Let’s solve the rational equation, step-by-step… 

simple inspection shows the denominator on the left would equal the two denominators on the right when the two are multiplied together. I.E. (3c)(c-2) = (3c^2 - 6c) Therefore... first step is to add the two algrebraic fractions on right side which yields the SAME denominator as left side. Next the same denominator for both sides is multiplied against the numerators on both sides. This eliminates this denominator on both sided. Then all common elements are collected together into an equation that = 0. Then that equation needs to be factored into: (c )(c ) = 0 Then inspection yields the roots of the solution FINALLY... Try each root in original equation to verify correct results..... with all of those denominators there's good chances a divide by 0 is possible. Divide by zero => means an invalid root. (-1-) 1/3c + 2/(c-2) = {(c-2)×1 + (3c)(2)}/(3c)(c-2) = {c -2 + 6c}}/(3c^2-6c) = (7c-2)/(3c^2-6c) now the equation is: 19/(3c^2-6c) = (7c-2)/(3c^2-6c) (19)-(7c-2)/(3c^2-6c) = 0 (21-7c)/(3c^2-6c) = 0

Love these I hate to say how old I am (retirement age) and finally have had the time to go back and learn the mechanics of working these equations. Love it

Nice explanation! Of a complex problem.

I used the Bow tie method on one side of the equation and ended up with c=3. But in general, is it wrong to do that or will you end up in trouble???

Thank you

c=3, but I'm sleepy, so it took about 20 seconds to do in my head. Sorry about that. Nice problem, though!! Keep it up! This is good practice for student seeking to improve their math solving abilities.

Let's get it let's go 🚀🥳🤸💯💫🤧

My trouble with these problems is copying them down wrong. C=3

C=3

Kewl