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Is the system of[br]linear equations below

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consistent or inconsistent?

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And they give us x[br]plus 2y is equal to 13

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and 3x minus y is[br]equal to negative 11.

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So to answer this[br]question, we need

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to know what it means to be[br]consistent or inconsistent.

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So a consistent[br]system of equations.

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has at least one solution.

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And an inconsistent system of[br]equations, as you can imagine,

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has no solutions.

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So if we think about[br]it graphically,

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what would the graph of a[br]consistent system look like?

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Let me just draw a[br]really rough graph.

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So that's my x-axis,[br]and that is my y-axis.

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So if I have just two[br]different lines that intersect,

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that would be consistent.

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So that's one line, and[br]then that's another line.

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They clearly have[br]that one solution

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where they both[br]intersect, so that

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would be a consistent system.

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Another consistent[br]system would be

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if they're the same[br]line, because then they

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would intersect at[br]a ton of points,

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actually at an infinite[br]number of points.

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So let's say one of the[br]lines looks like that.

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And then the other line is[br]actually the exact same line.

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So it's exactly[br]right on top of it.

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So those two intersect at[br]every point along those lines,

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so that also would[br]be consistent.

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An inconsistent system[br]would have no solutions.

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So let me again draw my axes.

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Let me once again draw my axes.

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It will have no solutions.

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And so the only way[br]that you're going

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to have two lines[br]in two dimensions

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have no solutions is if[br]they don't intersect,

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or if they are parallel.

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So one line could[br]look like this.

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And then the other line[br]would have the same slope,

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but it would be shifted over.

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It would have a[br]different y-intercept,

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so it would look like this.

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So that's what an inconsistent[br]system would look like.

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You have parallel lines.

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This right here is inconsistent.

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So what we could do is[br]just do a rough graph

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of both of these lines[br]and see if they intersect.

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Another way to do it is,[br]you could look at the slope.

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And if they have the same slope[br]and different y-intercepts,

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then you'd also have[br]an inconsistent system.

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But let's just graph them.

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So let me draw my x-axis[br]and let me draw my y-axis.

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So this is x and then this is y.

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And then there's a couple[br]of ways we could do it.

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The easiest way is really[br]just find two points

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on each of these that satisfy[br]each of these equations,

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and that's enough[br]to define a line.

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So for this first one, let's[br]just make a little table

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of x's and y's.

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When x is 0, you have[br]2y is equal to 13,

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or y is equal to 13/2, which[br]is the same thing as 6 and 1/2.

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So when x is 0, y is 6 and 1/2.

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I'll just put it[br]right over here.

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So this is 0 comma 13/2.

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And then let's just see[br]what happens when y is 0.

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When y is 0, then[br]2 times y is 0.

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You have x equaling 13.

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x equals 13.

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So we have the point 13 comma 0.

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So this is 0, 6 and[br]1/2, so 13 comma 0

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would be right about there.

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We're just trying to[br]approximate-- 13 comma 0.

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And so this line right[br]up here, this equation

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can be represented by this line.

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Let me try my best to draw it.

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It would look[br]something like that.

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Now let's worry about this one.

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Let's worry about that one.

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So once again, let's make a[br]little table, x's and y's.

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I'm really just looking for[br]two points on this graph.

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So when x is equal to[br]0, 3 times 0 is just 0.

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So you get negative y[br]is equal to negative 11,

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or you get y is equal to 11.

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So you have the point 0, 11, so[br]that's maybe right over there.

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0 comma 11 is on that line.

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And then when y is 0,[br]you have 3x minus 0

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is equal to negative 11, or[br]3x is equal to negative 11.

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Or if you divide[br]both sides by 3,

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you get x is equal[br]to negative 11/3.

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And this is the exact same[br]thing as negative 3 and 2/3.

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So when y is 0, you have x[br]being negative 3 and 2/3.

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So maybe this is about[br]6, so negative 3 and 2/3

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would be right about here.

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So this is the point[br]negative 11/3 comma 0.

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And so the second equation will[br]look like something like this.

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Will look something like that.

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Now clearly-- and I might have[br]not been completely precise

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when I did this hand-drawn[br]graph-- clearly these two

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guys intersect.

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They intersect right over here.

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And to answer[br]their question, you

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don't even have to find the[br]point that they intersect at.

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We just have to[br]see, very clearly,

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that these two lines intersect.

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So this is a consistent[br]system of equations.

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It has one solution.

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You just have to have at least[br]one in order to be consistent.

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So once again, consistent[br]system of equations.

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