Finding two numbers given their sum and their product 
Which two numbers when added together yield $16$, and when multiplied together yield $55$.

I know the $x$ and $y$ are $5$ and $11$ but I wanted to see if I could algebraically solve it, and found I couldn't. 
In $x+y=16$, I know $x=16/y$ but when I plug it back in I get something like $16/y + y = 16$, then I multiply the left side by $16$ to get $2y=256$ and then ultimately $y=128$. Am I doing something wrong?  

We are trying to solve the system of equations $x+y=16$, $xy=55$. Here are a couple of systematic approaches that work in general.
Approach $1$: We will use the identity $(x+y)^2-4xy=(x-y)^2$.  In our case, we have $(x+y)^2=256$, $4xy=220$, so $(x-y)^2=36$, giving $x-y=\pm 6$.
Using $x+y=16$, $x-y=6$, we get by adding that $2x=22$, and therefore $x=11$. It follows that $y=5$. 
The possibility $x+y=16$, $x-y=-6$ gives nothing new. Adding, we get $2x=10$, so $x=5$, and therefore $y=11$.
Approach $2$: From $x+y=16$, we get $y=16-x$. Substitute for $y$ in $xy=55$. We get $x(16-x)=55$. Simplification gives $x^2-16x+55=0$.  The quadratic factors as $(x-5)(x-11)$, so our equation becomes $(x-5)(x-11)=0$, which has the solutions $x=5$ and $x=11$.
But we cannot necessarily rely on there being such a straightforward factorization. So in general after we get to the stage $x^2-16x+55=0$, we would use the Quadratic Formula. We get
$$x=\frac{16\pm\sqrt{(-16)^2-4(55)}}{2}.$$
Compute. We get the solutions $x=5$ and $x=11$. The corresponding $y$ are now easy to find from $x+y=16$. 
Remarks: $1,$ Recall that the Quadratic Formula says that if $a\ne 0$, then the solutions of the quadratic equation $ax^2+bx+c=0$ are given by
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
Your approach was along reasonable lines, but things went wrong in the details. From $xy=55$ we get $x=\frac{55}{y}$. Substituting in the formula $x+y=16$, we get
$$\frac{55}{y}+y=16.$$
A reasonable strategy is to multiply through by $y$, getting $55+y^2=16y$, or equivalently $y^2-16y+55=0$. Now we have reached a quadratic equation which is basically the same as the one we reached above. 
$2.$ The first approach that we used (presented as an algorithm, and stripped of algebraic notation) goes back to Neo-Babylonian times. The "standard" problem was to find the dimensions of a door, given its perimeter and area.
