How do you evaluate #\frac { 7} { x ^ { 2} } + \frac { 1} { x } - \frac { 5} { x ^ { 2} } + \frac { 3} { x }#?

Question

534 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-09T03:20:29

#=>\frac {2+4x} {x ^2} #

#\frac { 7} { x ^ { 2} } + \frac { 1} { x } - \frac { 5} { x ^ { 2} } + \frac { 3} { x }#

#=>\frac{7}{x^2}-\frac{5}{x^2}+\frac{1}{x}+\frac{3}{x}# -----grouping like terms.

#=>\frac {7-5} {x ^2} + \frac {1+3} {x} #

#=>\frac {2} {x ^2} + \frac {4} {x} #

#=> \frac {2} {x ^2} + \frac {4} {x}(x/x) # ------ making denominators equal

#=>\frac {2} {x ^2} + \frac {4x} {x^2} #

#=>\frac {2+4x} {x ^2} #