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

Answer 2

Answer:

The real numbers $x,y,z$ satisfy the equation $(3x+5y+7z-9)2+(5x+4y+3z-2)2=0$ .The value of $x+y+z$ is 1.

Step-by-step explanation:

Given:

The given equation is $(3x+5y+7z-9)2+(5x+4y+3z-2)2=0$ and the real numbers $x,y,z$ satisfy the equation.

To Solve:

The equation has to be solved to get the value of $x, y$ and $z$ .

$(3x+5y+7z-9)2+(5x+4y+3z-2)2=0\\(3x+5y+7z-9)2=- (5x+4y+3z-2)2$

From the both sides of the equation,eliminate $2$ .

$(3x+5y+7z-9)=- (5x+4y+3z-2)\\(3x+5y+7z-9)= -5x -4y-3z+2$

Take all the terms in the LHS of the equation.

$3x+5x +5y+4y +7z+3z = 2+9\\8x+9y+10z= 11$

Consider the value of $x$ is $-1$ , $y$ is $1$ and $z$ is $1$

Put the value in the LHS of the equation.

$8(-1) + 9 \times 1 + 10 \times 1 = -8+9+10 = 11$

Therefore the value of  $x$ is $-1$ , $y$ is $1$ and $z$ is $1$ .

$x+y+z=-1+1+1=1$

Therefore, the value of x+y+z is $1$ .

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