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Hello.
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Today i'm going to help solve the problem of this kind of unique algebraic problem where we have this grid system such that every node, so every little corner here, as i show at the black dot, labeled 1, 2, 3, 4 can be written as the sum of all of the things it's connected to.
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So for example, they label these nodes as t.
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So for example, t1 is a number.
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The 20 from the top, that's this guy, followed by the 10, plus 10, plus t4, plus t2, all that divided by four.
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So we take the four corners of each one, and we write them as such.
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This is the definition of the problem.
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So the first part is, can you write this for all four nodes, for nodes 1, 2, 3, and 4? so we just did one.
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So let's see if we can write the other ones.
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So if we could look at node two, above it is 20, plus to the right is 40 going clockwise, plus node three plus node one again, all divided by four.
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Okay, let's repeat for node three.
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We start at the top.
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So we look up first, and then we'll go around in a clockwise fashion.
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So we get t2 plus 40 plus 30 plus t4 over 4.
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And then finally for the fourth node, we have again starting at the top, t1 plus t3 plus 30 plus 10, all divided by four.
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So the first part of this problem asks just for these set of equations.
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So now we have a system of equations.
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We can combine all of the numbers here.
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So this becomes 60, for example, this becomes 30, this becomes 70, and this becomes 40, just to make it a little cleaner, we can rewrite them as such.
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So that's part a.
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So that is now done.
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Part b is much more complicated.
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Part b, we are asked to solve for all of the ts.
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So what is t1 equal? what is t2 equal, etc? so to do that, we will use substitution here.
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So we're going to, i'm going to start with t4, and i'm going to plug in the equation for t1.
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I'm going to plug in the equation for t3.
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And so what we get when we do that is t4 equals r40, what i haven't read up here, plus i'm going to substitute the equation of t1.
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So 30 plus t2 plus t4, all over 4.
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So now i've put this guy in and i've taken our 40.
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Now i have the t3 left.
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So i add then our t3 70 plus t2 plus t4 over four.
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And this whole thing goes over four.
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Now notice there's a lot of fours in the denominator floating around here.
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It's going to be easier if we convert this to something over four.
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So if we do, if we get that common denominator of four, this becomes 160.
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And so now we can add everything on the top and put them all under four, such that we end up with 160 plus 30 plus 70.
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So that's 260 plus now we have two t2s, two t2s and two t4s, all that over four.
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And that gets divided again by four, but remember when you divide, you can write them as a fraction over a fraction.
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The insides and the outsides get multiplied.
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And so what we end up with, if i take all this away, is 16 here on the bottom.
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All right.
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Now we can further simplify this, since we have a t4 on the right side.
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First, we can get rid of a factor of two everywhere, if we wish.
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Up to you however you wish to solve for it.
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But we need to get this t4 to the other side.
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So we can rewrite this, let's see, 260 over 16, plus t2 over 8, plus t4 over 8.
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So i'm going to move this guy to the other side...