The real numbers x,y,z satisfy the equation (3x+5y+7z-9)^2+(5x+4y+3z-2)^2=0 calculate the value of x+y+z
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Step-by-step explanation:
(3x+5y+7z-9)2+(5x+4y+3z-2)2=0
(3x+5y+7z-9)2 = -(5x+4y+3z-2)2
(3x+5y+7z-9) = -(5x+4y+3z-2)
3x+5x +5y+4y +7z+3z = 2+9
8x + 9y + 10z = 11
by putting,
x= -1
y = 1
z= 1
we get LHS = RHS
so ( X + Y + Z) = (-1+1+1) = 1
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