$\frac{4}{x+16}-\frac{3}{x-16}=\frac{5 x}{x^2-256}$
a. Write the value or values of the variable that make a denominator zero.
$x = \square$ -16,16 (Use a comma to separate answers as needed.)
b. What is the solution of the equation? Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. The solution set is $\{-4\}$. (Use a comma to separate answers as needed.)
$\square$
B. The solution set is $\{x \mid x$ is a real number $\}$.
C. The solution set is $\varnothing$.
Answer (2)
Find the values of x that make the denominators zero: x = − 16 , 16 .
Multiply both sides of the equation by the LCD, ( x + 16 ) ( x − 16 ) , to eliminate the denominators.
Simplify the equation and solve for x , which gives x = − 28 .
Check for extraneous solutions. Since x = − 28 is not equal to 16 or − 16 , the solution is valid: − 28 .
Explanation
Problem Analysis We are given the equation x + 16 4 − x − 16 3 = x 2 − 256 5 x . We need to find the values of x that make the denominator zero and then solve the equation for x .
Finding Values That Make Denominators Zero First, let's find the values of x that make the denominators zero. The denominators are x + 16 , x − 16 , and x 2 − 256 . Setting each equal to zero, we have: x + 16 = 0 ⇒ x = − 16 x − 16 = 0 ⇒ x = 16 x 2 − 256 = 0 ⇒ x 2 = 256 ⇒ x = ± 16 So, the values of x that make the denominators zero are − 16 and 16 .
Solving the Equation Now, let's solve the equation. The least common denominator (LCD) is ( x + 16 ) ( x − 16 ) = x 2 − 256 . Multiply both sides of the equation by the LCD: ( x + 16 ) ( x − 16 ) ( x + 16 4 − x − 16 3 ) = ( x + 16 ) ( x − 16 ) ( x 2 − 256 5 x ) 4 ( x − 16 ) − 3 ( x + 16 ) = 5 x 4 x − 64 − 3 x − 48 = 5 x x − 112 = 5 x − 112 = 4 x x = − 28
Checking for Extraneous Solutions Now, we need to check for extraneous solutions. We found that x = − 28 . Since − 28 is not equal to 16 or − 16 , it is not an extraneous solution. Therefore, the solution set is { − 28 } .
Final Answer The solution to the equation is x = − 28 .
Examples
Rational equations are used in many real-world applications, such as calculating the time it takes to complete a task when multiple people are working together, or determining the concentration of a substance in a mixture. For example, if two pipes are filling a tank, and one pipe can fill the tank in x hours and the other pipe can fill the tank in y hours, the equation x 1 + y 1 = t 1 can be used to find the time t it takes for both pipes to fill the tank together. Solving such equations helps in optimizing processes and resource allocation.
The values of x that make the denominators zero are − 16 and 16 . The solution to the given equation is x = − 28 , which is valid and does not make any denominator zero.
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