Solving Rational Equation

Introduction

Sometimes in math, we come across equations with fractions in them. These are called rational equations. A rational equation is just an equation that has one or more fractions with variables in the denominators.

To solve a rational equation, we want to get rid of the fractions to make it easier to work with. We do this by finding something called the Least Common Denominator (LCD) - that's the smallest number that all the denominators can divide into. Then, we multiply every part of the equation by the LCD. This makes the fractions disappear!

After that, we solve the new equation just like any other equation. But there's one last important step: check your answers! Sometimes, your solutions might make a denominator in the original equation equal to zero - and can't divide by zero! If that happens, we call that extraneous solution, and we have to leave it out.

Solve for x: \bf \dfrac{1}{6x^2} = \dfrac{1}{3x^2} - \dfrac{1}{x}

Step q: Identify the LCD of 6x^2, \space 3x^2, \space \text{ and } x \text{ which is } 6x^2.

Step 2: Multiply both sides of the equations by 6x^2 and simplify to solve for x .

Step 3: Check for extraneous solutions, if it exists.

To verify if x= -\dfrac{1}{6} is a solution to the rational equation, we will substitute it the LCD, and the result must not be zero.

Since \dfrac{1}{6} \neq 0, \text{ then } x= -\dfrac{1}{6} is not an extraneous solution, so it is valid.

Final Answer: \boxed{x=-\dfrac{1}{6}}

Solve for x: \bf \dfrac{1}{x-1}= \dfrac{3}{(x-1)(x+1)}

Step 1: Identify the LCD of x +1 \text{ and } (x-1(x+1) \text{ which is } (x-1)(x+1).

Step 2: Multiply both sides of the equation by (x-1)(x+1) and simplify to solve for x.

Step 3: Check for extraneous solutions, if it exists.

To verify if x=4 is a solution to the rational equation, we will substitute it to the LCD, and the result must be zero.

Since 15 \neq 0, then x=4 is not an extraneous solution, so it is valid.

Final Answer: \boxed{x=4}

Solve for x: \bf \dfrac{1}{x^2} + \dfrac{1}{x} = \dfrac{1}{2x^2}

Step 1: Identify the LCD of x62, \space x, \text{ and } 2x^2, \text{ which is } 2x^2.

Step 2: Multiply both sides of the equation by 2x^2 and simplify solve for x .

Step 3: Check for extraneous solutions, if it exists.

To verify if x= -\dfrac{1}{2} is a solution to the rational equation, we will substitute it to the LCD, and the result must not be zero.

Since \dfrac{1}{2} \neq 0, \text{ then } x= -\dfrac{1}{2} is not an extraneous solutions, so it is valid.

Final Answer: \boxed{x=-\dfrac{1}{2}}

Solve for x: \bf \dfrac{1}{15} + \dfrac{1}{20} = \dfrac{1}{x}

Step 1: Identify the LCD of 15, 20, and x which is 60x.

Step 2: Multiply both sides of the equation by 60x and simplify to solve x.

Step 3: Check for extraneous solutions, if it exists.

To verify if x=\dfrac{60}{7} is a solution to the rational equation, we will substitute it to the LCD, and the result must not be zero.

Since \dfrac{3600}{7} \neq 0, \text{ then } x = \dfrac{60}{7} is not an extraneous solution, so it is valid.

Final Answer: \boxed{x= \dfrac{60}{7}}

Solve for x: \bf \dfrac{200}{x+5}=\dfrac{150}{x-5}

Step 1: Identify the LCD of x+5 \text{ and }x-5 which is (x+5)(x-5).

Step 2: Multiply both sides of the equation by (x+5)(x-5) and simplify to solve for x.

Step 3: Check for extraneous solutions, if it exists.

To verify is x = 35 is a solution to the rational equation, we will substitute it to the LCD, and the result must not be zero.

Since 1200\neq0. \text{ then } x=35 is not an extraneous solution, so it is valid.

Final Answer: \boxed{x=35}

Solve for x: \bf \dfrac{1}{x-2} + \dfrac{1}{(x-2)(x+5)} = \dfrac{6}{x-2}

Step 1: Identify the LCD of x-2 \text{ and } (x-2)(x+5) \text{ which is } (x-2)(x+5).

Step 2: Multiply both sides of the equation by (x-2)(x+5) and simplify to solve for x.

Step 3: Check for extraneous solutions, if it exists.

To verify if x=-\dfrac{24}{5} is a solution to the rational equation, we will substitute it to the LCD, and the result must not be zero.

Since -\dfrac{34}{25}\neq 0, \text{ then } x=-\dfrac{24}{5} is not an extraneous solution, so it is valid.

Final Answer: \boxed{x=-\dfrac{24}{5}}