Okay to solve this problem. First, we need to start by finding the palladium so apart, eh? Eh? So let's start with the first equation. Because this is only an equation of one variable. So we'll set that equal to zero. So here, see, one is equal to five. So we have five p one, uh, times one minus p one minus m one. Em one is equal to three. So three P one is equal to zero. So let's, um, work this out. So we have five p one minus five p one squared, minus three P one is equal to zero. So we have, uh, two p one minus five p one squared is equal to zero. Let's factor Rdp one. So we get p one times two minus five p one equal zero. Let's set each factory with zero. Then we get either p one equals zero or P one is equal to 2/5. All right, now, let's look at equation too. All right. We're gonna try to simplify equation to first as much as possible. Yes, we have 30 p too. Um, minus four times one minus P. One minus p too. And then minus I am, too. Is equal to three. So three p, too. And then minus soc one is five. So minus five p. One p. Too busy with zero. Now, we can factor out a P two from all of it. So we get 30 minus 30 p one minus 30 p too. Okay, then. Just minus three p too. Minus five p. One equal zero. So again, let's keep simplifying. We have P too. Times 30 minus 35 p one. 35 p one minus 33 p two equals zero. Whoops. Sorry. I, um Ms. Calculated here. So that should be minus three. Just minus three and then minus three minus five p one, like so is equal to zero. So then now we're left with this is gonna be 27 minus 35 p one minus just 30 p too. Okay, so we can break that into our two factors. So we have p two equals zero, or, um, we can say pee too, You know, 30 p, too is equal to 35 p one minus 27 minus 27. Like so. Okay, so now we plug in these two solutions from the first equation into both of our, um, so called this 13 This 14 rights, we plug in three and four into both of these and for so first, we have p two equals zero and p one equal zero. That's fine. Then we have p two equals zero and p two equals 2/5. So, uh, we have that. Our equilibrium points. First, we have 00 uh, 00 hips and then 0 2/5 or sorry, um, tooth ifs comma zero like so. Okay, so those aren't too. Now, let's do the, uh, second half second set. So we have P too is going to be equal to 35 times. 1st 0 minus, um zero minus 27. Divided by 30. So we just have negative 27/30 there. So we have that pee, too, is equal to 9/10. Now, if we plug in for the next one, p two is equal to 35 times 2/5 minus 27. Divided by 30. We get that pee too. P two is equal to and here we get 13/30. Okay? We get 13/30. So are two equilibrium points. are, uh, our other two. So we have four total. We have zero comma, 9/10 and then we also have to fifth comma, 13/30. So these are our four equilibrium points for part A. All right, so now for the Jacoby in, it'll be easiest to write out. Um, the equation in, um, you got that equation? Expanded form. So we have. For the first equation, we have five p one. Okay, minus five p. One squared and then minus three p one. So this is equal to F one. Um, now for the second equation, let's expand that out. I'll, um, So we can use this aversion here, Okay? Except we just multiply a P two throughout. So we have 27 p two minus 35 p one times p too, and then minus 30. Key to square. Okay. And this is able to r f, too. So to find the f one the, uh, p one, this is equal to five minus 10 p one minus three er 10 p one, minus three. So this is equal to two minus 10 p. One now. 40 f ah. One D p too. This is just going to be equal to zero. Now, let's look at the f. Uh huh. I mean, actually, put that, um, up here instead. Okay? Now for D F too. The P one. There's only one term with P one here. So is gonna be minus 35 p too. Okay. Now, for D F to the P two, this is equal to 27 minus 35 uh, 35 p one minus 60 p, too. So for part B, our answer the Jacoby in P one p two is equal to, uh so to minus 10 p one 10 p. One zero negative, 35 p too. And then 27 minus 35 p one minus 60 p too. So that's our answer for part B. Now, for part C, we need to plug in all of our equilibrium points. So I see. So we needed to take J of 00 was our first equilibrium. Point zero comma zero. So this is equal to to minus 10 times zero. Okay. Uh, zero negative. 35 times zero men 27 minus 30 was a 35 p. Once a 35 times zero, minus 20 right or minus 60 times zero times zero. Hey, this simplifies to two zero 0 27 Okay, so since this is triangular matrix, all of them will be trying to the matrices. Actually, we actually only care about these two numbers here to find the stability. Okay, so these are our, um, Aiken values here. Since they're both greater than zero, we have ah, um, unstable. We have an unstable equilibrium at 00 Okay, So unstable at, um 00 Okay, Now, for the next equilibrium point, that was 2/5 comma, zero. So to over five comma zero. Now, this is going to be equal to so to minus 10 times to fifth over. Five here. Zero negative. 35 times zero. Um, who's 27? Minus 35 times to fifth. So two divided by, uh, five and then minus 60 times zero. Simplify that. We have. So to minus 24 5 um, to minus 20/5. So to minus four. That's gonna be negative, too. Zero. Oh, see here, zero 200 And then again, um, so too minus 27 we have. Okay, So here are Aiken. Values are too and negative. 27. So, again, this is gonna be unstable, Dad. Zero common. 9/10. Okay, now our last equilibrium point. So that's J ads, Um, to fifth 2/5, comma 13/30. Okay, so here we have. So to minus, uh, 10 times. 2/5 Negative. 35 times 13/30. 27. Minus 35 times two over five. Minus 60 times searching over 30 zero. So 13/30. 13/30. Plug in our 2/5. Two very five. Right. So this gives us a kn answer of, um, here. This gives us a negative, too. Zero negative. 91/6, and then negative 13. So here are Aiken. Values are negative, too. And negative. 13. Therefore, um, this is actually going to be stable. Uh, so stable, Dad. 2/5. Um, 2/5. 13/30. So, for part B, yes, we will have, um, a part where or there will be able to exist at a stable equilibrium at, um, that last one. That last equilibrium