Question

If (x+3)²+(y-2)²=0, find x+y= ?

Answer 1

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Step-by-step explanation:

(x+2)2+(y+3)2+(x+2)(y+3)=741

Rearranging

(x+2)2+2(x+2)(y+3)+(y+3)2−(x+2)(y+3)=741

or

(x+y+5)2−(x+2)(y+3)=741

Let x+y+5=n, x+2=s, y+3=t

so we have

s+t=n

and

n2−st=741

st=n2−741

The solution for s and t in terms of n is

s=n±2964−3n2√2

for the solution to have real values 2964−3n2≥0

or 988−n2≥0

or −988−−−√≤n≤988−−−√

or −988−−−√≤x+y+5≤988−−−√

Answer 2

Step-by-step explanation:

x^2 + 9 +6x + y^2 + 4 -4y = 0

x^2 + y^2 +6x - 4y +13 = 0