00:01
Okay, so here we have the parametric equations, x equals t over t minus one, and y is equal to t minus two over t plus one, and t is strictly between one, negative one and one.
00:21
So the easiest thing to do here is to just solve for t and then substitute in, because we're trying to just get y in terms of x.
00:28
So if we look, you know, if i wanted to say get t by itself, well, i need to, i could first multiply through by t minus one, say.
00:46
So x times t minus 1 equals t.
00:50
And i have xt minus x equals t.
00:54
And i can add x and subtract t.
01:00
So in other words, i can factor out a t at x minus 1.
01:06
So i get the t is x minus 1.
01:07
So i get the t is x.
01:08
Over x minus 1, which i can then just plug in for y.
01:14
So y will be x over x minus 1 minus 2 over x over x minus 1 plus 1.
01:27
And i can kind of clear these denominators by just multiplying the top and bottom by x minus 1.
01:33
So let's do that.
01:36
And so i'll get x minus 2 times x minus 1.
01:45
Over x plus x minus 1, which is, okay, x minus 2x, that's minus 2x, and then minus 2 times negative 1, that's 2...