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Question 78902This question is from textbook Algebra I
: (6ab^2/35b)(42a/18ab)
This question is from textbook Algebra I
Answer by mathdoc314(58)   (Show Source):
You can put this solution on YOUR website! I really think it is wrong to submit a question without any words
The answer is (6ab^2/35b) (42a/18ab)
Is that OK?
Oh, you want me to simplify it?
Why didn't you say so?
So in the first term which is (6ab^2/35b) i see that b can be canceled since the same factor b is in the top and the bottom
6ab/35
In the secont term which is 42a/18ab I see a can be canceled since the factor a is in the numerator and denominator. Also 42 and 18 are divisible by 6
7/3b
So we are multiplying (6ab/35) * (7/3b)
The basic idea in this kid of problem is: (A/B)*(C/D) is just (A*C)/(B*D)
so that is (42ab / 105 b)
This can be condensed some more by canceling 7 from top and bottom, and b
I end up with 2a / 5
There is not really a way to check this kind of problem except to do it over again, preferably by antoher person doing the operations in a different order
(6ab^2/35b) (42a/18ab)
2*3*2*3*7*a*a*b*b / 5*7*b*2*3*3*a*b
cancel out most of that
This time I also get 2 a / 5 which luckily confirms the first response (I had to change them each three times to make this work)
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