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My 9th grade son had this math problem, which seemed unsolvable to me:

$$2x - 2y = 11$$ $$x = y - 2$$

So we can use substitution to come up with: $$2(y - 2) - 2y = 11$$

Now distribute: $$2y - 4 - 2y = 11$$

Reduce $2y$ and $-2y$ cancel each other out

Answer $$-4 = 11$$

Did we do something wrong here, or is this math problem unsolvable?

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    $\begingroup$ What you’ve done is fine: that system is inconsistent. $\endgroup$ – Brian M. Scott Nov 18 '14 at 19:57
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    $\begingroup$ The set of equations has no solution. Whether or not the problem is unsolvable depends on the wording of the problem. It's common to give students sets of equations where they have to find the solution if there is one or explain why there isn't one. $\endgroup$ – Randy E Nov 18 '14 at 20:05
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    $\begingroup$ And, unfortunately, it's not exactly unheard of for teachers to make up numbers and not actually solve the problem first and end up with impossible solutions when they didn't mean to. $\endgroup$ – Loren Pechtel Nov 19 '14 at 5:49
  • $\begingroup$ Maybe the teacher was expecting the student to assume that $x$ and $y$ are $GF(3)$. $\endgroup$ – DanielV Nov 19 '14 at 8:23
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This system is what is called inconsistent. That means there is no solution. You can see this from the beginning by rewriting the equations as:

$$2(x-y) = 11$$ $$x-y = - 2$$

The first equation says $x-y$ should be $11/2$. The second believes $x-y = - 2$ which is obviously not compatible. This is why we call it inconsistent.

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I can't believe that nobody has said that the coordinates that satisfy these equations represent a pair of parallel lines: $$y=x+2\text{ and }y=x-\frac{11}{2}.$$ Intersections are points that are on both lines and so satisfy both equations at the same time. However as the lines are parallel there is no intersection and therefore no solution of the simultaneous equations.

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    $\begingroup$ This is most likely what was expected from the teacher $\endgroup$ – Jean-Sébastien Nov 18 '14 at 20:01
  • $\begingroup$ Does it make a difference that this was Algebra class, not geometry? $\endgroup$ – Scottie Nov 18 '14 at 20:01
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    $\begingroup$ @Scottie: In the U.S. at least, one is more likely to deal with graphing lines in a coordinate system in high school algebra than in high school geometry. It's probably going to be in the geometry class as well, but the main focus there would be on synthetic geometry rather than on analytic geometry. $\endgroup$ – Dave L. Renfro Nov 18 '14 at 20:12
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    $\begingroup$ @Scottie It matters only if the teacher makes it matter. Fundamentally, if the goal is to solve the equations or demonstrate that there's no solution, any technique that works, works. $\endgroup$ – David Richerby Nov 18 '14 at 22:50
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You are right. You can easily rearrange the second equation into $2x-2y = -4$, giving $11 = -4$. Therefore there is no solution.

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Equation 1:

2x - 2y = 11 ==> 2 (x - y) = 11 ==> x - y = 11 / 2 ==> x = (11 / 2) + y

Now, submit the value of x in the same Equation 1 gives:

2 (11/2 + y) - 2y = 11 / 2 Gives y = 0

Equation 2:

Put this is y in eq 2

x = y - 2 ==> x = 0 - 2 ==> x = -2

Solution set:

X , Y = { -2 , 0 }

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  • $\begingroup$ 2 (11/2 + y) - 2y = 11 / 2 is not equivalent to y = 0. It is equivalent to 0 = -11/2. $\endgroup$ – DudeOnRock Nov 19 '14 at 6:55
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    $\begingroup$ In fact the second line should read: 2 (11/2 + y) - 2y = 11, which gives 0=0. Note that substituting a rearranged equation into the original form of that same equation almost always results in a tautology. $\endgroup$ – IanF1 Nov 19 '14 at 7:06

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