Following rational equation has denominators that contain variables. For this equation,
$\frac{x-2}{4 x}+1=\frac{x+2}{x}$
a. What is/are the value or values of the variable that makes the denominators zero?
b. Solve the equation. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. The solution set is {}.
B. The solution set is {$x \mid x$ is a real number }.
C. The solution set is $\varnothing$.
Answer (1)
Find values of x that make the denominators zero: x = 0 .
Multiply both sides of the equation by the LCD, 4 x : x − 2 + 4 x = 4 ( x + 2 ) .
Simplify and solve for x : 5 x − 2 = 4 x + 8 , which gives x = 10 .
Check for extraneous solutions: x = 10 is not an extraneous solution, so the solution set is 10 .
Explanation
Identify values that make denominators zero First, let's identify the values of x that make the denominators zero. The denominators in the equation 4 x x − 2 + 1 = x x + 2 are 4 x and x .
Solve for x To find the values that make the denominators zero, we set each denominator equal to zero and solve for x . So, we have 4 x = 0 and x = 0 . In both cases, x = 0 . Therefore, x = 0 makes the denominators zero.
Multiply by LCD Now, let's solve the equation 4 x x − 2 + 1 = x x + 2 . To do this, we first multiply both sides of the equation by the least common denominator (LCD), which is 4 x . This gives us: 4 x ⋅ ( 4 x x − 2 + 1 ) = 4 x ⋅ x x + 2
Simplify the equation Distributing 4 x on the left side, we get: ( 4 x ⋅ 4 x x − 2 ) + ( 4 x ⋅ 1 ) = 4 x ⋅ x x + 2 Simplifying, we have: x − 2 + 4 x = 4 ( x + 2 )
Combine like terms Combining like terms on the left side, we get: 5 x − 2 = 4 ( x + 2 ) Expanding the right side, we have: 5 x − 2 = 4 x + 8
Solve for x Now, we solve for x . Subtracting 4 x from both sides, we get: 5 x − 4 x − 2 = 4 x − 4 x + 8 x − 2 = 8 Adding 2 to both sides, we get: x − 2 + 2 = 8 + 2 x = 10
Check for extraneous solutions We need to check if our solution x = 10 is an extraneous solution. Since x = 0 is the value that makes the denominator zero, and our solution is x = 10 , it is not an extraneous solution. Therefore, the solution set is 10 .
Examples
Rational equations are useful in various real-world scenarios, such as calculating work rates. For example, if one person can complete a task in x hours and another person can complete the same task in y hours, the rational equation x 1 + y 1 = t 1 can be used to find the time t it takes for them to complete the task together. Understanding how to solve rational equations allows us to solve problems involving rates, proportions, and inverse relationships, which are common in everyday life and various fields of study.
