Are the equations $2x - 2y = 11, x = y - 2$ unsolvable?

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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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4 Answers

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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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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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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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