Question

Reduce this algebraic fraction
4x^2-y^2/8x^2+10xy+3y^2*4x^2-9xy-9y^2/2x^2-5xy-3y^2

Answer 1

add like terms

Answer 2

then do the equation in order of multiplication, division, addition, and subtraction

Answer 3

what do you get?

Answer 4

do you understand or no?

Answer 5

(4x^2 - y^2) / (x^2 - 6xy + 9y^2) ???

Answer 6

no?

Answer 7

here let me recruit lol I have confused myself

Answer 8

@campbell_st

Answer 9

wait now take what you have and complete polynomial division otherwise know as long division

Answer 10

factor the numerator of the 1st fraction, its the difference of 2 squares

\[4x^2 - y^2 = (2x - y)(2x + y)\]

the denominator of the 2nd fraction can be factored to

\[2x^2 -5xy - 3y^2 = (2x + y)(x - 3y)\]

substitute these 2 factored forms and then remove any common factors.

Answer 11

take it away Campberll I figured I was wrong

Answer 12

the numerator of the 2nd fraction can also be factored to

\[4x^2 - 9xy - 9y^2 = (4x + 3y)(x - 3y)\]

see if you can factor the denominator of the 1st fraction

Answer 13

so your problem now looks like

\[\frac{(2x - y)(2x + y)}{8x^2 + 10xy + 3y^2} \times \frac{(4x+3y)(x - 3y)}{(2x + y)(x - 3y)}\]

you just need to factor the denominator of the 1st fraction then remove common factors for the solution

hope it helps

Answer 14

wouldn't it just be 2x-y/x-3y then?

Answer 15

well what did you get when you factored the denominator of the 1st fraction...

and I think your solution is incorrect...

Answer 16

(4x+5+1y^2)

Answer 17

no that's not close..

Answer 18

I have no idea then

Answer 19

well 3y^2 is prime so the factors are y and 3y

find the factors of 8x^2 that when multiplied by y and 3y give the middle term
10xy