Question

10(x+1)^2 + (x+1) (y-2)-3(y-2)^2=0

Answer 1

Solution:

Given

10(x+1)²+(x+1)(y-2)-3(y-2)²=0

Splitting the middle term, we get

=> 10(x+1)²+6(x+1)(y-2)-5(x+1)(y-2)-3(y-2)² = 0

=> 2(x+1)[5(x+1)+3(y-2)]-(y-2)[5(x+1)+3(y-2)]=0

=> [5(x+1)+3(y-2)][2(x+1)-(y-2)]=0

=> [5x+5+3y-6][2x+2-y+2]=0

=> (5x+3y-1)(2x-y+4) = 0

Therefore,

5x+3y-1= 0 ---(1)

2x-y+4 = 0

=> y = 2x+4 ---(2)

Substitute (2) in equation (1) , we get

=> 5x+3(2x+4)-1 = 0

=> 5x+6x+12-1 = 0

=> 11x + 11 = 0

=> 11x = -11

=> x = (-11)/11

=> x = -1

Now ,

put x = -1 in equation (2), we get

y = 2(-1)+4

=> y = -2 +4 = 2

Therefore,

x = -1 , y = 2

••••