How can I find x and z if: $\sqrt{(x-20)^{2} + (5-30)^{2} + (z-40)^{2}} = 100$ and $x \sqrt\frac{1}{6} + 5\sqrt\frac{1}{3} + z \sqrt\frac12= 0$?

Question

How can I find x and z if:

$\sqrt{((x-20)^{2} + (5-30)^{2} + (z-40)^{2})} = 100$ and $\left(x\times \sqrt\frac{1}{6} + 5\times \sqrt\frac{1}{3} + z\times \sqrt\frac{1}{2}\right) = 0$ ?

Answer 1

Since we have $$(x-20)^2+(z-40)^2=100^2-25^2$$ $$x=-\sqrt 3z-5\sqrt 2$$ we have $$(-\sqrt 3z-5\sqrt 2-20)^2+(z-40)^2=100^2-25^2.$$ Expand these to get the form $Az^2+Bz+C=0$ and use $$z=\frac{-B\pm\sqrt{B^2-4AC}}{2A}.$$

P.S. Multiplying the both sides of $$x\times\sqrt{\frac 16}+5\times\sqrt{\frac 13}+z\times\sqrt{\frac 12}=0$$ by $\sqrt 6$ gives us $$x\times\frac{\sqrt 6}{\sqrt 6}+5\times\frac{\sqrt 6}{\sqrt 3}+z\times\frac{\sqrt 6}{\sqrt 2}=0$$ i.e. $$x+5\sqrt 2+\sqrt 3z=0\Rightarrow x=-\sqrt 3z-5\sqrt 2.$$