Simplify to have your streak!

Simpler solution:

$x+x^2+xy=14$

$y+y^2+xy=28$

Now add them to get $(x+y)+(x+y)^2=42$ . Solving this we get $x+y=6,\;-7$ .

Probably the problem poser should have mentioned the sum is positive.

First of all, the claim that none of the numbers can equal $0$ shouldn't be there... Of course, it is quite obvious that none of them is $0$ , since otherwise we would have $0=14$ and/or $0=28$ .

And second of all, $x+y$ can be equal to two different numbers, as we will see later on, so the answer to this question is not unique as you imply.

We have a system of equations: $\begin{cases}a^2+a+ab=14 \iff a(a+b+1)=14\\b^2+b+ab=28\end{cases}$

Subtract the first equation from the second: $(b^2-a^2)+(b-a)=14$ $\iff (a+b)(b-a)+(b-a)=14$ $\iff (b-a)(a+b+1)=14$ $\implies (b-a)(a+b+1)=a(a+b+1)$ $\iff (a+b+1)(b-2a)=0$ $\iff \begin{cases}b=-1-a & \text{or}\\b=2a\end{cases}$

If $b=2a$ , by simply substituting this to the initial system of equations we have $\begin{cases}\begin{cases}a=2\\b=4\end{cases} & \text{or}\\\begin{cases}a=-2\frac{1}{3}\\b=-4\frac{2}{3}\end{cases}\end{cases}$

By checking these solutions, we see that both of them are correct.

If $b=-a-1$ , we have $a^2+a+a(-a-1)=a^2+a-a^2-a=0=14$ . Hence we know that $b\neq -a-1$ .

Therefore the only solutions are: $\begin{cases}\begin{cases}a=2\\b=4\end{cases} & \text{or}\\\begin{cases}a=-2\frac{1}{3}\\b=-4\frac{2}{3}\end{cases}\end{cases}$

Which leads us to $2$ possible answers: $\boxed{6}$ and $\boxed{-7}$ .

We have our system of equations:

x + $x^{2}$ + xy = 14

y + $y^{2}$ + xy = 28

Now we can add them together, and switch the order around a bit to get:

$x^{2}$ + $y^{2}$ + 2xy + x + y= 42

And we can then factor the $x^{2}$ + $y^{2}$ in our equation to get :

$(x+y)^{2}$ + (x + y) = 42

And if we replace (x+y) with a to make it simpler and easier to understand, we get

$a^{2}$ + a = 42

And then we can subtract 42 on both sides to get 0 on one side

$a^{2}$ + a - 42 = 0

Then, we can factor $a^{2}$ + a - 42 into (a + 7)(a - 6) and get:

(a + 7)(a - 6) = 0

And through the Zero Product Property, we can conclude that

a + 7 = 0 or a - 6 = 0

So if we solve for a, that means

a = -7 or a = 6

And since a = x + y, and from the question we know that x + y cannot be negative, that means a = $\boxed{x + y = 6}$

This was my second problem today but why didn't my streak go up?