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

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Answer 1 · 2017-10-11T04:28:10

$\frac{15 b - 8}{3 b (3 b - 4)}$ $(15b-8)/(3b(3b-4))$

Since $L C M (6 b; 3 b - 4) = 6 b (3 b - 4)$ $LCM(6b;3b-4)=6b(3b-4)$ , 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:

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

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

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