Jesset C.

asked • 03/11/20

Can anyone please explain how I got this question wrong?

Simplify: (xb-3y3b)/(xb/3yb+3)


(A) x(4/3)(b-3)y4b-3

(B) x(3b-9)/(b)y(3b)/(b+3)

(C) (xyb^2 -9)b^2

(D) x-3+(2b)/(3)y-3+2

(E) None of these


For this I got; (E) None of these, but the answer is incorrect. How did I get this wrong?



*Edit/Update:* Choice (D) did have a b in the end, I just didn’t see it.


Al P.

I suspect answer D is what they expect, but they likely have omitted a final 'b' (the exponent of y is shown as "-3+2" but was probably intended to be printed as "-3+2b"). Just a guess (who writes an exponent as -3+2 without simplifying it to -1?) but thats how it looks to me.
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03/11/20

2 Answers By Expert Tutors

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Arturo O. answered • 03/12/20

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By the rules of exponentiation,


(xb-3y3b) / (xb/3yb+3) = xb-3-b/3 y3b-(b+3) = x(2/3)b-3 y2b-3


I agree with you and with John M. that none of the choices are correct. What did you get as the solution? Where did this problem come from? There could be typos in the original problem or in the choices.

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Edward A.

Arturo, Looks like answer D would be accurate if the exponent of y were “-3+2b” instead of “-3+2”, so I think you’ve got it: one mistaken choice in the problem statement.
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03/12/20

Arturo O.

I agree with you, Edward. Answer (D) happens to have the correct exponent for x, so it must be a typo in the exponent for y. Note that I asked the student where this question came from. If it was copied correctly, the textbook (or online lesson) has problems.
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03/12/20

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