How do you add #\frac { 5} { 6u x ^ { 2} } + \frac { 3} { 2u ^ { 2} x }?#

2 Answers
harsh s. · EZ as pi
Aug 16, 2017

see below

Explanation:

#\frac { 5} { 6u x ^ { 2} } + \frac { 3} { 2u ^ { 2} x }" "larr LCM = 6u^2x^2#

#=>(\frac { 5} { 6u x ^ { 2} }*1) + (\frac { 3} { 2u ^ { 2} x }*1)#

#=>(\frac { 5} { 6u x ^ { 2} }*u/u) + (\frac { 3} { 2u ^ { 2} x }*3/3*x/x)#

#=>(5u)/(6u^2x^2)+(9x)/(6u^2x^2)#

#=>(5u+9x)/(6u^2x^2)#

Jumbotron
Aug 22, 2017

#(5u + 9x)/(6u²x²)#

Explanation:

Also by using this method as well..

#5/(6ux^2) + 3/(2u^2x)#

Multiplying both terms by the LCM added with all over the LCM

L.C.M = 6u²x²

e.g #rArr (a xx LCM + b xx LCM)/(LCM)#

#rArr (5/(6ux^2) xx 6u²x² + 3/(2u^2x) xx 6u²x² )/(6u²x²)#

#rArr (5/(cancel6cancelu cancel(x^2)) xx cancel6cancel(u^2)^1 cancel(x²) + 3/(cancel2cancel(u^2)cancelx) xx cancel6^3 cancel(u²) cancel(x²)^1 )/(6u²x²)#

#rArr (5 xx u + 3 xx 3 xx x)/(6u²x²)#

#rArr (5u + 9x)/(6u²x²)#

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