<s>[INST] Find $y$ in $\sqrt{4+(y-6)^2}+\sqrt{16+(y-3)^2}=3\cdot\sqrt{5}$ I would like to solve this equation for $y$. Any tips? It seems like you cant really do it analytically?

$$\sqrt{4+(y-6)^2}+\sqrt{16+(y-3)^2}=3\cdot\sqrt{5}$$

 [/INST] Set $y-3=z,$ then square both sides and simplify to get $$z^2-3z-8=-\sqrt{4+(z-3)^2}\sqrt{16+z^2}.$$ Squaring again and simplifying the right hand side gives $$(z^2-3z-8)^2=(z^2-6z+13)(16+z^2).$$ Now let $16+z^2=w,$ then expand and simplify to get $$z^2+16z+64=5w.$$ Substituting back for $w,$ simplifying and factoring gives the quadratic $$(z-2)^2=0.$$ Thus, $z=y-3=2,$ so that if there is a solution, then it has to be $y=5.$ Substituting this in the original equation assures us that this is indeed the only solution.
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