Question

((x-4)(x+4))/((9-x))*(((x-10)(x+9))/((x-10)(x-4)))
how do you find the excluded values?

Answer 1

I know its to find any zeros cancelling out the denominators (4, 10, and 9) but I wanted to make sure I was doing it right

Answer 2

Kind of hard to tell where fractions begin and end. Is this it?
\[\frac{ (x-4)(x+4) }{ (9-x) }\cdot \frac{ (x-10)(x+9) }{ (x-10)(x-4) }\]

Answer 3

yes sorry I don't know how to put up diagrams but yes.

Answer 4

You can just hit the equation button below and get some practice messing with it. Putting that kind of text into the question itself is more complicated, though.

But yes, if I have the denominator in factored form, like you do, you would just set each factor equal to 0

9-x = 0
x-10 = 0
x-4 = 0

this gives

x = 9, 10, 4. So you're right, those are the values that must be excluded from the domain :)

Answer 5

OK thanks... can you help me with one other problem too?

Answer 6

Whatcha got?

Answer 7

Sorry it might be out of low confidence but I don't know how to simplify-

Answer 8

Are we using the same expression to simplify?

Answer 9

http://www.virtualnerd.com/tutorials/?id=Alg1_8_1_1 use this link to help you :)

Answer 10

\[((10x ^{2} -20x)/40x ^{3} -80x ^{2} )/ (16x ^{3} +80x ^{2}) / 6x+30\]

Answer 11

Wow, this is to advanced for me :)

Answer 12

so do I simplify as in the x^2s to cancel into 70x^2 in the denominator, or do i subtract an exponent out of xfrom x^2 and x^3?

Answer 13

(if that makes sense at all)

Answer 14

\[\frac{ 10x^{2}-20x }{ 40x^{3}-80x^{2} }\cdot \frac{ 16x^{3}+80x^{2} }{ 6x+30 } = \frac{ 10x(x-2) }{ 40x^{2}(x-2) }\cdot \frac{ 16x^{2}(x+5) }{ 6(x+5) }\]

Can you see how I did those factorings?

Answer 15

pfft I was wrong either way thx for the clarification. could you please stay a bit longer and tell me if I mess something up?

Answer 16

Im around for a bit, yeah.

Answer 17

so If I am correct, do I cancel out (x-2) and (x+5)? or do I combine the 2 equations first before cancelling them out?

Answer 18

Because it is all groups of multiplication, you can cancel now.

But just to show you an example of when you CANNOT cancel:
\[\frac{ 10x(x-2)+16x^{2}(x+5) }{ 40x^{2}(6)(x-2)(x+5) }\]

Some of the numbers can simplify, but I have a plus sign up top in this example. That interrupts the groups of multiplication I mentioned, so I cannot just cancel (x-2) and (x+5), I cannot have + and - in between multiplcation. If that makes sense.

Answer 19

\[((10x)/(40x^2))*((16x^2)/(6))\]
is that right?

Answer 20

Yep, that's good so far. It can be reduced further, though :)

Answer 21

Ya it does thanks. So in this case, (after reducing more)I just combine right?
ultimately, can I go either ways for this equation as combining before simplifying, or is simplifying the only way to get the right answer?

Answer 22

You eventually have to simplify. But what exactly do you mean by combining? You're allowed to write it all as one fraction, but DO NOT distribute and foil, cancel first.

Answer 23

(sorry I hd to reload the page)

Answer 24

\[\frac{ (10x)(40x^{2}) }{ (16x^{2})(6) }\]

Something like that would be fine to do, one fraction, but try to avoid distributing and foiling until you've cancelled and reduced all you could. And even then, it's probably best not to.

Answer 25

The ad was too big to let me post, but ya thanks so I ended up simplifying to
((x)/(4x^2))*((8x^2/(3))

Answer 26

If I combine now, it would get me the same answer as you up at top right? or am I unable to simplify 10x with 40x^2?

Answer 27

\[\frac{ (10x)(40x^{2}) }{ (16x^{2})(6) } = \frac{ (10x)(40) }{ (16)(6) } = \frac{ (5x)(5) }{ (2)(3) }= \frac{ 25x }{ 6 }\]

I can show in more detail some of that reducing if you would like, but that would be the full simplification.

Answer 28

Oh so you combined the two then cancelled out the common x exponents, but in what way did you simplify the numbers on the numerators and denominators?

Answer 29

wait nvm i see now

Answer 30

Myself, I simplified 40/16 by reducing by 8. The remaining numbers, 10 and 6, I reduced by 2.

Answer 31

:P ya sorry I saw that after I asked...

Answer 32

:)

Answer 33

so how do I find the excluded digits? sorry thats the last question so is it any of the numbers we started with? like 6x+30=0
6x=-30
x=-5?

Answer 34

and so on with the other starting denominators?

Answer 35

Yeah, you're looking at the factored form (not simplified) of the denominators you started with. So I'd be looking at

\[\frac{ (10x)(x-2) }{ 40x^{2}(x-2) }\cdot \frac{ (16x^{2})(x+5) }{ 6(x+5) }\]

So each factor gets set to zero.

40x^2 = 0
x-2 = 0
x+5 = 0 which gives us

x = 0
x = 2
x = -5

Answer 36

Thanks so the answer is any real number except for 0, 2, -5?
Thank you so so so so much

Answer 37

Yep, that's it :3

Answer 38

YUS thank you for the help. I just became your fan if you didn't mind

Answer 39

Nah, I dont mind, lol.

Answer 40

sweet! I love this website.welp I guess ill go back to work then :) thank you again

Answer 41

You're welcome :)