When there are 3 variables i with 2 ( independent linear )equations , we can reduce the the equations to a two variable equation and the solutions are not unique but many.
If the equations are of the form ax+by + cz - d = 0 and k(ax+by-d) = 0, then also the solutions are many and any solution of ax+by cz-d = 0 is also a solution of k(ax+by+cz-d)
If the two equation are of the forms, ax+by+cz- d = o and k(ax+by +cz -e) = 0 , where d and e are different, then the equations are inconsistent and so there can be no solutions.
In case of higher degree equations of 3 variables we can eliminate one variable and get a single relation (or a single equation of ) two variables, which has many solutions.
If we are given 2 equations with three variables, they can be of the form
a1x+ b1y+ c1z+d1=0… (1)
a2x+ b2y+ c2z+d2=0… (2)
Now as we have 3 variables we can only eliminate one of the variables using the two equations, so we have to express two of the variables as factors of the third variable.
For example, if we perform a2*(1) - a1*(2)
We can eliminate x we are left with a2b1y+ a2c1z+a2d1- a1b2y-a1c2z-a1d2=0
=> y= [a1d2- a2d1 +a1c2z – a2c1z]/ (a2b1- a1b2)
Similarly, x also can be written as an expression with z in it. It is not possible to find numeric values of x, y and z if only two equations are given.
To solve for variables in a system of equations, we need as many unique equations as we have variables. Therefore, given 3 variables and only 2 equations its impossible to determine an exact solution for all the variables. You can only solve for a variable in terms of the other two variables.