Can the equation $5x-3y=2$ be derived, by a sequence of Gaussian reduction steps, from the equations in this system?

\begin{array}{*{2}{rc}r}2x&+&2y&=&5\\ 3x&+&y&=&4\end{array}

This is taken from the Linear Algebra textbook on Wikibooks (problem 10, 1.1 Linear Systems)

The worked solution given is "No. The given equation is satisfied by the pair (1,1). However, that pair does not satisfy the first equation in the system."

Can someone please explain why? The way the solution is given, it's as if any ($x,y$) that satisfies $5x-3y=2$ must also satisfy any of the two equations in the given system of equations for there to be a sequence of Gaussian reduction steps to reach $5x-3y=2$. But I don't see that.

What I do know is that the theorem in the same page explains that any system of equation derived from another through Gaussian operations must have the same set of solutions. But in this case, there is only 1 equation, not a system of equations.

  • $\begingroup$ They do appear to have gotten it backwards. You must find a solution to the original system that doesn’t satisfy the new equation, not the other way around. $\endgroup$
    – amd
    Jul 4 '18 at 0:21
  • $\begingroup$ @chabra Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$
    – user
    Aug 6 '18 at 21:49

Gaussian elimination corresponds to linear combination of the equations in order to simplify a given linear system, in you example


we can multiply the first equation by 3 and the second by 2 and then subtract the first equation from the second to obtain


and then


and the latter system has the same solutions of the original one.

  • $\begingroup$ Yes I follow what you are saying but it doesn't answer the question. The question is asking if you can derive specifically 5x-3y=2 by a sequence of steps from the starting systems of equations (2x+2y=5 & 3x+y=4) $\endgroup$
    – chabra
    Jul 3 '18 at 18:38
  • $\begingroup$ @chabra No of course we can't indeed the system has solution $y=7/4$ and $x=3/4$ but $$5x-3y=15/4-21/4=-3/2\neq 2$$ $\endgroup$
    – user
    Jul 3 '18 at 18:46
  • $\begingroup$ Ah, I see. Yes, you are right. Very helpful. I think a more complete answer to the question could be: Fact 1:2x+2y=5 & 3x+y=4 => x=3/4,y=7/4. Fact 2: If 5x-3y=2 is to be obtained as a row after a sequence of Gaussian operations on the system, then the final system containing 5x-3y=2 would have the same solution set as the original (from the Gauss's theorem) Fact 3: Since x=3/4,y=7/4 is not a solution to 5x-3y=2, but IS a solution to the original system, it cannot be obtained through a sequence of Gaussian operations. (solutions sets are not same). QED. $\endgroup$
    – chabra
    Jul 3 '18 at 18:49
  • $\begingroup$ Yes exactly, that's correct! $\endgroup$
    – user
    Jul 4 '18 at 5:17

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