How do you simplify #(2)/(x) + (2)/(x-1) - (2)/(x-2)#?

1 Answer
Jun 19, 2018

Answer:

See a solution process below:

Explanation:

To add fractions the fractions must be over a common denominator. First, we need to multiply each fraction by the appropriate form of #1# to put them over a common denominator of:

#x(x - 1)(x - 2)#

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

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

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

Next, we can expand the numerators for each fraction:

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

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

#(2x^2 - 6x + 4)/(x(x - 1)(x - 2)) + (2x^2 - 4x)/(x(x - 1)(x - 2)) - (2x^2 - 2x)/(x(x - 1)(x - 2))#

Then, we can add the numerators over the common denominator:

#((2x^2 - 6x + 4) + (2x^2 - 4x) - (2x^2 - 2x))/(x(x - 1)(x - 2)) =>#

#((2x^2 - 6x + 4 + 2x^2 - 4x - 2x^2 + 2x))/(x(x - 1)(x - 2)) =>#

#((2x^2 + 2x^2 - 2x^2 - 6x - 4x + 2x + 4))/(x(x - 1)(x - 2)) =>#

#((2 + 2 - 2)x^2 + (-6 - 4 + 2)x + 4)/(x(x - 1)(x - 2)) =>#

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

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

If you need to simplify the denominator it would be:

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

#(2x^2 - 8x + 4)/(x^3 - 2x^2 - x^2 + 2x ) =>#

#(2x^2 - 8x + 4)/(x^3 + (-2 - 1)x^2 + 2x ) =>#

#(2x^2 - 8x + 4)/(x^3 + (-3)x^2 + 2x ) =>#

#(2x^2 - 8x + 4)/(x^3 - 3x^2 + 2x )#