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January 30, 2015

January 30, 2015

Posted by **jasmine20** on Saturday, December 23, 2006 at 12:51pm.

8x –4y = 16

y = 2x –4

My answer:

This problem does not have a unique solution.

This problem therefore is consistent and dependent

These equations are the same. If you solve the first one for y, you will see that the two equations end up equal.

Does that help?

Amy :)

Amy is correct.

Jasmine, you are correct when you say it does not have a UNIQUE solution. There are an infinite number of solutions, and they lie along the line y = 2x –4

The other equation amounts to the same thing, and graphs as the same line.

see so should i just leave it as the answer as i have it or the result of solving the equation which was:

0 = 0

The answer should be one of the two equations, e.g.

y = 2x –4

and not 0 = 0.

The 0 = 0 is obtained because the two equations are equivalent. Suppose that one equation had been a bit different. Then, instead of the 0 = 0, you could have obtained, say, x = 5. But to find the value of y you still need one of the two equations and plug in x = 5 in there. So, after eliminating variables by combining the two equations you still need either one of the two original equations.

In this case this means that the original system of the two equations is equivalent to:

y = 2x –4 AND 0 = 0

But the statement 0 = 0 is always true and thus doesn't constrain x and y any more than the first equation:

y = 2x –4 AND 0 = 0

is equivalent to just:

y = 2x –4

The 0 = 0 is just the maths telling you that the extra information you were searching for that would have specified the value for x isn't there.

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