Please solve correctly this equation of both these sides given... in terms of x and y.
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(x+1)2+(y−5)2=0
Adding −(y−5)2 on both sides:
(x+1)2+(y−5)2−(y−5)2=0−(y−5)2
(x+1)2+0=0−(y−5)2
(x+1)2=−(y−5)2⟶ Equation E
The square of any number is always non-negative. So, we have
(x+1)2≥0
and
(y−5)2≥0
So, the equation E is valid if and only if
(x+1)2=0
and
(y−5)2=0
Solving the equations, we have
(x+1)2=0⟹x+1=0
⟹x=−1
and
(y−5)2=0⟹y−5=0
⟹y=5
So, we have
x+y=y+x=5+(−1)=5−1=4
x+y=4
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♥*♡∞:。.。Answer by Shreyas。.。:∞♡*♥
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