# 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?

• @mrsamy, what do you mean? The system that I am talking about is only with $u$ as a variable. – Medi Feb 1 at 16:09
• If we arrive at a contradiction, the system of equations have no solution. – player3236 Feb 1 at 16:09
• I mean which previous question are you referring to? – user9464 Feb 1 at 16:09
• @mrsamy, I am talking about the second case when $v=u+1$. Substituting into the original system gives:... – Medi Feb 1 at 16:09
• That's really not the point. I think my question is clear even if we don't have the context. I just thought it would be fine if I wrote it. – Medi Feb 1 at 16:14

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.

• Thanks for the downvote. – Yves Daoust Feb 1 at 16:18
• I mean, I saw that it were only 4 seconds you wrote the answer and it got downvoted, but I upvoted it :). – Light Yagami Feb 1 at 16:19
• @LightYagami: I can accept downvotes, provided they are explained. – Yves Daoust Feb 1 at 16:20
• Yes, true, people don't explain why they downvote, maybe because there were already 5 answers written, so you didn't add anything useful (maybe). – Light Yagami Feb 1 at 16:22
• @IDoktorova: whatever you do with the equations ! – Yves Daoust Feb 1 at 16:36

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.

• Can you clarify for me why by substracting these 2 equations and getting $u=0$, this means only $0$ is a possible solution? – Medi Feb 1 at 16:17
• Essentially, taking the sum or difference of two equations is an operation that is always valid, but might "forget" some information about their solutions. That is, if $u$ is a solution to two equations, then $u$ must also be a solution to their difference — but if $u$ is a solution to the difference of two equations then $u$ need not be a solution to each equation individually. So your manipulations have shown that $u=0$ is the only possible option, but not that $u=0$ is an actual solution.. – Micah Feb 1 at 17:33

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 $$20$$.

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...

• I am not sure I understand why by subtracting the two equations I never used the fact they originally equaled $20$. I am substracting $u^2+3u-19=0$ and $u^2+u-19=0$, right? – Medi Feb 1 at 16:12
• @IDoktorova: Yes, you are. But In that process, did those equations included the information that they equalled $20$? No, right? So, the only information that was contained was that $u^2+3u-19$ and $u^2+u-19$ are equal, which is true for $u=0$. – Light Yagami Feb 1 at 16:14
• @IDoktorova: If however, somehow you plugged the value of $u$ by solving one equation into the other equation, you would get no solutions directly, because now both things have been used... (one that $u^2+u-19$ and $u^2+3u-19$ are equal and also that they equal to $20$ simultaneousy. – Light Yagami Feb 1 at 16:17
• Thank you!!!!!!! – Medi Feb 1 at 16:24

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.

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.

• Thank you! Can you explain to me why by doing subtraction, I am assuming that there is a solution $u$? – Medi Feb 1 at 16:25
• @IDoktorova: you are welcome. When you "subtract" something from another you need to have two things in the first place. In the simpler example above, when you say subtract the equation $u+1=3$ from $2u+1=3$, you mean that there is a $u$ such that both equations hold in the first place; otherwise, you have nothing to do the subtraction. To give you another example, consider two equations: $u^2+1=0$ and $2u^2+1=0$. Both equations have no solution. But if you do subtractions, you get $u^2=0$ which has a solution. – user9464 Feb 1 at 16:28