(k-5)/(2k+3)=4/6

Solution for (k-5)/(2k+3)=4/6 equation:

(k-5)/(2k+3)=4/6

We move all terms to the left:

(k-5)/(2k+3)-(4/6)=0


Domain of the equation: (2k+3)!=0

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

2k!=-3

k!=-3/2

k!=-1+1/2

k∈R


We add all the numbers together, and all the variables

(k-5)/(2k+3)-(+4/6)=0

We get rid of parentheses

(k-5)/(2k+3)-4/6=0

We calculate fractions


(6k-30)/(12k+18)+(-8k-12)/(12k+18)=0

We multiply all the terms by the denominator


(6k-30)+(-8k-12)=0

We get rid of parentheses

6k-8k-30-12=0

We add all the numbers together, and all the variables

-2k-42=0

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

-2k=42

k=42/-2

k=-21