How do you simplify #\frac { 4} { 6b } + \frac { 3} { 3b - 4}#?

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Answer 1 · 2017-10-11T04:28:10

#(15b-8)/(3b(3b-4))#

Since #LCM(6b;3b-4)=6b(3b-4)#, the sum is:

#(4(3b-4)+3(6b))/(6b(3b-4))=(12b-16+18b)/(6b(3b-4))=(30b-16)/(6b(3b-4))=(cancel2(15b-8))/(cancel6^3b(3b-4))=(15b-8)/(3b(3b-4))#

Answer 2 · 2017-10-11T04:36:15

=#\frac {15b-8} {9b^2-12b} #

#\frac { 4} { 6b } + \frac { 3} { 3b - 4}#

We need to write the expression with common denominator, so we will first make the denominators equal:

=#\frac { 4} { 6b }\times ((3b-4)/(3b-4))+ \frac { 3} { 3b - 4}\times((6b)/(6b))#

=#\frac {12b-16} { (6b)(3b-4) } + \frac {18b} { (3b - 4)(6b)}#

=#\frac {12b-16 +18b} { (6b)(3b-4) } #

=#\frac {30b-16} { (6b)(3b-4) } #

=#\frac {30b-16} { (18b^2-24b) } #

We can take 2 common from numerator and denominator:

=#\frac {cancel2(15b-8)} { cancel2(9b^2-12b) } #

=#\frac {15b-8} {9b^2-12b} #