So you're going to get some combination of the i and j unit vectors.

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So, minus 2 3 x, you go down 2 for every 3 that you go out.

We'd also want to multiply the right hand side by x.

You multiply that times negative 8

Someone has to say, well, when x is 2, y is 3.

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And so, if we just put this on the other side of the equation, what do we get?

And then we distribute again.

And now I evaluate this at 1 and 0.

let's just divide by a again.

So this was c plud d i, so this is its conjugate.

We multiply the denominators 3 times 2 is 6.

We can see that we have one text node.

3 minus nothing is 3.

Transform of any arbitrary function very, very quickly.

Is greater than 3 minus 7 which is negative 4.

And if we factor out a 5, what do we get?

That's what multiplying times 10 to the negative 8 did for us.

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We have AC and DC, get 12 volts DC, so that could be used for certain applications.

4 and negative 21 look tantalizing, but when you add them, you get negative 17.

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We're multiplying our f of t times e to the 1t.

And we can verify. Let's try all of the different scenarios again.

The latency distance, though, is quite high. It's 2098 kilometers.

But this is hypothesis generating.

Now let's go to the Interactive Developer Guide.

This 10 times 5 which is 50 and you would have gotten the same result.

Let's factor out the x plus 7.

If you divide negative 16 by 8, you get negative 2, times 3 is a negative 6.

5 times 9 is 45.

Now if I multiply this whole thing times dy,

The next time through the loop, i is now going to be 2.

Got it.

Here we go the other way.

And then we can essentially just add and subtract them from the center.

So we could say a is equal to negative 2 same idea there and then you square it.

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I'm going to subtract 12 from it, but I could get to any negative value.

4 times 9 is 36.

And you press 7 and click on log, it'll give you this number.

And then you'd get x could be equal to negative 8, or x could be equal to negative 10.