Sunday, July 14, 2013

# How to solve an equation that has a variable in the denominator?

2:20 AM
The equations that have variables in the denominator are termed as rational equations. Even though this may be a new name for you, it uses the same concepts used in solving regular algebra equations.
Here are some e xamples of rational equations:
$\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \frac{3}{x}=x+4$
$\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \frac{5+y}{y}=8$

To solve rational equations, you should first convert them into a standard linear or quadratic form. This can be done using one or more of the following steps. Usually the simpler rational equations may not involve all of these steps:
1. Isolate the terms with variable on one side and constants on the other side
2. Add or subtract terms with variable
3. Cross-multiply both sides to get a standard linear or quadratic equation
Let’s take an example involving a simple rational equation that converts to a linear form:
Example:

What is the value of  $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} z$ if $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \frac{3z}{5-z}=12$?
A) 1
B) 2
C) 3
D) 4
E) 5
Solution:

Step 1: Here we see that all variable terms are already on one side. Therefore we first use the cross-multiplication to simplify the equation:
\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \begin {align*} \frac{3z}{5-z}&=12 \\ 3z&=12\times (5-z) \end{align*}
Step 2: Next we get rid of the parentheses:
\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \begin {align*} 3z&=60-12z \\ \end{align*}
Step 3: Finally isolate the variable to solve the equation
\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \begin {align*} 3z&=60-12z \\ 15z&=60 \\ z&=\frac{60}{15} \\ z&=4 \\ \end{align*}
D is the correct answer.

Sometimes you can also simplify the equation into a linear or quadratic form a lot quicker by multiplying both sides of the equation with the denominator containing the variable term. Let’s look at a grid-in example:
Example:

If  $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \frac{2}{x} - 3 = -x$ what is a possible value of $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} x$?
Solution:

This question can be easily solved by plugging in the values if you have the luxury of answer choices. But alas! This seems to be a fill in the blank type of question. Sadly you will have to solve this question using math concepts.
Step 1: To solve the question we need to eliminate the $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} x$ in the denominator. This can be done by multiplying both sides of the equation by $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} x$
$\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} 2-3x=-x^2 \Rightarrow x^2-3x+2=0$
Step 2: The next step is to factorize the quadratic equation:
$\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \Rightarrow x^2 - x - 2x + 2 = 0$
$\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \Rightarrow (x-1)(x-2)=0$
$\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \Rightarrow x=1, x=2$
Either 1 or 2 could be the correct answer.