#Algebra
35 messages · Page 1 of 1 (latest)
is it correct then
cant be minsu cuz x>0 and y> 0
answer is 5:2
but
idk why
what did i just find out
i got no clue
let me sub x =2 int othe equation
got 5:2 by subing in a random value
ig thats allowed then
you're meant to just factorise it
6x^2 - 7xy - 20y^2 = 0
(3x + 4y)(2x - 5y) = 0
3x = -4y
2x = 5y
both are positive so it has to be 2x = 5y
2x = 5y
x/y = 5/2
x:y = 5:2
Yeah this is what I did
the quadratic formula method is correct its just a bit scuffed witht the y variables
I wouldn’t do trial and error unless it was something like iteration
oh wait is that trial and error
i didnt see
if there is trial and error included in that then ye dont do that
oh at the time i got stumped because I didn't know how to factorise it
Well it isn't that i didn't know its just that i never done it with 2 terms
but i did it like that and got the same as you so ye
ty
How do u factorise when theres 2 letters
just forget about y and write a quadratic in terms of x then factorise that
Then just plug in y into ur two factors
idk i did 6*-20 to get -120 then i got 2 numbers -15 and 8
but its xy so i did -15xy and +8xy then i just factorised
prolly this is better
@waxen horizon