Question

Solve 6 over x minus 6 equals the quantity of x over x minus 6, minus six halves. for x and determine if the solution is extraneous or not.

Answer 1

First, let us express it into a mathematical equation:

$\frac{6}{x} -6=( \frac{x}{x-6} )- \frac{6}{2}$

Then we simplify the equation by combining like terms

$\frac{6}{2} -6=( \frac{x}{x-6} )- \frac{6}{x}$

$-3 = \frac{ x^{2} -6(x-6)}{x(x-6)}$

$-3 = \frac{ x^{2} -6x+36}{x(x-6)}$

$-3 = \frac{ x^{2} -6x+36}{x(x-6)}$

$-3x(x-6) = x^{2} -6x+36$

$-3 x^{2} +18x = x^{2} -6x+36$

$-3 x^{2} - x^{2} +6x-36+18x =0$

$-4 x^{2} +24x-36 =0$

$(-4 x^{2} +24x-36 =0)/-4$

$x^{2}-6x+9 =0$

$(x-3)^{2} =0$

x = 3

There is a solution so it is not extraneous.