If $x,y\in \mathbb N$ is s.t. $$x+xy+y=55,$$ find $x+y$ ?
I tried to write $x+xy+y=65$ as an equation of $x+y$, but can't go anywhare :
but I just transposed the problem with $x-y$.
Other things I tried is to replace $xy$ by $$\frac{1}{2}((x+y)^2-x^2-y^2)$$ but still, I can't do anything with $x^2+y^2$.
any idea ?
Disclaimer: I am going to assume that $\mathbb{N}$ includes $0$ since @joshua did not specify.
$x+xy+y=55 \implies x+xy+y+1=(x+1)(y+1)=56$
Since $x, y\in\mathbb{N}$, $x+1, y+1\in\mathbb{Z}^{+}$.
Listing out all the cases, we get that $$\begin{align*} x+y&=0+55=55 \text{ or}\\ &=1+27=28 \text{ or}\\ &=3+13=16 \text{ or}\\ &=6+7=13 \end{align*}$$
Hence, $\boxed{x+y=55, 28, 16 \text{ or } 13}$.
Since the OP did not state exactly what he means by the symbol $\,\Bbb N\,$ (positive integers or nonnegative integers), in my answer I am assuming that $\,x\,$ and $\,y\,$ are nonnegative integers.
As Gerry pointed out, $\,x+xy+y=55\,,\,$ is equivalent to
$\color{blue}{x+1}+\color{red}{xy+y}=56\quad$ which is equivalent to
$\color{blue}{1(x+1)}+\color{red}{y(x+1)}=56\quad$ which is equivalent to
$(x+1)(1+y)=56\quad$ which is equivalent to
$(x+1)(y+1)=56\,,\quad$ consequently,
$x+1=1\quad$ and $\quad y+1=56$
or
$x+1=2\quad$ and $\quad y+1=28$
or
$x+1=4\quad$ and $\quad y+1=14$
or
$x+1=7\quad$ and $\quad y+1=8$
or
$x+1=8\quad$ and $\quad y+1=7$
or
$x+1=14\quad$ and $\quad y+1=4$
or
$x+1=28\quad$ and $\quad y+1=2$
or
$x+1=56\quad$ and $\quad y+1=1$
Hence,
$x+y\in\big\{55,28,16,13\big\}\,.$
In the case that the OP refers to $\,\Bbb N\,$ as the set of all positive integers, it results that, $\,x+y\neq55\,,\,$ consequently,
$x+y\in\big\{28,16,13\big\}\,.$
$x+y+xy=55$
$x+y+xy+1=56$
$(x+1)(y+1)=56$
$56=1×56=2×28=4×14=7×8=8×7=4×14=28×2=56×1$
Possible values of $(x,y)$ and $x+y:$
We find four unique values for the sum: $$x+y=13\text{ or }16\text{ or }28\text{ or }55.$$
