Solving a system of equations

[instawin]

Hello everyone. I'm having some issues finding all of the solutions to this system:

108z - 4yz - 2z^2 = 0
108y - 2y^2 - 4yz = 0

I've been able to solve it for the 3 of the solutions:
y = 0 and z = 0 (written as the pair (0, 0))
y = 54 and z = 0 (written as the pair (54, 0))
y = 0 and z = 54 (written as the pair (0, 54))

but there is one more solution I haven't been able to get: y = 18 and z = 18 (or (18, 18)). I'm guessing that I'm seriously overthinking this, but I have been working on this for a long time and I don't see how to get that 4th answer (18, 18). Help or some advice for finding this one would be greatly appreciated.

 

[instawin]

My apologies, I'm new here. I don't see how I can edit my original post, so I'm just going to attach my work here if needed. Solving this system is part of a larger calc 3 problem

 

[Dr.Peterson]

Hello everyone. I'm having some issues finding all of the solutions to this system:

108z - 4yz - 2z^2 = 0
108y - 2y^2 - 4yz = 0

I've been able to solve it for the 3 of the solutions:
y = 0 and z = 0 (written as the pair (0, 0))
y = 54 and z = 0 (written as the pair (54, 0))
y = 0 and z = 54 (written as the pair (0, 54))

but there is one more solution I haven't been able to get: y = 18 and z = 18 (or (18, 18)). I'm guessing that I'm seriously overthinking this, but I have been working on this for a long time and I don't see how to get that 4th answer (18, 18). Help or some advice for finding this one would be greatly appreciated.

If y and z are nonzero, you can divide by them and get

108 - 4y - 2z = 0​

108 - 2y - 4z = 0​

That is a simple linear system to solve.

 

[instawin]

If y and z are nonzero, you can divide by them and get

108 - 4y - 2z = 0​

108 - 2y - 4z = 0​

That is a simple linear system to solve.

yeah, that was the part that tripped me up. I wasn't sure if I was allowed to divide by y and z since I thought y or z may or may not be zero. But for the problem I'm doing, y and z represent dimensions of a box in this optimization problem I'm doing. So they better be non-zero. As always, over thought it. Makes sense, thank you so much!

 

[Dr.Peterson]

yeah, that was the part that tripped me up. I wasn't sure if I was allowed to divide by y and z since I thought y or z may or may not be zero. But for the problem I'm doing, y and z represent dimensions of a box in this optimization problem I'm doing. So they better be non-zero. As always, over thought it. Makes sense, thank you so much!

Even apart from context, you can eliminate zero because you already found the solutions where y or z is zero.

Properly, I would solve the system by factoring:

z(108 - 4y - 2z) = 0​

y(108 - 2y - 4z) = 0​

Thus, either z = 0 or 108 - 4y - 2z = 0, and either y = 0 or 108 - 2y - 4z = 0. You found that if z = 0, then y = 0 or 54, and that if y = 0, then z = 0 or 54. The remaining combination is the linear system I stated.

 

[HallsofIvy]

yeah, that was the part that tripped me up. I wasn't sure if I was allowed to divide by y and z since I thought y or z may or may not be zero.

That was why Dr. Peterson said "if y and z are not 0". Once you have got those solutions, you need to look at what happens if one or the other is 0.

If y= 0 then the first equation becomes 108z - 2z^2 = 0 and the second equation is just 0= 0. That second equation tells you that y= 0 is a solution and the first equation is 2z^2- 108z= 2z(z- 54)= 0. One solution is y= 0, z= 0 and another is y= 0, z= 54.

If z= 0 then the first equation is just 0= 0 and the second equation becomes
108y - 2y^2= 0 or y^2- 54y= y(y- 54). We already had y= 0, z= 0 as a solution but now we know that y= 54, z= 0.

But for the problem I'm doing, y and z represent dimensions of a box in this optimization problem I'm doing. So they better be non-zero. As always, over thought it. Makes sense, thank you so much!