We subtract the green expression from the light blue expression and we're left with this.

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Well just on this simple diagram we have 8 million

We can't just erase them, but we can forward them back into the surviving matrix by the following simple operation we take the surviving matrix to be called omega prime and the surviving vector to be called X prime, and you can get omega prime and X prime by using the function take that is in your own matrix library.

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This will be a day long remembered.

These cancel out.

Well 141 (And it's a little bit more mental computation than you might be used to.) But 141 9 is going to be 132.

He'd still, of course, be left with 5.

And that and that cancel out, and the negatives cancel out.

The right hand side, I got rid of that 6y.

And we are left with

Now if you divide this term by 3y, all you're

If I had 3 of something, and I take away 1 of them, if I take away 1 of them, I'm left with 2 of that something.

The answer is 1 over here, 2 here, 3 here, 4 here.

On the left hand side, the x plus one's cancel out.

If you took 1 away, then of course, he'd only have 4 left.

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We have 2 times 1 2, that cancels out.

Then we can divide a squared divided by a, so you're just

For example, if the input was aaa111 , we've used the a , so now we want it to be just aa111 .

Well, this x plus two and this x plus two will cancel out.

Down here we can factor out a 2x let me make it in that same color. We can factor out a 2x plus 5.

This is 1 million, this is 750,000.

So, it's logical that you'd have one there, on the minor road entering the major.

E can be rewritten by INT. E can be rewritten by INT.

And then something very dramatic happened.

But most the time people just realize they like leaving things in terms of pi.

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These guys cancel out.

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Now we could subtract x from both sides.

If we subtract 2 from both sides, we are going to be left with on the left hand side we're going to be left with negative 7x.

You might know that fat has memory.

So, our answer is 2x plus 1 goes into 8x to the third minus 7x squared plus 10x minus 5 it goes into it 4x squared minus 11 2 x plus 31 4 times.

This is really a great way to program. It's very declarative. You just say what you want, and the system makes it for you.

Now you can kind of call it our intermediate remainder is what's left.

I'm going to be left with 5 ys

We see 2 goes into 13 six times, 6 times 2 is 12, you subtract that from that, you get a remainder of 1.

So let's say I have just a bunch of I'll just make another A, G, and it goes down, down, down, and you have a T and then a C and a C, and let's say I have 4,500 of these.

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If we canceled these out, you're just left with

LADY CAPULET We follow thee. Exit Servant.

So when y is equal to negative 1, you're just left with x minus 1 squared over 16 is equal to 1.