How do you solve #\frac { x + 1} { x - 1} + \frac { 2} { x } = \frac { x } { x + 1}#?

1 Answer
George C.
Aug 5, 2017

#x = -1/10+-sqrt(41)/10#

Explanation:

Given:

#(x+1)/(x-1)+2/x=x/(x+1)#

Multiply both sides of the equation by #x(x-1)(x+1)# to get:

#x(x+1)^2+2(x-1)(x+1)=x^2(x-1)#

Multiply out to get:

#x^3+2x^2+x+2x^2-2=x^3-x^2#

Add #x^2-x^3# to both sides and combine terms to get:

#5x^2+x-2 = 0#

This is in standard quadratic form:

#ax^2+bx+c = 0#

with #a=5#, #b=1# and #c=-2#

It has solutions given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(x) = (-1+-sqrt(1^2-4(5)(-2)))/(2*5)#

#color(white)(x) = (-1+-sqrt(41))/10#

#color(white)(x) = -1/10+-sqrt(41)/10#

Finally note that these are valid solutions of the original rational equation since neither of them causes any denominator to be zero.

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