How would I work this out algebracally

[noelo2014]

These 2 equations are part of a calculus problem:
6z-2yz-z^2=0
6y-y^2-2yz=0

I have already done the problem, and I can see from some inspection that the solutions to these equations are z=0,y=0 z=6,y=0 z=0,y=6 and z=y=2

My question is how can this be worked out in a formal algebraic way? Thanks for any help

 

[lurflurf]

Such problems can be difficult in general. This one can be factored.

6z-2yz-z^2=0
6y-y^2-2yz=0

z(6-2y-z)=0
y(6-y-2z)=0

We see there are four solutions and they are easy to work out. By zero product property

z=0
y=0

z=0
(6-y-2z)=0

(6-2y-z)=0
y=0

(6-2y-z)=0
(6-y-2z)=0

 

[noelo2014]

Thanks,
(6-y-2z)=0
-2z=-6+y
2z=6-y
z=2.1 , y=1.8

but that won't satisfy 6-2y-z=0...

So it's a case of guesswork? These sort of equations come up a lot in the calculus I'm doing at the moment, in my textbook they never show how they solve these equations , they just say "y is 2", or whatever, without giving the algebra behind it.

While I'm on this subject I had a similar problem recently that involved 2 quadratic curves, one in terms of x and the other in terms of y,

Quick example: y=3x^2-20,x=y^2+x-10

If anyone knows some clues or tricks for solving these I'd love to hear them, the only way I can solve them atm with my limited time and knowledge is Wolfram Alpha

Cheers

 

[lurflurf]

^I don't follow your work.
There is no guessing.
You need to solve two equations at a times as in my above post.
Maybe drawing the graph will help.

 

[noelo2014]

Sorry the equations I gave were wrong, I'll post them another time :)