Math, asked by Anonymous, 1 year ago

Show that the equation,

 {x}^{2}  - 3xy + 2 {y}^{2}  - 2x - 3y - 35 = 0

for every real value of 'x' there is a real value of 'y' , and for every real value of 'y' there is a real value of 'x'.


❌❌NO SPAMMING❌❌​

Answers

Answered by rahman786khalilu
3

Hope this solutio will help you

Attachments:

mkrishnan: question is ok bro please make correction
mkrishnan: your product is not -9y it is 9y
mkrishnan: my answer is correct bro
rahman786khalilu: in question it is -3y but we get +9y bro
mkrishnan: in my answer is -3y bro please check
Answered by mkrishnan
3

Answer:

Step-by-step explanation:

x^2 -3xy +2y^2 = [x-2y] [x-y]

x^2 -3xy +2y^2 -2x-3y -35 = [x-2y+m ] [x-y+n]

now

             m+n = -2

             -m-2n = -3

adding        -n = -5

                    n=5

then             m= -7

x^2 -3xy +2y^2 -2x-3y -35 = [x-2y-7 ] [x-y+5]

this is a pair of st lines [ two lines ]

so all real x  we find 2  real values  y   [except the point of intersection]

and converse also  true


rahman786khalilu: once check product bro
rahman786khalilu: ok write
mkrishnan: oh thank u
Similar questions