Question

http://gyazo.com/ebc75ca83755299f7b889645720d91ee

Answer 1

@jim_thompson5910

Answer 2

what do you have so far

Answer 3

nothing i get confused on the simplifying

Answer 4

When you divide two fractions, you flip the second fraction (in the bottom) and multiply like so

\[\Large
\frac{\frac{15xy^2}{x^2-5x+6}}
{\frac{5x^2y}{2x^2-7x+3}}
=
\frac{15xy^2}{x^2-5x+6}*\frac{2x^2-7x+3}{5x^2y}
\]

Answer 5

notice how \[\Large \frac{5x^2y}{2x^2-7x+3}\] flipped

Answer 6

What's next?

Answer 7

cross multiply?

Answer 8

no that only works if you have 2 fractions set equal to one another

Answer 9

focus on

\[\Large
\frac{15xy^2}{x^2-5x+6}*\frac{2x^2-7x+3}{5x^2y}
\]

Answer 10

factor everything as much as possible

Answer 11

then cancel common factors

Answer 12

luike where do i start im confused

Answer 13

try factoring x^2 - 5x + 6

Answer 14

idk confused

Answer 15

find two numbers that multiply to 6 (the last number) and add to -5 (middle coefficient)

Answer 16

3

Answer 17

what about 3?

Answer 18

3x3=6

Answer 19

3 times 3 is 9

Answer 20

2*

Answer 21

3 times 2 = 6, good

Answer 22

but 3 + 2 = 5

we want -5 and not +5

Answer 23

not possible then

Answer 24

-3 x -2

Answer 25

yes

-3 times -2 = +6
-3 plus -2 = -5

Answer 26

you got it

Answer 27

so x^2 - 5x + 6 factors to (x-3)(x-2)

Answer 28

now to factor 2x^2 - 7x + 3, you'll have to factor by grouping

Answer 29

add 7x and 2x^2?

Answer 30

first coefficient is 2
last term is 3

2*3 = 6

you need to find two numbers that multiply to +6 and add to -7 (middle coefficient)

Answer 31

those two numbers are -6 & -1

Answer 32

so break up the -7x into -6x - 1x

Answer 33

2x^2 - 7x + 3

2x^2 - 6x - 1x + 3

Answer 34

now you factor 2x^2 - 6x - 1x + 3 by grouping

Answer 35

2x^2 -5x+3?

Answer 36

no

Answer 37

2x^2 - 6x - 1x + 3

(2x^2 - 6x) + (-1x + 3)

2x(x - 3) - 1(x - 3)

I'll let you do the last step

Answer 38

2x^2-5x - x-3

Answer 39

no

Answer 40

factor out x-3

Answer 41

2x -1

Answer 42

read this article

http://www.regentsprep.org/regents/math/algtrig/atv1/revfactorgrouping.htm

Answer 43

do you factor with the other side by cancling the x-3

Answer 44

you factor out x-3

Answer 45

basically follow the distributive property in reverse

Answer 46

so you should get (x-3)(2x-1)

Answer 47

So if you factor everything as much as possible, you get this

\[\Large \frac{15xy^2}{x^2-5x+6}*\frac{2x^2-7x+3}{5x^2y}\]
\[\Large \frac{3*5*x*y*y}{x^2-5x+6}*\frac{2x^2-7x+3}{5x^2y}\]
\[\Large \frac{3*5*x*y*y}{(x-3)(x-2)}*\frac{2x^2-7x+3}{5x^2y}\]
\[\Large \frac{3*5*x*y*y}{(x-3)(x-2)}*\frac{(x-3)(2x-1)}{5x^2y}\]
\[\Large \frac{3*5*x*y*y}{(x-3)(x-2)}*\frac{(x-3)(2x-1)}{5*x*x*y}\]

Answer 48

now cancel out common terms between the numerator and denominator

Answer 49

-3 and 3 cancel out

Answer 50

no you can't do that cancellation

Answer 51

3 does NOT cancel with the -3 in the "x-3". It doesn't work like that

Answer 52

however, the two "x-3" terms cancel

there are other cancellations

Answer 53

5 and 5

Answer 54

yes, what else

Answer 55

x and ys

Answer 56

so we have this going on....

\[\Large \frac{3*\cancel{5}*x*y*y}{(x-3)(x-2)}*\frac{(x-3)(2x-1)}{\cancel{5}*x*x*y}\]
\[\Large \frac{3*x*y*y}{(x-3)(x-2)}*\frac{(x-3)(2x-1)}{x*x*y}\]
\[\Large \frac{3*\cancel{x}*y*y}{(x-3)(x-2)}*\frac{(x-3)(2x-1)}{\cancel{x}*x*y}\]
\[\Large \frac{3*y*y}{(x-3)(x-2)}*\frac{(x-3)(2x-1)}{x*y}\]
\[\Large \frac{3*\cancel{y}*y}{(x-3)(x-2)}*\frac{(x-3)(2x-1)}{x*\cancel{y}}\]
\[\Large \frac{3y}{(x-3)(x-2)}*\frac{(x-3)(2x-1)}{x}\]
\[\Large \frac{3y}{\cancel{(x-3)}(x-2)}*\frac{\cancel{(x-3)}(2x-1)}{x}\]
\[\Large \frac{3y}{x-2}*\frac{2x-1}{x}\]

Answer 57

Now multiply straight across and rearrange terms

\[\Large \frac{3y}{x-2}*\frac{2x-1}{x}\]
\[\Large \frac{3y(2x-1)}{(x-2)x}\]
\[\Large \frac{3y(2x-1)}{x(x-2)}\]

Answer 58

wow so just simplification

Answer 59

That means

\[\Large
\frac{\frac{15xy^2}{x^2-5x+6}}
{\frac{5x^2y}{2x^2-7x+3}}
\]

simplifies to

\[\Large \frac{3y(2x-1)}{x(x-2)}\]

Answer 60

yes that is what is stated in the instructions

Answer 61

and just one more http://gyazo.com/85c14cd14916032fb0641771878f1487

Answer 62

@jim_thompson5910