Solve for x,y,and z
2x - y -z = 4
-x + 2y + z = 1
3x + y + z = 16

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Ответ:

Niaax7803

27.04.2023

Mathematics

Step-by-step explanation:

the standard equation of a circle will be given in the form of

$(x-x_{1} )^2+(y-y_{1} )^2 = r^2$ , where r is the radius and (x1, y1) is the center point of the circle

in this situations r^2 = 16

r = +4 or -4, but since length cannot be negative, r = 4

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Solve for x,y,and z 2x - y -z = 4 -x + 2y + z = 1 3x + y + z = 16​