At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
[3/(x-1)] - [(2x+10)/(x^2+2x-3)] = 1/3
what is x?
this question is solved yours other subtittle.
we say x=0
lcd = 3(x^2 +2x-3) = 3(x-1)(x+3) multiplying through by this gives: 9(x+3) - 3(2x+10) = (x-1)(x+3) 9x +27 - 6x - 30 = x^2 +2x - 3 x^2 +x = 0 x(x+1)=0 x = 0 or x = -1
yeah cnsu90, thanks for ur help :)
oops look like an error in last two steps! 9x +27 - 6x - 30 = x^2 +2x - 3 -x^2 +x = 0 x (1 - x) = 0 x = 0 or x = 1
@jimmy..denominator cant be 0..so u cant say x=1..
yes - you are right
jimmy, whats lcd? lowest common denominator?
this is given me a real headache - cant understand how i got erroneous root
the anwser is x=0
yes - i know thats correct - i just trying to find out why my method also gave x=1
your solution is true jimmy.but then you must try your root in function.
there must be some reason for this
at x=1 function is undefined.
yea i know
- i just dont like mysteries...
ahh - i'd forgot a very important fact about multiplying the whole equation by the lcd!! - this will sometimes give extraneous roots - when x = 1 you are effectively multiplying by zero. you guys solved it differently - you simplified the lhs and left rhs alone - thats why you didn't get the extra root
its ok when you multiply by a constant - it only applies when you you multiply by a variable
hey jimmy, so which method is better? :)