Dr. Lalitha Subramanian
About the Author
Aksharam
Equations involving rational expressions are called rational equations. These are of the form $\frac{p(x)}{q(x)}$, where $p(x)$ and $q(x)$ are polynomials. A rational equation may contain one or more rational expressions, either on one side of on both sides of the equality sign. To solve such equations, we first find the LCD of denominators of all the rational expressions in the equation, and then multiply each term on both side by that LCD. This will help eliminate the denominators and form a linear or higher degree equation, which can then be solved.
Whenever we multiply or divide an equation by an expression containing variables, the resulting equation may have solutions that are not actually solutions of the original equations. Such solutions are called extraneous solutions. For this reason, when we solve rational equation, we must always check the resulting solutions by substituting each value in the original equations in order to spot any extraneous solutions, and exclude these from the solution set.
Solve: $\frac{6}{x}+x=5$
Solution:LCD of all terms is $x$. Multiplying each term on both sides by $x$: \begin{eqnarray} x \ast \frac{6}{x} + x \ast x &=& x \ast 5\\ x^2+6 &=& 5x\\ x^2-5x+6 &=& 0\\ (x-2)(x-3) &=& 0\\ x = 2 \quad or \quad x = 3\\ \end{eqnarray} Check these values on the original equation: \begin{eqnarray} x = 2 \Rightarrow \frac{6}{2} + 2 = 3 + 2 = 5\\ x = 3 \Rightarrow \frac{6}{3} + 3 = 5\\ \end{eqnarray} So, there are no extraneous values. solution set is ${2, 3}$.
Solve: $ x + \frac{3}{x-1} = 5$
Solution:LCD of all terms on both sides is $(x-1)$. Multiplying each term on both sides by $(x-1)$: \begin{eqnarray} (x-1) \ast x + (x-1) \ast \frac{3}{(x-1)} &=& (x-1) \ast 5\\ x^2-x + 3 &=& 5x - 5\\ x^2 - 6x + 8 &=& 0\\ (x-4)(x-2) &=& 0\\ x = 4 \quad or \quad x = 2\\ \end{eqnarray} checking for extraneous solutions: \begin{eqnarray} x = 4 \Rightarrow 4 + \frac{3}{3} = 4 + 1 = 5\\ x = 2 \Rightarrow 2 + \frac{3}{1} = 2 + 3 = 5\\ \end{eqnarray} No extraneous solutions. So the solution set is ${2, 4}$.
Solve: $2 - \frac{1}{x+1} = \frac{1}{x^2+x}$
Solution:Rewriting the equation with factorized denominators would give $2 - \frac{1}{x+1} = \frac{1}{x(x+1)}$ LCD is $x(x+1)$. Multiplying each term on both sides by this LCD and solving the resulting equations: \begin{eqnarray} 2 \ast x(x+1) - 1 \ast x &=& 1\\ 2x^2 + 2x - x -1 &=& 0\\ 2x(x+1) - 1(x + 1) &=& 0\\ (2x-1)(x+1) &=& 0\\ x = -1 \quad or \quad x = \frac{1}{2}\\ \end{eqnarray} Checking for extraneous solutions: \begin{eqnarray} x = -1 \Rightarrow LHS &=& 2 - \frac{1}{0}\\ RHS &=& \frac{1}{0}\\ x = \frac{1}{2} \Rightarrow LHS &=& 2 - \frac{2}{3} &=& \frac{4}{3}\\ RHS &=& \frac{1}{\left(\frac{1}{4}+\frac{1}{2}\right)} &=& \frac{4}{3}\\ \end{eqnarray} We find that the value $x = -1$ results in a meaningless expression. So, $x = -1$ is an extraneous solution. Hence the only solution for this equation is $x=\frac{1}{2}$.
Solve: $\frac{3x}{x+5} - \frac{7}{x^2+3x-10} = \frac{-1}{x-2}$
Solution:Rewriting the equation with factored denominators would give $\frac{3x}{x+5} - \frac{7}{(x+5)(x-2)} = \frac{-1}{x-2}$.
Multiplying each term on both sides by the LCD and simplifying would result in $3x(x-2) -7 = -1(x+5)$ Simplifying and solving: \begin{eqnarray} 3x^2 - 6x - 7 &=&
-x-5\\ 3x^2 -5x - 2 &=& 0\\ (x-2)(3x+1) &=& 0\\ x = 2 \quad or \quad x = -\frac{1}{3}\\ \end{eqnarray}
Checking $x = 2$ in the original equation results in a meaningless expression. So, this is an extraneous solution. Checking $x = -\frac{1}{3}$, we get:
\begin{eqnarray} LHS \rightarrow -\frac{3}{14} + \frac{9}{14} &=& \frac{3}{7}\\ RHS \rightarrow \frac{6}{14} &=& \frac{3}{7}\\ \end{eqnarray}
Answer: Solution is $x = \frac{3}{7}$
Solve: $\frac{3}{x+2} + \frac{6}{x^2+2x} = \frac{3-x}{x}$
Solution: The denominators of this equations are already factored, and we note that the LCD is $x(x+2)$. Multiplying each term on both sides by the LCD and solving, we get: \begin{eqnarray} 3x + 6 &=& (3-x)(x+2)\\ 3x+6 &=& 3x + 6 - x^2 - 2x\\ 2x+x^2 &=& 0\\ x(x+2) &=& 0\\ x &=& 0 \quad or \quad x = -2\\ \end{eqnarray}
$x = \frac{3}{2}$ or $x = \frac{2}{4}$
${-1, \frac{1}{2}}$. $x = -1$ is an extraneous solution. So, final solution is $x = \frac{1}{2}$.
$x =-6$ or $x = -1$
$x = \frac{-3}{4}$
$x = -4 \pm \sqrt{13}$
$x = -4$ or $x = 1$
$x = \frac{1 \pm \sqrt{13}}{6}$
$x = 2$; $x = -1 \pm \sqrt{5}$
$x = -2$; $x = -1$; $x = 3$
No Solution.
$x = -3$; $x = 4$
$x = \frac{2-y}{y-1}$
$x = \frac{3-2y}{2-y}$
$x = -\frac{1}{2}$; $x = 0$
$x = 1$