How do you simplify \frac { 4} { 6b } + \frac { 3} { 3b - 4}46b+33b4?

2 Answers
Oct 11, 2017

(15b-8)/(3b(3b-4))15b83b(3b4)

Explanation:

Since LCM(6b;3b-4)=6b(3b-4)LCM(6b;3b4)=6b(3b4), 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))

Oct 11, 2017

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

Explanation:

\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}