How do you evaluate #\frac { 2} { x } - \frac { 5} { 2x + 1}#?

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Answer 1 · 2018-01-23T23:19:52

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

#(2)/(x) - (5)/(2x+1)#

#x# and #2x + 1# have no common factors.

to find a common denominator, both denominators must be multiplied together.

this gives:

#(2(2x+1))/(x(2x+1)) - (5x)/((2x+1)x)#

after expanding brackets:

#(4x+2)/(2x^2+x) - (5x)/(2x^2+x)#

#4x+2 - 5x = -x + 2#, or #2-x#

so the final fraction is:

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

Answer 2 · 2018-01-23T23:27:23

# 2/x - 5/(2x+1)#
= #(4x+2)/(2x^2+x) - (5x)/(2x^2+x)#
= #(4x+2-5x)/(2x^2+x)#
= #(2-x)/(2x^2+x)#
= #-(x-2)/(x(2x+1))#