(3k)/(2k-5)=7/3

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Solution for (3k)/(2k-5)=7/3 equation: 



(3k)/(2k-5)=7/3

We move all terms to the left:

(3k)/(2k-5)-(7/3)=0



Domain of the equation: (2k-5)!=0

We move all terms containing k to the left, all other terms to the right

2k!=5

k!=5/2

k!=2+1/2

k∈R



We add all the numbers together, and all the variables

3k/(2k-5)-(+7/3)=0

We get rid of parentheses

3k/(2k-5)-7/3=0

We calculate fractions


9k/(6k-15)+(-14k+35)/(6k-15)=0

We multiply all the terms by the denominator


9k+(-14k+35)=0

We get rid of parentheses

9k-14k+35=0

We add all the numbers together, and all the variables

-5k+35=0

We move all terms containing k to the left, all other terms to the right

-5k=-35

k=-35/-5

k=+7

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