How do you solve #x/(x-2)+1/(x-4)=2/(x^2-6x+8)# and check for extraneous solutions?

1 Answer
Aug 26, 2016

Answer:

#x = -1#

Explanation:

We have an equation with fractions. This means we can get rid of the denominators.

Factor the right side denominator first.
#x^2-6x+8 = (x-4)(x-2)#

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

#x !=4, x!=2#

To make it fit on one line, let #color(red)((x-4) = a) and color(blue)((x-2) = b)#

The LCM = #color(red)((x-4)color(blue)((x-2)) = color(red)(a)) xx color(blue)(b)#

Multiplying all the terms by this LCM .

#(color(red)(a) xx color(blue)(cancelb)xx x)/(cancel(x-2)) + (color(red)(cancela) xx color(blue)(b)xx1)/(cancel(x-4)) = (color(red)(cancela) xx color(blue)(cancelb)xx2)/(cancel(x-2)cancel(x-4))#

Much better! #x(x-4) + x-2 = 2#

#x^2 -4x +x -2 = 2" make" = 0#

#x^2 -3x-4 = 0" find factors"#

#(x-4)(x+1) = 0#

#x = 4, or x = -1#

However, #x=4# is not permissible.

#x = -1#