Question

check?

Answer 1

@Yttrium

Answer 2

You got this :)

Answer 3

yay thank you!!!

Answer 4

check? ^^^

Answer 5

Very good!! :))

Answer 6

check^^^

Answer 7

Recheck your work :p

Answer 8

ohh d sorry

Answer 9

Right :)

Answer 10

yay!!!

Answer 11

They're too many, let me get a paper hahaha

Answer 12

okay lol

Answer 13

oops recheck your work :p

Answer 14

its not 112/5x(x+5)(5-x) ?

Answer 15

no no no come back please come back!!!!

Answer 16

i neeeeedddddd you nooooooooooooo

Answer 17

holy crap is it a???

Answer 18

@johnweldon1993 yay you're hereee help please?!?!

Answer 19

I dont get 'a' either :P

Answer 20

?

Answer 21

lol did you give up on that one?

Answer 22

this one

Answer 23

and noooo never!!!

Answer 24

Well the lets keep working on the one you had before :)

Answer 25

this iis the one we were working on her was checking it

Answer 26

Oh okay lol...let me get on the same page lol....okay we're on the

\[\large \frac{3x^2}{4y} \div\frac{xy}{9}\] right?

Answer 27

yea lets do this one

Answer 28

Check your work again :)

Answer 29

i got 27x/4y^2

Answer 30

grrr okay

Answer 31

Remember you can turn this into

\[\large \frac{3x^2}{4y} \times \frac{9}{xy}\]

Answer 32

Yeah 27x/4y^2 is correct

Answer 33

okay okay

Answer 34

yay!!!!!

Answer 35

this one this one this one then d

Answer 36

Nope :P

Answer 37

Wanna do it together?

Answer 38

neither?!?!?!?!

Answer 39

and yes

Answer 40

\[\large \frac{4x - 12}{x^2 -9} \times \frac{x + 3}{5x^2 - 25x} \times \frac{x^2 - 25}{28}\]

notice how I already changed the last fraction to multiplication okay?

Answer 41

okay

Answer 42

Focus on the very first fraction....

\[\large \frac{4x - 12}{x^2 - 9}\]

This can be factored....what can we factor out of the numerator?

Answer 43

(2x + 3) (2 - 4)

Answer 44

*hint looks like both numbers are divisible by 4*

Answer 45

No not quite...in this case...I just meant we can take out a 4 from there...

\[\large 4(x - 3)\]

right? since both 4x and 12 are divisible by 4...we can factor that out...

Answer 46

4

Answer 47

4(x-3)/(x+3)(x-3)

Answer 48

is that right?

Answer 49

Perfect! but do you see something that can cancel?

Answer 50

(x-3)

Answer 51

so then 4/(x+3)

Answer 52

Right...that cancels and all we have left is

\[\large \frac{4\cancel{(x - 3)}}{\cancel{(x - 3)}(x + 3)} = \frac{4}{x + 3}\]

Answer 53

Alright...that was the first fraction...now for the second

\[\large \frac{x + 3}{5x^2 - 25x}\] can you simplify it?

Answer 54

yes the bottom right?

Answer 55

right...and it will simplify into...

Answer 56

so then is it 3x/-5x ???

Answer 57

Uhh no not quite...not too sure how you got that...

so we know the bottom can be simplified...

\[\large 5x^2 - 25x\]

*hint...looks like both numbers are divisible by 5...and both have at least 1 'x' in it

Answer 58

thats what i did so i got 3+x/-5x am i wrong

Answer 59

Well when we factor out a 5...and an 'x' from that bottom...we have

\[\large 5x^2 - 25x \rightarrow 5x(x - 5)\]

right? because we want to WRITE the 5x that we factored out...and also what is left over when we do that

so that fraction will be

\[\large \frac{x + 3}{5x(x - 5)}\]

Answer 60

does that make sense?

Answer 61

ohh im stupid

Answer 62

lol yes

Answer 63

no you're not :)

Answer 64

Okay...so we have to focus on that third fraction too!

\[\large \frac{x^2 - 25}{28}\]

what can be simplified?

Answer 65

(x+5)(x-5)/28

Answer 66

???

Answer 67

Perfect!

Answer 68

yay!!!

Answer 69

Alright...so altogether...we have

\[\large \frac{4}{(x + 3)} \times \frac{(x + 3)}{5x(x - 5)} \times \frac{(x + 5)(x - 5)}{28}\]

see anything that can be simplified?

Answer 70

yes (x+3) and (x+3) can be crossed out and (x-5) and (x-5) can be crossed out

Answer 71

Perfect!

\[\large \frac{4}{\cancel{(x + 3)}} \times \frac{\cancel{(x + 3)}}{5x\cancel{(x - 5)}} \times \frac{(x + 5)\cancel{(x - 5)}}{28}\]

so all we have left is

\[\large \frac{4(x + 5)}{5x \times 28}\]

we can see that both 4...and 28...are both divisible by 4 right? so what happens if we factor a 4 out of both the numerator and denominator?

Answer 72

i really dont know

Answer 73

That's fine...well actually lets do 1 more step...it might make it easier to see...

do the evaluation of the denominator...what is 28 times 5x?

Answer 74

umm hold on

Answer 75

*holding* :P

Answer 76

140x

Answer 77

right!...so now we have

\[\large \frac{4(x + 5)}{140x}\]

Now....4 and 140 are divisible by 4....so if we were to factor out a 4 from the entire fraction...that would look like

\[\large \frac{4(x + 5)}{4(35x)}\]

right? all we did was divided the bottom (denominator) by 4...now we have a 4 factored out...just like on the top...hint*

\[\large \frac{\cancel{4}(x + 5)}{\cancel{4}(35x)}\]

Answer 78

omg come to south bend i freaking love you lol thank you hold on im like writing this down at the same time

Answer 79

hahaha okay be there in a day ;P

Answer 80

really yay!!!

Answer 81

haha nope :P

Answer 82

damn dude

Answer 83

i got excited lol

Answer 84

hahaha :P

and the least common denominator...is an expression that both denominators can go into evenly

Can the first denominator go into (k - 3) evenly? Nope....try again :)

Answer 85

grrrr 3-k???

Answer 86

ohh no a

Answer 87

noo no no no dont you leave me to @johnweldon1993