Danny's system of equations

Adding the three equations together yields: $x^2-4x+y^2+6y+z^2+12z = -49$

We can now complete the squares on the left side: $x^2-4x+4+y^2+6y+9+z^2+12z+36 = -49+4+9+36$ $(x-2)^2+(y+3)^2+(z+6)^2 = 0$

For the sum of squares of real numbers to be $0$ , each number must be $0$ .

Hence, $x-2 = y+3 = z+6 = 0 \leadsto (x,y,z) = (2,-3,-6)$ .

Therefore, $|x+y+z| = |2-3-6| = \boxed{7}$ .

Add all three equations, rearrange, and factor to obtain $(x-2)^{2}+(y+3)^{2}+(z+6)^{2}=0$ By the trivial inequality, $(x,y,z)=(2,-3,-6)$ so $|x+y+z|=\boxed{007}$ .

Add up the given eqns

x^2+6y+y^2+12z+z^2-4x=-49

(x^2-4x)+(y^2+6y)+(z^2+12z)+49=0

(X^2-4x+4)+(y^2+6y+9)+(z^2+12z+36)=0

(X-2)^2+(y+3)^2+(z+6)^2=0

Sum of three squared terms zero implies all terms are zero

X=2, y=-3, z=-6

X+y+z=-7

Note that $(x^2+6y)+(y^2+12z)+(z^2-4x)=(-14)+(-63)+(28)$ , so $(x^2-4x)+(y^2+6y)+(z^2+12z)=-49$ . Thus $(x^2-4x+4)+(y^2+6y+9)+(z^2+12z+36)=-49+4+9+36=0$ , which could be restated as $(x-2)^2+(y+3)^2+(z+6)^2=0$ . Since a square is never negative, it is $0$ if and only if its square root is $0$ , and the sum of $(x-2)^2$ , $(y+3)^2$ and $(z+6)^2$ is $0$ , all of $x-2$ , $y+3$ , and $z+6$ is $0$ . Therefore $x=2$ , $y=-3$ , and $z=-6$ . Note that $x+y+z=-7$ , so $|x+y+z|=\boxed{7}$ .

Adding the three equations we get

x^2 +y^2+z^2 -4x+6y+12z+49=0 or,

(x-2)^2+(y+3)^2+(z+6)^2=0

sum of squares of real numbers adding to zero ??

since square of a real number is always positive and they are adding to zero hence the solution.

x=2 , y=-3 , z=-6 . therefore |x+y+z|=7.

I did this by trying different values for x and y, starting with 1. Using 1 for x, we get (6y = -15), which doesn't work as x, y and z are all integers. Then trying 2 for x we get (6y = -18, so y = -3.) Substituting this into the second equation, we get 1(2z = -72), so (z = -6). Checking these in the third equation, we get (36 - 8 = 28), which is correct. Therefore (x +y + z = -7), so the answer is 7.

Add the equations and we get $x^2-4x+y^2+6y+z^2+12=-49$

Through completing the square, we get $\left(x-2\right)^2 -4 + \left(y+3\right)^2 -9 + \left(z+6\right)^2 -36 = -49$

$\left(x-2\right)^2 + \left(y+3\right)^2 + \left(z+6\right)^2 = 0$

The only way for this to be true is for every bracket to be equal to $0$ , and so $x = 2, y = -3, z=-6$

Therefore $|x+y+z| = |-7| = 7$

We add the 3 equations and factor to get

$(x-2)^2+(y+3)^2+(z+6)^2 = 0$

Since $a^2 \ge 0$ for all real values of $a$ we must have

$(x-2)^2 = 0$

$(y+3)^2 = 0$

$(z+6)^2 = 0$

$x = 2, y = -3, z = -6$

$|x+y+z| = 7$

The equation is equivalent to $$(z+6)^2+(x-2)^2+(y+3)^2=0$$ Thus, $z=-6,x=2,y=-3$. We get $|x+y+z|=7$.

first eqn =(1), second eqn =(2), third =(3)

$(1)+(2)+(3)=x^{2}-4x+y^{2}+6x+z^{2}+12z=-49$

$=>(x-2)^{2}+(y+3)^{2}+(z+6)^{2}=0$ $\therefore x=2, y=-3, z=-6.$

X²+y²+z²-4x+6y+12z+49=0 (x-2)²+(y+3)²+(z+6)²=0 x=2 y=-3 z=-6 |x+y+z|=|2-3-6|=7

By adding the three equations and completing the squares, we obtain the equation $(x-2)^2-4+(y+3)^2-9+(z+6)^2-36 = -49$ , or in other words, $(x-2)^2+(y+3)^2+(z+6)^2 =0$ , which has exactly one real solution: $(x,y,z)=(2,-3,-6)$ . The requested value then is $|x+y+z| = |-7|=7$ .

adding the three equations and completing the square we get (x-2)^2+(y+3)^2+(z+6)=0.

Sum of squares of real numbers is zero only when they are all zero. Therefore x=2 ,y=-3,z=-6.

Hence |x+y+z|=7

Add the equations and complete the square for each variable. The result is $(x-2)^2 + (y+3)^2 + (z+6)^2 = 0$ . Clearly, each term must equal zero! So $x=2, y=-3, z=-6$ , and $|x+y+z|=7$ .