Question

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

Answer 1

Answer :

1

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