Is the book wrong again? Algebra

einstein

New member
Joined
Sep 14, 2010
Messages
46
This is the expression EXACTLY as displayed in the book:

\(\displaystyle 5a^2b \times (-6a^5b^7) \div 2a^2b^2\)

It asks you to simplify the expression.

The book's answer is:

\(\displaystyle -15a^5b^6\)

Is this right or wrong.

I did think this answer, the book's answer, was right, but after my last post in this forum i am confused and now think the answer is \(\displaystyle -15a^5b^1^0\)

thanks
 
Looks to me like the book's answer is right. When you multiply you add the exponents, and when you divide you subtract them.
 
Someone in another forum wrote that the answer is:

\(\displaystyle 5a^2b \times \left ( -6a^5b^7 \right ) \div 2a^2b^2\)

\(\displaystyle \frac{-30a^7b^8}{2}\cdot a^2b^2=-15a^9b^{10}\)

These questions are really confusing me and it seems everyone else too!

Can anyone chime in and state the right answer and why it is the right answer?
 
einstein said:
Someone in another forum wrote that the answer is:

\(\displaystyle 5a^2b \times \left ( -6a^5b^7 \right ) \div 2a^2b^2\)

\(\displaystyle \frac{-30a^7b^8}{2}\cdot a^2b^2=-15a^9b^{10}\)

These questions are really confusing me and it seems everyone else too!

Can anyone chime in and state the right answer and why it is the right answer?

\(\displaystyle User \ einstein, \ 5a^2b \times \left (-6a^5b^7 \right ) \div 2a^2b^2 \ = \ \frac{5a^2b \times \left (-6a^5b^7)}{2a^2b^2} \ = \\)

\(\displaystyle \ \frac{-30a^7b^8}{2a^2b^2} \ = \ -15a^5b^6 \ ,*but*\)


\(\displaystyle 5a^2b \times \left (-6a^5b^7 \right) \div2a^2b^2 \ \ne \ 5a^2b \times \left (-6a^5b^7)/2a^2b^2\)


The answer given by you, taken from the book, in your first post is correct.

\(\displaystyle The \ expression \ on \ the\ left-hand \ side \ equals\)

\(\displaystyle \ 5a^2b \ \times \ \left (-6a^5b^7)/(2a^2b^2)\)


\(\displaystyle . \ . \ . \ . \ Note \ the \ needed \ grouping \ symbols.\)


Edit: The margin cut off two of my lines.

The exponent typo of mine was fixed, and einstein was corrected
about using a wrong username for me.
 
lookagain said:
einstein said:
Someone in another forum wrote that the answer is:

\(\displaystyle 5a^2b \times \left ( -6a^5b^7 \right ) \div 2a^2b^2\)

\(\displaystyle \frac{-30a^7b^8}{2}\cdot a^2b^2=-15a^9b^{10}\)

These questions are really confusing me and it seems everyone else too!

Can anyone chime in and state the right answer and why it is the right answer?

\(\displaystyle User \ einstein, \ 5a^2b \times \left (-6a^5b^7 \right ) \div 2a^2b^2 \ = \ \frac{5a^2b \times \left (-6a^5b^7)}{2a^2b^2} \ = \ \frac{-30a^7b^8}{2a^2b^2} \ = \ -15a^2b^6 \ ,*but*\)


\(\displaystyle 5a^2b \times \left (-6a^5b^7 \right) \div2a^2b^2 \ \ne \ 5a^2b \times \left (-6a^5b^7)/2a^2b^2\)


The answer given by you, taken from the book, in your first post is correct.

\(\displaystyle The \ expression \ on \ the\ left-hand \ side \ equals \ 5a^2b \ \times \ \left (-6a^5b^7)/(2a^2b^2)\)


\(\displaystyle . \ . \ . \ . \ Note \ the \ needed \ grouping \ symbols.\)

Okay i think i've got it now. Please tell me if this is right:

a/bc = a divided by b, then times c

a/(bc) = a divided by the answer of b times c

\(\displaystyle a \div bc = \frac{a}{bc} =\) a/(bc) = a divided by the answer of b times c

BUT following this logic it seems like youre saying that all the terms AFTER a \(\displaystyle \div\) sign always get put into brackets. Which cant be right
 
Here is how the problem can be equivalently written:

\(\displaystyle \frac{5a^{2}b\cdot -6a^{5}b^{7}}{2a^{2}b^{2}}\)

\(\displaystyle \frac{-30a^{7}b^{8}}{2a^{2}b^{2}}\)

\(\displaystyle -15a^{5}b^{6}\)

The poster on the other site made read the problem incorrectly, perhaps by misinterpreting the order of operations. It's easy to do.
 
galactus said:
\(\displaystyle \frac{5a^{2}b\cdot -6a^{5}b^{7}}{2a^{2}b^{2}}\)
Albert Jr., look at that carefully; the "horizontal line" makes it quite clear what the divisor is, right?

But if you post that all in one line this way: 5 a^2 b(-6 a^5 b^7) / 2 a^2 b^2,
then it is NOT the same thing: the "/" does not replace the "hozizontal line".
BUT this makes it the SAME: 5 a^2 b(-6 a^5 b^7) / (2 a^2 b^2)

An alternate way to "play safe" is assign a variable to the divisor:
then post as: 5 a^2 b(-6 a^5 b^7) / k , where k = 2 a^2 b^2.
 
Top