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QuestionSolve the equation 2xx+4−8x−4=2x2+32x2−16\frac{2 x}{x+4}-\frac{8}{x-4}=\frac{2 x^{2}+32}{x^{2}-16}. Is it an identity, conditional, or inconsistent?

Studdy Solution

STEP 1

Assumptions1. The given equation is xx+4−8x−4=x+32x−16\frac{x}{x+4}-\frac{8}{x-4}=\frac{x^{}+32}{x^{}-16} . We are asked to solve for x and determine the nature of the equation (identity, conditional, or inconsistent)

STEP 2

First, we need to simplify the equation. Notice that the denominator of the right side of the equation, x2−16x^{2}-16, can be factored as (x+4)(x−4)(x+4)(x-4).
2x2+32x2−16=2x2+32(x+4)(x−4)\frac{2x^{2}+32}{x^{2}-16} = \frac{2x^{2}+32}{(x+4)(x-4)}

STEP 3

We can also factor out a2 from the numerator of the right side of the equation.
2x2+32(x+)(x−)=2(x2+16)(x+)(x−)\frac{2x^{2}+32}{(x+)(x-)} = \frac{2(x^{2}+16)}{(x+)(x-)}

STEP 4

Now, let's rewrite the left side of the equation with a common denominator of (x+4)(x−4)(x+4)(x-4).
2xx+4−8x−4=2x(x−4)(x+4)(x−4)−8(x+4)(x+4)(x−4)\frac{2x}{x+4}-\frac{8}{x-4} = \frac{2x(x-4)}{(x+4)(x-4)} - \frac{8(x+4)}{(x+4)(x-4)}

STEP 5

implify the left side of the equation.
2x(x−4)(x+4)(x−4)−8(x+4)(x+4)(x−4)=2x2−8x−32(x+4)(x−4)\frac{2x(x-4)}{(x+4)(x-4)} - \frac{8(x+4)}{(x+4)(x-4)} = \frac{2x^{2}-8x-32}{(x+4)(x-4)}

STEP 6

Now we have the equation in the form2x2−8x−32(x+4)(x−4)=2(x2+16)(x+4)(x−4)\frac{2x^{2}-8x-32}{(x+4)(x-4)} = \frac{2(x^{2}+16)}{(x+4)(x-4)}

STEP 7

Since the denominators are the same, we can set the numerators equal to each other and solve for x.
2x2−x−32=2(x2+16)2x^{2}-x-32 =2(x^{2}+16)

STEP 8

Expand the right side of the equation.
2x2−8x−32=2x2+322x^{2}-8x-32 =2x^{2}+32

STEP 9

Subtract 2x22x^{2} from both sides of the equation.
−8x−32=32-8x-32 =32

STEP 10

Add32 to both sides of the equation.
−8x=64-8x =64

STEP 11

Divide both sides of the equation by -8.
x=−8x = -8The solution set is {−8}\{-8\}.

STEP 12

Now, let's check if the equation is an identity, a conditional equation, or an inconsistent equation. An identity is true for all values of x, a conditional equation is true for some values of x, and an inconsistent equation is never true.
Since our equation is only true for x=−8x = -8, it is a conditional equation.

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