Question

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'.


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Answer 1

Hope this solutio will help you

Answer 2

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