The Purplemath Forums

by **Tennise** on Mon Jun 08, 2009 8:58 pm

I understand the numerical ones, but am having trouble with the variable ones...

It works in theory, and I can see how it would work, but when I substitute numbers for the variables, it doesn't work:
x = 2, y = 3
This was the expression: x²y³+xy
4*27+6 = 114

This was the answer: xy(xy² + 1)
6(2*9) = 12*54 = 648
or...
6(18) = 108

Did I evaluate the equations wrong? I can't seem to get the same answer for them both...

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by **jaybird0827** on Mon Jun 08, 2009 9:01 pm

Tennise wrote:I understand the numerical ones, but am having trouble with the variable ones...

It works in theory, and I can see how it would work, but when I substitute numbers for the variables, it doesn't work:
x = 2, y = 3
This was the expression: x²y³+xy
4*27+6 = 114

This was the answer: xy(xy² + 1)
6(2*9) = 12*54 = 648
or...
6(18) = 108

Did I evaluate the equations wrong? I can't seem to get the same answer for them both...

Let's look at
$xy(xy^2\,+\,1)$

Are you sure that you plugged everything in, correctly? We know that $xy = (2)(3)\,=\,6$ is correct.
Do you think that $114$ is the correct evaluation of $x^2y^3\,+\,xy$ ?
Also, what is the next multiple of 6 after 108?
Assuming that the answer to the first question is "yes", the answer to the second question should tell you that the value of the second factor $(xy^2\,+\,1)$ should be
$(next\,multiple\,of\,6\,after\,108)\div 6$ .
Now, go back and re-evaluate
$xy(xy^2\,+\,1)$ .

If you're still stuck, please reply and show your all the steps you attempted to correct your work.

by **Tennise** on Mon Jun 08, 2009 9:10 pm

jaybird0827 wrote:
Tennise wrote:I understand the numerical ones, but am having trouble with the variable ones...

It works in theory, and I can see how it would work, but when I substitute numbers for the variables, it doesn't work:
x = 2, y = 3
This was the expression: x²y³+xy
4*27+6 = 114

This was the answer: xy(xy² + 1)
6(2*9) = 12*54 = 648
or...
6(18) = 108

Did I evaluate the equations wrong? I can't seem to get the same answer for them both...

Let's look at
$xy(xy^2\,+\,1)$

Are you sure that you plugged everything in, correctly? We know that $xy = (2)(3)\,=\,6$ is correct.

ah I see why: I ignored the +1 in the second expression! It's so easy to forget...

by **jaybird0827** on Mon Jun 08, 2009 9:15 pm

Tennise wrote:
ah I see why: I ignored the +1 in the second expression! It's so easy to forget...

You obviously got that before I finished editing. Awesome job!

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simple factoring: (x^2)(y^3)+xy

simple factoring: (x^2)(y^3)+xy

Re: simple factoring: (x^2)(y^3)+xy

Re: simple factoring: (x^2)(y^3)+xy

Re: simple factoring: (x^2)(y^3)+xy