It's a rectangle.

Here is what I think we need to do here.

So this right here, there's no constraints.

So let's do something very similar to what we did in the past.

So 32 minus 8z squared.

And I'll get x is equal to 3.

We've now factored the numerator and the denominator.

So let's look at what that means.

And so what can we factor out of that group right there?

But we say, look, both of these things have to be true.

Let's say that they're functions of t.

When x is 0 here, 0 plus 3 is equal to 3.

You could get fairly sophisticated with your mathematics.

Negative two ( 2) times negative seven ( 7).

Here it's minus 2 as well.

And that's actually the y intercept.

And you might have been able to solve that in your head.

The absolute value of negative 11, only 11 away from 0.

So we can already see the bottom of our rectangle that they're talking about, right there.

Now, there's two ways to do this.

There weren't x's that satisfy these two constraints.

You go 1 2 3 minus 3, minus 1.

Let's just say what the area squared is equal to.

But instead, what we can do is, we can say, hey look, this

There's three ways.

So I've written these two expressions.

So they would define, could define this line right over here but both of these points, and in fact this entire line, exists on both of these planes that I just drew and I could keep rotating these planes

We have x squared on both sides of this, we subtract x squared from both sides of the equation.

We can say v1 one is equal to ac, and we could write that v2 is equal to bd.

It actually doesn't take too long.

Prototyping is a strategy for efficiently dealing with things that are hard to predict.

look something like that.

Because you could see 10 goes into the 20, 10 goes into 10.

So I could have just as easily drawn my vector like that.

But maybe we can if we subtract 2 from both sides of this equation.

625 is equal to r squared.

And you just say, OK, I see.

You'll get 2x plus 63 is less than negative 5.

He wasn't down for dinner. No, and he wasn't down for lunch.

The graphs have the same shape with different vertices.

We know that these are both divisible by 2.

If we subtract 4 from both sides of this equation let me switch colors if we subtract 4 from both sides of this equation, we get y minus 4 is equal to 2x, and then if we divide both sides of this equation by 2, we get y over 2 minus 2 4 divided by 2 is 2 is equal to x.

Now, what if I were to ask you, is there an x and y pair that satisfies both of these equations?

The biggest number that divides into both of these guys is five.

We're doing this case right here.