Question

Find the value of x+y+z if x^2+ y^2+z^2=18 and xy +yz +x,=9

Answer 1

Answer:

x+y+z=6

Step-by-step explanation:

Given,

         $x^{2} +y^{2}+z^{2}$=18

            xy+yz+zx = 9

We know,

              $(x+y+z)^{2}$= $x^{2} +y^{2}+z^{2}$+ 2(xy+yz+zx)

By substituting the values, we get

$(x+y+z)^{2}$= (18) + 2(9)

$(x+y+z)^{2}$= 18 + 18

$(x+y+z)^{2}$= 36

      x+y+z = $\sqrt[]{36}$

      x+y+z = 6

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