Question

Help?

Answer 1

@whpalmer4

Answer 2

Am I right?

Answer 3

wait.. no.. x+1?

Answer 4

uh, don't think so...

\[\frac{1}{x} + \frac{x}{x+3} = \frac{1}{x}*\frac{(x+3)}{(x+3)} + \frac{x}{(x+3)}*\frac{x}{x} = \frac{x+3+x^2}{x(x+3)} = \]

Answer 5

So the answer would be x+3+x^2?

Answer 6

you're supposed to find the numerator, right?

I would write it in descending exponent order: \(x^2+x+3\)

Answer 7

Okay got it..

Answer 8

How do I figure out this one?

Answer 9

Same procedure:

\[\frac{3}{2x+1}-\frac{x}{x-1}= \frac{3}{2x+1}*\frac{x-1}{x-1} - \frac{x}{x-1}*\frac{2x+1}{2x+1} \]\[\qquad= \frac{3(x-1)-x(2x+1)}{(2x+1)(x-1)} = \]

Answer 10

Then what?.. Don't I subtract somewhere?

Answer 11

expand the numerator and simplify...

Answer 12

-2x^2+4x-3?

Answer 13

close, but not quite...

Answer 14

check your algebra carefully on the \(-x(2x+1)\) part...

Answer 15

\[-x(2x+1) = -x*2x -x*1 = -2x^2-x\]

Answer 16

ohhh okay oops.
So.. -2x^2+2x-3?

Answer 17

yes

Answer 18

that's a common mistake, learn to avoid it :-)

sometimes replacing the - with -1 in such places helps remember it

Answer 19

I think x = 0 would be an easy value to plug in to see if it is true or not

Answer 20

False..

Answer 21

yes, it turns out x = 0 is a bad test case because it might make you think it is true :-)

you could look at it like this:

\[\frac{x-2}{x-1}*\frac{x-1}{x+2} = \frac{x+2}{x-2}\]\[\frac{x-2}{\cancel{x-1}}*\frac{\cancel{x-1}}{x+2} = \frac{x+2}{x-2}\]\[\frac{x-2}{x+2}\ne\frac{x+2}{x-2}\]except as it turns out at \(x = 0\)!

Answer 22

False?

Answer 23

if you cancel the matching terms, what are you left with?

Answer 24

\[\frac{ 2 }{ -1}\]\[\frac{ -2 }{ 2 }\]

Answer 25

no, you can't cancel part of the expression like that! look at what I did in the previous problem

Answer 26

\[\frac{(x+2) (x-2)}{(x-1) (x+2)}= \frac{\cancel{(x+2)} (x-2)}{(x-1) \cancel{(x+2)}} = \frac{x-2}{x-1}\]

Answer 27

Ohhh I get it.. So it is true got it.

Answer 28

what you did was similar to simplifying a fraction like this:

\[\frac{14}{12} = \frac{\cancel{1}4}{\cancel{1}2} = \frac{4}{2}\]:-)

Answer 29

we just did that...

Answer 30

Oops wrong one.

Answer 31

very good

Answer 32

I'm sure we did that one already, because I see it on my scratch paper :-)

Answer 33

yeah, we did it here: http://openstudy.com/users/xxxbambygirlxxx#/updates/5373d1afe4b09fb937cbc759