Question

Help needed, i'm stuck. What is the simplified form of the quantity x squared minus 5x plus 6 over 15 x y squared all over the quantity 2x squared minus
7x plus 3 over 5 x squared y?

Answer 1

Sounds like you're dealing with this?
\[\Large \frac{x^2-5x+6}{15xy}\div\frac{2x^2-7x+3}{5x^2y}\]
Anyways, I'll let @Lord_Box answer this one

Answer 2

\[\frac{ x^2-5x+6 }{ 15xy^2} \div \frac{ 2x^2-7x+3 }{ 5x^2y }\] Just with a square on the first y.

Answer 3

Alright when dividing a fraction by another fraction, how is this accomplished?

Answer 4

Cross multiplication?

Answer 5

Yes, multiplying the top of one by the bottom of the other, and vice versa.
Can you show me what that would look like?

Answer 6

Cross multiplication only happens if you set two fractions equal to each other

Answer 7

6x2-5x+y+6?

Answer 8

Alright, I'll clear it up. When you divide one fraction by another, you "flip the second fraction and multiply.\[\frac{ x^2-5x+6 }{ 15xy^2 } \times \frac{ 5x^2y }{ 2x^2-7x+3 }\]

Answer 9

okay

Answer 10

hint: try to factor `x^2-5x+6` and `2x^2-7x+3`

Answer 11

the first one is (x-2)(x-3) I couldn't figure out the second

Answer 12

You can factor the second one by grouping. First, multiply the coefficients of the first and last terms, 2 * 3 = 6. What two numbers multiply to 6 but add to - 7?

Answer 13

-6 and -1

Answer 14

Right, so you can rewrite this as 2x^2 - 6x - x +3. Think of these as two parts, 2x^2 - 6x, then -x + 3, can you identify the common factor that goes into both of these?

Answer 15

I'll help you out, -x + 3 fits into both of these. Once you find this greatest common factor, you need to take it out of both:\[2x^2-6x-x+3 = 2x(x-3) -1(x-3) = (2x-1)(x-3)\]

Answer 16

the beginning should read x-3 instead of -x+3, sorry

Answer 17

okay

Answer 18

so right now we have \[\frac{ (x-2)(x-3) }{ 15xy^2 } \times \frac{ 5x^2y }{ (2x-1)(x-3) }\] What all could you cancel out?

Answer 19

(x-3)

Answer 20

see the attached text document to see another way to factor `2x^2-7x+3`

Answer 21

\[\frac{ (x-2) }{ 15xy^2 } \times \frac{ 5x^2y }{ (2x-1) }\] Good, you cancelled the (x-3), but there's still more that can be simplified. HINT: top right, bottom left

Answer 22

Thank you @jim_thompson5910

Oh, okay XD

Answer 23

one is 5x^2y and the other is 5xy^2

Answer 24

Hint:

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

\[\Large \frac{5x^2y}{15xy^2}=\frac{{\color{red}{5}}*{\color{blue}{x}}*x*{\color{purple}{y}}}{3*{\color{red}{5}}*{\color{blue}{x}}*{\color{purple}{y}}*y}\]

Answer 25

oh

Answer 26

That's a detailed way to show that the '5's will pair up and cancel out, a pair of 'x's will cancel out. So will a pair of 'y's. What is left over when all that canceling is done?

Answer 27

5x and 3y

Answer 28

the '5's will cancel

Answer 29

oh

Answer 30

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

\[\Large \frac{5x^2y}{15xy^2}=\frac{{\color{red}{5}}*{\color{blue}{x}}*x*{\color{purple}{y}}}{3*{\color{red}{5}}*{\color{blue}{x}}*{\color{purple}{y}}*y}\]

\[\Large \frac{5x^2y}{15xy^2}=\frac{{\color{red}{\cancel{5}}}*{\color{blue}{x}}*x*{\color{purple}{y}}}{3*{\color{red}{\cancel{5}}}*{\color{blue}{x}}*{\color{purple}{y}}*y}\]

Answer 31

yeah, i got confused XD

Answer 32

I now have x and 3y

Answer 33

Yes, so
\[\Large \frac{5x^2y}{15xy^2} = \frac{x}{3y}\]
which means
\[\Large \frac{ (x-2) }{ 15xy^2 } \times \frac{ 5x^2y }{ (2x-1) }\]
turns into
\[\Large \frac{ (x-2) }{ 3y } \times \frac{ x }{ (2x-1) }\]

Answer 34

Good stuff. Let's recap: \[\frac{ (x-2) }{ 3y } \times \frac{ x }{ (2x-1) }\]

Answer 35

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

Answer 36

okay, thank you for the help!