Solve $\begin{cases}u+(u+1)^2=20\\u^2+(u+1)=20\end{cases}$ This question is related to the previous question I posted today. So I am trying to solve $$\begin{cases}u+(u+1)^2=20\\u^2+(u+1)=20\end{cases}.$$
The system is equivalent to $$\begin{cases}u^2+3u-19=0\\u^2+u-19=0\end{cases}.$$ We can actually solve the two equations for $u$ and see if they have common solution(s). But I tried to substract the the two equations to get $$2u=0\\ \Rightarrow u=0.$$ Actually $u=0$ isn't root to any of the equations. Is this a contradiction? What do we actually get when we substract the equations? (or when we substract 2 equations in general) What does this mean?
 A: You've showed that the only possible solution to your system of equations is $u=0$. But $u=0$ is not in fact a solution. So there are no solutions.
Since this is a system of two equations in only one variable, you should not find this fact surprising! You'd have to get very lucky in order for the system to have any solutions.
A: Visually, here's what's happening:
$\hskip{3cm}$
The red curve is $y=x+(x+1)^2$, and the green is $y=x^2+(x+1)$.
The curves do intersect, namely at $x=0$. But the value of $y$ at that point of intersection is not equal to $20$.
A: From the equations you immediately draw $u=u^2$, but neither $0$ nor $1$ are solutions of the system.
Substracting the equations does not introduce new solutions, but if you drop the other equations, you do.
A: By subtracting the two equations you  never used the fact they originally equaled $20$. You merely used the fact the expressions on LHS are equal, which is indeed true for $u=0$. The constant number on RHS got cancelled while subtracting...
A: Not a contradiction: not every system of equations has a solution.
Your calculation shows that if the system has a solution, then $u=0$. And since $u=0$ is not a solution, it follows that your system has no solution.


But I tried to subtract the two equations to get...

When you do subtraction, you are assuming that there is a solution $u$ that satisfies both equations, which is not necessarily true.
Consider a much simpler example.
You try to solve
$$
2u+1=3,\quad u+1=3
$$
You can solve the two equations for $u$ to see if they have a common solution.
You can also subtract the second equation from the first one to get
$$
u=0
$$
which is not a solution to both equations.
A: If we introduce the variable $v$ and the constraint $v=u+1$, then the system of equations may be written as
\begin{align}u+v^2&=20\\u^2+v&=20\\ v-u&=1 \end{align}
Plotting these equations (below) we obtain two parabolas and a line. The parabolas happen to intersect in four points, none of which happen to land on the line. As such there's no solution to all three equations.

