How do you solve # (x-4)/(x-2) = (x-2)/(x+2) + (1)/(x-2)#?

Question

Eva Adamson · Accepted Answer

#x-4 = ((x-2)(x-2))/(x+2) + 1#
#(x-4)(x+2) = (x-2)(x-2) + (x+2)#
#x^2 - 2x - 8 = x^2-4x+4 + x + 2#

Simplify
#2x - 12 = x + 2#
#x = 14# The easiest way to do this type of calculation is to get rid of all the denominators. To do this, first times every single term by #x-2#, then do the same, but with #x+2#. At that point, it becomes a matter of simplification.

Jameson Allen · Answer

undefined To solve the equation (x-4)/(x-2) = (x-2)/(x+2) + (1)/(x-2), we can start by simplifying both sides of the equation.

First, we can simplify the right side by finding a common denominator for (x-2)/(x+2) and (1)/(x-2). The common denominator is (x+2)(x-2).

Next, we can rewrite the equation as follows:

(x-4)/(x-2) = [(x-2)(x+2) + 1]/[(x+2)(x-2)].

Expanding the numerator on the right side, we get:

(x-4)/(x-2) = (x^2 - 4 + 1)/(x^2 - 4).

Simplifying further, we have:

(x-4)/(x-2) = (x^2 - 3)/(x^2 - 4).

To eliminate the fractions, we can cross-multiply:

(x-4)(x^2 - 4) = (x-2)(x^2 - 3).

Expanding both sides, we get:

x^3 - 4x - 4x^2 + 16 = x^3 - 3x^2 - 2x^2 + 6.

Combining like terms, we have:

x^3 - 4x - 4x^2 + 16 = x^3 - 5x^2 + 6.

Rearranging the equation, we get:

x^3 - x^3 - 4x^2 + 5x^2 - 4x - 6 + 16 = 0.

Simplifying further, we have:

x^2 - 4x + 10 = 0.

At this point, we can either factor the quadratic equation or use the quadratic formula to find the solutions for x.

Factoring the quadratic equation, we have:

(x - 2)(x - 2) + 6 = 0.

Simplifying, we get:

(x - 2)^2 + 6 = 0.

Since (x - 2)^2 is always non-negative, there are no real solutions for x in this case.

Therefore, the equation (x-4)/(x-2) = (x-2)/(x+2) + (1)/(x-2) has no real solutions.