Question

need help with algebra

Answer 1

do you understand how this works:
\[\frac{ 2 * 3 }{ 5 * 2 } = \frac{ 3 }{ 5 }\] because the 2's cancel out?

Answer 2

so for 1
you can cross out a (b - 4), a (b - 3), and a (b) from the top and bottoms
so what would you be left with?

Answer 3

Option D?

Answer 4

yep :D

Answer 5

now for 2
you would have to understand the concept of:
\[\frac{ 2 }{ 5 } \div \frac{ 7 }{ 9 } = \frac{ 2 }{ 5 } * \frac{ 9 }{ 7 }\]

Answer 6

oh yeah I can kind of see how that works

Answer 7

then for 2 it would become:
\[\frac{ (3x ^{2} - 1) }{ 4x(x - 5) } * \frac{ (x - 5) }{ (3x ^{2}-1)(x - 1) }\] then you can cancel out a (3x^2 - 1), and a (x - 5)
so then what would you be left with?

Answer 8

Option b?

Answer 9

yep :)

Answer 10

yay you help me so much. If you don't mind, there are a few more questions.

Answer 11

sure :)

Answer 12

the same idea as before, expect now it's a little more involved
so like it would become:
\[\frac{ m + 3 }{ 4m } * \frac{ 32m ^{2}(m-3) }{ m ^{2}-9 }\] and then to cancel out properly you would have to know that
m^2 - 9 = (m + 3)(m - 3)
and 32m^2 = 8*4*m*m

Answer 13

Either A or C I'm not sure.

Answer 14

you're getting good at this :)
it would be A, do you need more explanation for it?

Answer 15

Glad to hear that I am haha :) No I'm good

Answer 16

When you just tell me about what needs to be canceled out it helps a lot.

Answer 17

for 4, you would have to factor first...
like 6a^2 - 30a = 6a(a - 5)
for as much as you can, then start crossing out...

Answer 18

It looks like it could be B or D but I think it might be B.

Answer 19

not, B
because once you finish crossing out you would be left with
\[\frac{ 3(a + 4) }{ a }\] then distribute the numerator :)

Answer 20

So C? Idk I'm a little confused

Answer 21

okay, so once you factored all of it, you would get
\[\frac{ 6a(a-5) }{ (a-2) } * \frac{ (a-2)(a+4) }{ 2a ^{2}(a-5) }\]then you would cancel out stuff and get
\[\frac{ 3(a+4) }{ a }\] and you would distribute so
D. \[\frac{ 3a + 12 }{ a }\]

Answer 22

Ohhh okay. That makes a lot more sense than me choosing B lol. I knew it had to have been B or D.

Answer 23

yep :) you're good at narrowing things down :) :P

Answer 24

haha yeah! now for #4 :)

Answer 25

oops I meant 5 and 6*

Answer 26

same idea, flip the part after the divided by
then factor. then cancel :)

Answer 27

Option A?

Answer 28

not A...
so after you flip and factor you would get:
\[\frac{ 12s ^{2}(4s+3) }{ 9s(s-5)(s+2) } * \frac{ (s-5)(s-2) }{ 4(4s+3) }\]

Answer 29

so it's between b and c and i think its c

Answer 30

C is correct :)

Answer 31

the factored is correct :)
umm and then I supposed you would solve for x from what you got, since that is what the question is asking?

then for (27x / x - 9) / (9x / x - 9)
\[\frac{ 27x }{ x-9 } * \frac{ x-9 }{ 9x }\] that look better? :)

Answer 32

yeah that's all i had to do for that one.

idk what im supposed to put for the final one

Answer 33

im kind of in a hurry right now

Answer 34

for the (27x / x - 9) / (9x / x - 9)
you would cross out a (x-9) and a (9x)
to be left with a 3

Answer 35

oh so just 3? thank you

Answer 36

you're welcome :)

Answer 37

thanks for all the help bye :) have a good day or night