# What am I doing wrong here?

Consider this system of equations:

$$\begin{cases} x+y=6\\x-y=5\\2x+3y=7 \end{cases}$$

This is an overdetermined system and doesn't have a solution (the 3 lines don't intersect). But by adding 2nd and 3rd equations we get

$$\begin{cases} x+y=6\\3x+2y=12 \end{cases}$$

which has the solution $(x,y)=(0,6)$. Obviously my logic is flawed as the answer is not correct. But where is my mistake?

• $\;6+7=13\;$ ...But this doesn't really matter: you get a solution for the system with the second and third equations...and now you must check that solution also satisfies the first equation, as your original system has three equations. No contradiction here. May 3, 2014 at 15:18
• when you add 2nd and 3rd equation you get 3x+2y=13. May 3, 2014 at 15:21
• Corrected. Thanks May 4, 2014 at 4:08

$6+7=13$, not $12$, but that doesn't solve the problem, the problem is that you cannot just throw away an equation,

$$\begin{cases} x+y=6\\x-y=6\\2x+3y=7 \end{cases}$$

is equivalent to

$$\begin{cases} x+y=6\\x-y=6\\3x+2y=13 \end{cases}$$

(which has no solutions as well).

A solution must satisfy all the 3 equations simultaneously, the "solution" $(0,6)$ satisfies only 2 of them, you can reduce your system to a system of 2 equations while doing calculations, but then you have to remember to check if the solutions you found in this way satisfy also the third equation

• Thanks. I know my solution is incorrect and I know the correct answer. But you are throwing away an equation as well?! May 4, 2014 at 4:10
• No, I'm adding two of them, but I still have 3 of them. Adding a nonzero multiple of an equation in the system to another equation in the system doesn't change the solutions. May 4, 2014 at 6:31
• Yeah that's exactly my question. Why does that operation not affect the system, even though we are basically replacing one line with another? May 6, 2014 at 5:30

You essentially have this picture.

Your equations are the the red, green and blue lines, and to solve we need to find the point where all three intersect. There is no such point.

When you combine the second and third equations, you get $3x+2y=13$ (not $12$) which is the purple line. This passes through the intersection of the green and blue lines, but otherwise is different. The "solution" you found is where the purple line intersects the red line, but since the red line does not include the intersection of the green and blue lines, this "solution" does not satisfy the second or third equations.

In the same way, by just looking at the first two equations, you would obtain $x=5.5$, $\>y=0.5$; but these values don't satisfy the third equation. But there is no flaw in the logic – neither in your example, nor in mine.

A system $\Psi$ of equations defines implicitly a solution set $S$: Any point ${\bf x}$ of the ground set $X$ agreed upon in advance can easily be tested whether it satisfies all the equations. When it does, then ${\bf x}$ belongs to $S$. Solving such a system $\Psi$ means converting this implicit representation of $S$ into an explicit representation in the form of a list $S=\{{\bf x}_1, {\bf x}_2,\ldots, {\bf x}_r\}$, or a parametric representation $S=\bigl\{{\bf x}(\iota)\>\bigm|\>\iota\in I\bigr\}$, where $I$ is a specified index set, and $\iota\mapsto {\bf x}(\iota)$ is a "function expression".

Usually a solving process consists in a sequence $${\bf x}\in \Psi\Rightarrow\ldots\Rightarrow\ldots\Rightarrow\ldots\Rightarrow {\bf x}\in S'\ ,$$ where $S'$ is a certain set of points ${\bf x}\in X$ resulting in the computation, as in our example. But this proves only the following: If a point ${\bf x}\in X$ aspires to belong to $S$, then it has to be a member of $S'$, in other words: that $S\subset S'$.

We now have to test each member ${\bf x}\in S'$ (hopefully finitely many) whether it actually satisfies all the given conditions. The ${\bf x}$ that pass this test belong to $S$, the others have to be thrown away, that's all. In the example at hand no ${\bf x}$ remains, which means that in fact $S=\emptyset$. There is no logic gone astray here.

• That's the sort of mathematical answer I like! Thanks! May 6, 2014 at 5:33

$$\begin{cases} x+y=6\\x-y=6 \end{cases}$$

$$2x+3y=\frac52(x+y)-\frac12(x-y)=\frac52\cdot6-\frac12\cdot6=12\ne7$$

so your system of equations has no solutons