##### Need algebra help with 2 x 2 system or linear equations that is inconsistent or consistent dependent

 Algebra Tutor: None Selected Time limit: 1 Day

Nov 19th, 2015

First lets roughly define what inconsistent and consistent independent mean.

Inconsistent would mean that no solution exists, i.e. both equations have no point of intersection.

Consistent Dependent means there is exactly one solution - There is exactly one point (x,y) where the two equations intersect.

Looking at the top problem...

First solve for x:

-4x=0  --->  x=0

Now find y by plugging in x=0 to both equations:

0-y=-2  --->  y=-2

0+y=-2 ---> y=-2

Since the y values are the same for x=0, we can say that this system is consistent dependent, with the one point of intersection being at (0,-2).

Now the bottom one...

Rearrange the system to get x and y on the same side (not necessary, but for clarity)

-2x+y=-6

2x-y=-6

Notice that when solving for x, you get 0=-12.

And similarly when solving for y.

This means there are no points of intersection, making this system inconsistent.

Nov 20th, 2015

So for the 1st problem the system has a unique solution which is (x,y) = 0, -2

Nov 20th, 2015

Yes. You can also graph it, or any other system of equations, to double check.

Nov 20th, 2015

Actually my apologies the 1st one would have two solutions of (0,2) and (0,-2).  This is consistent dependent where more than one solution exists. Consistent Independent is where there is only one solution.

Nov 20th, 2015

We got that marked wrong.  Thanks anyway was told correct answer is:  (x,y)= 1,0

Nov 20th, 2015

Yes I am completely wrong that would be correct I'm very sorry. Please keep the \$1 you set out for me to answer this.

Nov 20th, 2015

not necessary I still gave you a good review. Thanks for everything

Nov 20th, 2015

You are very welcome! I can assure you I will not make errors like this in the future.

Nov 20th, 2015

We are all human, that being said we all make mistakes every now and then

Nov 20th, 2015

Yes thanks for understanding. If I had the slightest doubt about not being able to answer this correctly then I would not have tried. You are the first one I have helped here and I hope I can display myself as a more reliable math tutor here on out.

Nov 20th, 2015

I am sure you will!!!!!!!!!!!!!!!!!

Nov 20th, 2015

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Nov 19th, 2015
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Nov 19th, 2015
Dec 3rd, 2016
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