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Question: Find whether the solution set of $$\begin{cases}2x = 1\\ y + 5 = x\\ x = y + 3\end{cases}$$ is a singleton.

My attempt: Rewriting the first equation will give us $x = \frac{1}{2}$.

The other two equations can be written as $x - y = 5$ and $x - y = 3$.

Now, the solution of these equations is $\emptyset$. So we can say that the solution set of $$\begin{cases}2x = 1\\ y + 5 = x\\ x = y + 3\end{cases}$$ is not singleton.

Kindly verify and throw some insights.

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    $\begingroup$ A singleton is a set with one element. However, as you pointed out, the solution set of your equation is the empty set, which by definition does not have elements. So, the only thing you can say is the solution set is $\emptyset$, or equivalent, that the problem does not have solutions. $\endgroup$
    – Darío G
    Mar 22, 2016 at 10:09
  • $\begingroup$ @Wore yeah I was thinking about that too because there is no value of $x$ and $y$ for which all three equations will be true. $\endgroup$
    – Heisenberg
    Mar 22, 2016 at 10:11

2 Answers 2

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The equations you gave represent three lines in $\mathbb R^2$. When we have three (or more) lines in a plane, we can have one of the following situations:

  1. They have infinitely many common points.
  2. They have one point in common.
  3. They have no points in common.

The lines $2x=1$ and $y+5=x$ are clearly not parallel (can you see this?). So they intersect at one point, which means we are not dealing with case one. The lines will intersect at the point $(x,y)=\left(\frac12;-4\frac12\right)$.

Now the three lines will have one point in common if $\left(\frac12;-4\frac12\right)$ lies on the third line and no points in common when it doesn't. I will leave it to you to figure out if it is on the third line or not.


After edit from OP: We can solve this even quicker if we just look at the bottom two equations: $$\begin{cases} x=y+5\\ x=y+3\end{cases}$$ These lines are obviously parallel (and don't coincide). You could also argue that this implies $y+5=y+3$, which is a contradiction.

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  • $\begingroup$ @gebruiker so it all comes down to that the equations will have a solution geometrically but not algebraically. And further the solution set of the given equations will not be sigleton either way. Am I right here? Correct me if I'm wrong or missing something. $\endgroup$
    – Heisenberg
    Mar 22, 2016 at 10:54
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    $\begingroup$ No. Using either method, you can find that they don't have any point in common. Looking at it geometrically is often just a nice way to convince yourself that what you're doing makes sense. In fact, if they answer you found algebraically is in contradiction with what you would find geometrically, then you probably made a mistake. $\endgroup$
    – gebruiker
    Mar 22, 2016 at 10:58
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A vertical line ( first equation ) cuts two parallel lines ( last two ) transversely, creating two points of intersection or two solutions.

If there were three (non-concurrent) lines then there would be three points of intersection enclosing a triangle.

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    $\begingroup$ But there is no point where all three lines will intersect. $\endgroup$
    – Heisenberg
    Mar 22, 2016 at 11:26
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    $\begingroup$ It is a very special case that was mentioned. $\endgroup$
    – Narasimham
    Mar 22, 2016 at 11:35

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