SOLUTION: Will someone please tell me if my calculation are corrct for the following problem? Thank you. (3x+3w+bx+bw)/(x^2-w^2 )�(9-b^2)/(6-2b) =(3(x+w)+ b(x+w))/((x+w)(x-w)) � ((3+b)(3

Question 292546: Will someone please tell me if my calculation are corrct for the following problem? Thank you.
(3x+3w+bx+bw)/(x^2-w^2 )�(9-b^2)/(6-2b)
=(3(x+w)+ b(x+w))/((x+w)(x-w)) � ((3+b)(3-b))/(2(3-b))
= (3+b)/(x-w) � (3+b)/2
= (6+2b)/(3x+xb-3w-bw)
= (2(3+b))/(x(3+b) � w(3-b))
= 2/((x-w)(3-b))

Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
2/(x-w)
You got lost after (3+b)/(x-w) � (3+b)/2
Remember that dividing is the same as multiplying by the reciprocal.
so we flip the second part and multiply
(3+b)/(x-w) *2/(3+b)
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