplify. | \(2(x-2)+4(x+2)=x-1 \nonumber \) |
|
Distribute. |
\(2 x-4+4 x+8=x-1 \nonumber \) |
| Solve. | \(\begin{aligned} 6 x+4&=x-1\\ 5 x&=-5 \\ x&=-1 \end{aligned}\) |
|
Check: We have no extraneous solutions. Check \(x=-1\) in the original equation. |
\(\begin{aligned} \dfrac{2}{x+2}+\dfrac{4}{x-2} &=\dfrac{x-1}{x^{2}-4} \\ \dfrac{2}{(-1)+2}+\dfrac{4}{(-1)-2} &\overset{?}{=} \dfrac{(-1)-1}{(-1)^{2}-4} \\ \dfrac{2}{1}+\dfrac{4}{-3} &\overset{?}{=} \dfrac{-2}{-3} \\ \dfrac{6}{3}-\dfrac{4}{3} &\overset{?}{=} \dfrac{2}{3} \\ \dfrac{2}{3} &=\dfrac{2}{3} \surd \end{aligned} \nonumber \) |
| Conclude. | The solution is \(x=-1\). |
Solve \(\dfrac{2}{x+1}+\dfrac{1}{x-1}=\dfrac{1}{x^{2}-1}. \nonumber \)
- Answer
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\(x=\dfrac{2}{3}\)
Solve \(\dfrac{5}{y+3}+\dfrac{2}{y-3}=\dfrac{5}{y^{2}-9}. \nonumber \)
- Answer
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\(y=2\)
In the next example, the first denominator is a trinomial. Remember to factor it first to find the LCD.
Solve \(\dfrac{m+11}{m^{2}-5 m+4}=\dfrac{5}{m-4}-\dfrac{3}{m-1}. \nonumber \)
Solution
| \(\dfrac{m+11}{m^{2}-5 m+4}=\dfrac{5}{m-4}-\dfrac{3}{m-1} \nonumber \) | |
| Note any value of the variable that would make any denominator zero. Use the factored form of the quadratic denominator. | \(\dfrac{m+11}{(m-4)(m-1)}=\dfrac{5}{m-4}-\dfrac{3}{m-1}, m \neq 4, m \neq 1 \nonumber \) |
| Find the least common denominator of all denominators in the equation. | The LCD is \((m-4)(m-1)\). |
| Clear the fractions by multiplying both sides of the equation by the LCD. | \((m-4)(m-1)\left(\dfrac{m+11}{(m-4)(m-1)}\right)=(m-4)(m-1)\left(\dfrac{5}{m-4}-\dfrac{3}{m-1}\right) \nonumber \) |
| Distribute. | \((m-4)(m-1)\left(\dfrac{m+11}{(m-4)(m-1)}\right)=(m-4)(m-1) \dfrac{5}{m-4}-(m-4)(m-1) \dfrac{3}{m-1} \nonumber \) |
| Remove common factors. | \(\cancel {(m-4)(m-1)}\left(\dfrac{m+11}{\cancel {(m-4)(m-1)}}\right)=\cancel {(m-4)}(m-1) \dfrac{5}{\cancel{m-4}}-(m-4)\cancel {(m-1)} \dfrac{3}{\cancel {m-1}} \nonumber \) |
| Simplify. | \(m+11=5(m-1)-3(m-4) \nonumber \) |
| Solve the resulting equation. | \(\begin{aligned} m+11&=5 m-5-3 m+12 \\ 4&=m \end{aligned} \nonumber \) |
| Check. | We see that substituting 4 for \(m\) in the original equation would make a denominator equal to zero. The only solution to the nearly equivalent linear equation is an extraneous solution. |
| Conclude. | There is no solution to the original equation. |
Solve \(\dfrac{x+13}{x^{2}-7 x+10}=\dfrac{6}{x-5}-\dfrac{4}{x-2}. \nonumber \)
- Answer
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There is no solution.
Solve \(\dfrac{y-6}{y^{2}+3 y-4}=\dfrac{2}{y+4}+\dfrac{7}{y-1}. \nonumber \)
- Answer
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There is no solution.
The equation we solved in the previous example had only one algebraic solution, but it was an extraneous solution. That left us with no solution to the equation. In the next example we get two algebraic solutions. Here one or both could be extraneous solutions.
Solve \(\dfrac{y}{y+6}=\dfrac{72}{y^{2}-36}+4. \nonumber \)
Solution
| \(\dfrac{y}{y+6}=\dfrac{72}{y^{2}-36}+4 \nonumber \) | |
|
Factor all the denominators, so we can note any value of the variable that would make any denominator zero. |
\(\dfrac{y}{y+6}=\dfrac{72}{(y-6)(y+6)}+4, y \neq 6, y \neq-6 \nonumber \) |
| Find the least common denominator. | The LCD is \((y-6)(y+6)\). |
| Clear the fractions. |
\((y-6)(y+6)\left(\dfrac{y}{y+6}\right)=(y-6)(y+6)\left(\dfrac{72}{(y-6)(y+6)}+4\right) \nonumber \) Note that the above equation has exactly the same solutions as the original except for, perhaps, \(y=-6\) and \(y=6\). |
| Simplify. | \((y-6) \cdot y=72+(y-6)(y+6) \cdot 4 \nonumber \) |
| Simplify. | \(y(y-6)=72+4\left(y^{2}-36\right) \nonumber \) |
| Solve the resulting equation. | \(\begin{aligned} y^{2}-6 y&=72+4 y^{2}-144\\ 0&=3 y^{2}+6 y-72 \\ 0&=3\left(y^{2}+2 y-24\right) \\ 0&=3(y+6)(y-4) \\ y&=-6, y=4 \end{aligned} \nonumber \) |
|
Check. \(y=-6\) is an extraneous solution. Check \(y=4\) in the original equation (not strictly necessary). |
\(\begin{aligned} \dfrac{y}{y+6} &=\dfrac{72}{y^{2}-36}+4 \\ \dfrac{4}{4+6} &\overset{?}{=}\dfrac{72}{4^{2}-36}+4 \\ \dfrac{4}{10} &\overset{?}{=} \dfrac{72}{-20}+4 \\ \dfrac{4}{10} &\overset{?}{=} -\dfrac{36}{10}+\dfrac{40}{10} \\ \dfrac{4}{10} &=\dfrac{4}{10} \surd \end{aligned} \nonumber \) |
| Conclude. | The solution is \(y=4\). |
Solve \(\dfrac{x}{x+4}=\dfrac{32}{x^{2}-16}+5. \nonumber \)
- Answer
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\(x=3\)
Solve \(\dfrac{y}{y+8}=\dfrac{128}{y^{2}-64}+9. \nonumber \)
- Answer
-
\(y=7\)
In some cases, all the algebraic solutions are extraneous.
Solve \(\dfrac{x}{2 x-2}-\dfrac{2}{3 x+3}=\dfrac{5 x^{2}-2 x+9}{12 x^{2}-12}. \nonumber \)
Solution
| \(\dfrac{x}{2 x-2}-\dfrac{2}{3 x+3}=\dfrac{5 x^{2}-2 x+9}{12 x^{2}-12} \nonumber \) | |
| We will start by factoring all denominators, to make it easier to identify extraneous solutions and the LCD. | \(\dfrac{x}{2(x-1)}-\dfrac{2}{3(x+1)}=\dfrac{5 x^{2}-2 x+9}{12(x-1)(x+1)} \nonumber \) |
| Note any value of the variable that would make any denominator zero. | \(\dfrac{x}{2(x-1)}-\dfrac{2}{3(x+1)}=\dfrac{5 x^{2}-2 x+9}{12(x-1)(x+1)}, x \neq 1, x \neq-1 \nonumber \) |
| Find the least common denominator. | The LCD is \(12(x-1)(x+1)\). |
| Clear the fractions. | \(12(x-1)(x+1)\left(\dfrac{x}{2(x-1)}-\dfrac{2}{3(x+1)}\right)=12(x-1)(x+1)\left(\dfrac{5 x^{2}-2 x+9}{12(x-1)(x+1)}\right) \nonumber \) |
| Simplify. | \(6(x+1) \cdot x-4(x-1) \cdot 2=5 x^{2}-2 x+9 \nonumber \) |
| Simplify. | \(6 x(x+1)-4 \cdot 2(x-1)=5 x^{2}-2 x+9 \nonumber \) |
| Solve the resulting equation. | \(\begin{aligned} 6 x^{2}+6 x-8 x+8&=5 x^{2}-2 x+9\\ x^{2}-1&=0 \\ (x-1)(x+1)&=0 \\ x&=1 \text { or } x=-1 \end{aligned} \nonumber \) |
| Check. | \(x=1\) and \(x=-1\) are extraneous solutions. |
| Conclude. | The equation has no solution. |
Solve \(\dfrac{y}{5 y-10}-\dfrac{5}{3 y+6}=\dfrac{2 y^{2}-19 y+54}{15 y^{2}-60}. \nonumber \)
- Answer
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There is no solution.
Solve \(\dfrac{z}{2 z+8}-\dfrac{3}{4 z-8}=\dfrac{3 z^{2}-16 z-16}{8 z^{2}+2 z-64}. \nonumber \)
- Answer
-
There is no solution.
Solve \(\dfrac{4}{3 x^{2}-10 x+3}+\dfrac{3}{3 x^{2}+2 x-1}=\dfrac{2}{x^{2}-2 x-3}. \nonumber \)
Solution
| \(\dfrac{4}{3 x^{2}-10 x+3}+\dfrac{3}{3 x^{2}+2 x-1}=\dfrac{2}{x^{2}-2 x-3} \nonumber \) | |
| Factor all the denominators, so we can note any value of the variable that would make any denominator zero. | \(\dfrac{4}{(3 x-1)(x-3)}+\dfrac{3}{(3 x-1)(x+1)}=\dfrac{2}{(x-3)(x+1)}, x \neq-1, x \neq \dfrac{1}{3}, x \neq 3\nonumber \) |
| Find the least common denominator. | The LCD is \((3 x-1)(x+1)(x-3)\). |
| Clear the fractions. |
\(\quad (3 x-1)(x+1)(x-3)\left(\dfrac{4}{(3 x-1)(x-3)}+\dfrac{3}{(3 x-1)(x+1)}\right)\) \(=(3 x-1)(x+1)(x-3)\left(\dfrac{2}{(x-3)(x+1)}\right) \nonumber \) |
| Simplify. | \(4(x+1)+3(x-3)=2(3 x-1) \nonumber \) |
| Distribute. | \(4 x+4+3 x-9=6 x-2 \nonumber \) |
| Simplify. |
\(7 x-5=6 x-2 \nonumber \) \(x=3 \nonumber \) But this is an extraneous solution. |
| Conclude. | There is no solution to this equation. |
Solve \(\dfrac{15}{x^{2}+x-6}-\dfrac{3}{x-2}=\dfrac{2}{x+3}. \nonumber \)
- Answer
-
There is no solution.
Solve \(\dfrac{5}{x^{2}+2 x-3}-\dfrac{3}{x^{2}+x-2}=\dfrac{1}{x^{2}+5 x+6}. \nonumber \)
- Answer
-
There is no solution.
Solve a Rational Equation for a Specific Variable (Optional)
When we solved linear equations, we learned how to solve a formula for a specific variable. Many formulas used in business, science, economics, and other fields use rational equations to model the relation between two or more variables. We will now see how to solve a rational equation for a specific variable.
When we developed the point-slope formula from our slope formula, we cleared the fractions by multiplying by the LCD.
\(\begin{aligned} m &=\frac{y-y_{1}}{x-x_{1}} \\ m\left(x-x_{1}\right) &=\left(\frac{y-y_{1}}{x-x_{1}}\right)\left(x-x_{1}\right) \quad \text{Multiply both sides of the equation by } x-x_1.\\ m\left(x-x_{1}\right) &=y-y_{1} \quad \text {Simplify.}\\ y-y_{1} &=m\left(x-x_{1}\right) \quad \text {Rewrite the equation with the y terms on the left.} \end{aligned} \nonumber \)
In the next example, we will use the same technique with the formula for slope that we used to get the point-slope form of an equation of a line through the point \((2,3)\). We will add one more step to solve for \(y\).
Solve \(m=\dfrac{y-2}{x-3}\) for \(y\).
Solution
| \(m=\dfrac{y-2}{x-3} \nonumber \) | |
| Note any value of the variable that would make any denominator zero. | \(m=\dfrac{y-2}{x-3}, x \neq 3 \nonumber \) |
| Clear the fractions by multiplying both sides of the equation by the LCD, \(x-3\). | \((x-3) m=(x-3)\left(\dfrac{y-2}{x-3}\right) \nonumber \) |
| Simplify. | \(x m-3 m=y-2 \nonumber \) |
| Isolate the term with \(y\). | \(x m-3 m+2=y \nonumber \) |
| Conclude. | \(y=x m-3 m+2\) |
Solve \(m=\dfrac{y-5}{x-4}\) for \(y\).
- Answer
-
\(y=m x-4 m+5\)
Solve \(m=\dfrac{y-1}{x+5}\) for \(y\).
- Answer
-
\(y=m x+5 m+1\)
Remember to multiply both sides by the LCD in the next example.
Solve \(\dfrac{1}{c}+\dfrac{1}{m}=1\) for \(c\).
Solution
| \(\dfrac{1}{c}+\dfrac{1}{m}=1 \text { for } c \nonumber \) | |
| Note any value of the variable that would make any denominator zero. | \(\dfrac{1}{c}+\dfrac{1}{m}=1, c \neq 0, m \neq 0 \nonumber \) |
| Clear the fractions by multiplying both sides of the equations by the LCD, \(cm\). | \(cm\left(\dfrac{1}{c}+\dfrac{1}{m}\right)=cm(1) \nonumber \) |
| Distribute. | \(cm\left(\frac{1}{c}\right)+cm \frac{1}{m}=cm(1) \nonumber \) |
| Simplify. | \(m+c=cm \nonumber \) |
| Collect the terms with \(c\) to the right. | \(m=cm-c \nonumber \) |
| Factor the expression on the right. | \(m=c(m-1) \nonumber \) |
| To isolate \(c\), divide both sides by \(m-1\). | \(\dfrac{m}{m-1}=\dfrac{c(m-1)}{m-1} \nonumber \) |
| Simplify by removing common factors. | \(\dfrac{m}{m-1}=c \nonumber \) |
| Conclude. | \(c=\dfrac{m}{m-1}\) |
Notice that even though we excluded \(c=0\) and \(m=0\) from the original equation, we must also now state that \(m \neq 1\).
Solve \(\dfrac{1}{a}+\dfrac{1}{b}=c\) for \(a\).
- Answer
-
\(a=\dfrac{b}{c b-1}\)
Solve \(\dfrac{2}{x}+\dfrac{1}{3}=\dfrac{1}{y}\) for \(y\).
- Answer
-
\(y=\dfrac{3 x}{x+6}\)
- What is the difference between an expression and an equation?
- What does it mean to solve an equation?
- Is it necessary to check your answer if you know you have not made a mistake? Explain.
Solve \(\dfrac{y}{y+3}+\dfrac{3}{y-3}=\dfrac{18}{y^2-9}\).
Key Concepts
Rational equation
Solultion to a rational equation
Extraneous solution to a rational equation
Solving a rational equation