Question

Can someone help me with math, please?

Answer 1

I need help with a buch of problems!

Answer 2

Just, let me get them.

Answer 3

A bunch?

Answer 4

Well, actually, not a lot. I just re-counted them

Answer 5

But those are the first 2

Answer 6

how many in total?

Answer 7

I need help doing it step-by-step and um let me look

Answer 8

8 right now, but I'll try and do the others myself, but I can't really do them that well.

Answer 9

i'll show you the first one
factor out 2 and 3,
\[\frac{2(a^2-2a+1)}{3(a^2-3)}\]
then both of the quadratics inside the brackets can be factored as well,
\[\frac{2\cancel{(a-1)}(a-1)}{3\cancel{(a-1)}(a+1)} \\ \\ \frac{2(a-1)}{3(a+1)}\]

Answer 10

the technique for other questions are basically similar, use this as a guide

Answer 11

I'm just asking for help, there is a total of 39, but...I'm sure you don't want to do that or help with that.

Uh, ok, when I finish the others do you mind checking my work?

Answer 12

@.Sam. Ok, I'm confused for the one you just did..

Answer 13

How did you get 1? Besides the +1 in the beginning?

Answer 14

Which step?

Answer 15

The 2nd

Answer 16

And basically all of them, I don't understand what you did.

Answer 17

do you understand from here?
\[\frac{2a^2-4a+2}{3a^2-6} \\ \text{\to} \\ \frac{2(a^2-2a+1)}{3(a^2-1)}\]

Answer 18

No, I don't.

So, like, the 3a and the 2 at the top is multiplying each other? Erg

Answer 19

I can't understand this

Answer 20

no, 2 is dividing by 3

Answer 21

OH! Hah, so..haha okay, dumb moment there.

Ok, I think I get it now

Answer 22

Before you go, can you check 9? I did it yesterday

Answer 23

ok

Answer 24

I don't think you can factor that

Answer 25

Oh how?

Answer 26

and don't simply cancel, cancelling terms in only for, when the numerator has same stuff as the denominator

Answer 27

what do you mean?

Answer 28

examples of cancelling are
\[\frac{\cancel a}{\cancel a}=1 \\ \\ \frac{3\cancel x(x-5)}{2\cancel x(x-3)}=\frac{3(x-5)}{2(x-3)} \\ \\ \frac{3\cancel{(x-2)}}{\cancel{(x-2)}}=3\]

Answer 29

brb

Answer 30

\[7.\frac{ 2a ^{2}-4a+2 }{ 3a ^{2} -3}\]
\[=\frac{ 2\left( a ^{2}-2a+1 \right) }{ 3\left( a ^{2}-1 \right) }\]
\[=\frac{ 2\left( a ^{2}-a-a+1 \right) }{ 3\left( a+1 \right)\left( a-1 \right) }\]
\[=\frac{ 2[a \left( a-1 \right)-1\left( a-1 \right)] }{\left( a+1 \right)\left( a-1 \right)}\]
\[=\frac{ 2\left( a-1 \right)\left( a-1 \right) }{ \left( a+1 \right)\left( a-1 \right) }\]
\[=\frac{ 2\left( a-1 \right) }{ \left( a+1 \right) }\]

Answer 31

\[9. \frac{ 4-c }{ 2c-8 }=\frac{ -\left( -4+c \right) }{2\left( c-4 \right)}=\frac{ -1 }{ 2 }\]

Answer 32

I forgot writing 3 in the denominator in last three steps.