Question

find value of x and y
$(x+y+10)^2+(2x-2y-8)^2=0$

Answer 1

ok bro here is ur answer

(x+y+10)^2+(2x-2y-8}^2=0

let (x+y+10) = a

    (2x-2y-8}=b

now put in the equation

a^2+b^2=0

therefore we can conclude that a=0 and b=0

x+y+10=0               ,                2x-2y-8=0

x+y=-10 eq1                      ,             2x-2y=8    or x-y=4eq2

now add eq 1 and 2

x+y+x-y=-10+4

2x=-6

x=-3

and put x=-3 in eq 1

x+y=-10

y=-10+3

y=-7

x=-3     and         y=-7

Answer 2

Given: x2−xy−10−2y2=0x2−xy−10−2y2=0;

Rearrange x2−y2−y2−xy−10=0x2−y2−y2−xy−10=0;

Apply a2−b2=(a−b)(a+b)a2−b2=(a−b)(a+b): (x−y)(x+y)−y2−xy−10=0(x−y)(x+y)−y2−xy−10=0 --> (x−y)(x+y)−y(y+x)−10=0(x−y)(x+y)−y(y+x)−10=0;

Factor out x+yx+y: (x+y)(x−y−y)−10=0(x+y)(x−y−y)−10=0;

Since given that x+y=2x+y=2, then we have that 2(x−2y)−10=02(x−2y)−10=0 --> x−2y=5x−2y=5.