1. Solve for a
a(a-1)x =  a²
where x is a given number.
Solution.  a = 0 is a solution for every x.  To find other solutions,  divide the equation by a.
Then we get a linear equation    (a-1)x =  a, or   (x-1)a= x in standard form.
So   a= x/(x-1) when  x =/ 1 and  there are  no nozero solutions when x =1.
Remark. After solving this quadratic equation, you should appriciate how easy  liner equations are.

2.  xz -> max, |x-1| +  |z+ 1|  <=  5.
Solution. The feasible region can be given by 4 linear constraints. It is is diamand with vertices
(x, z) = (6, -1), (1,4), (-4, -1), (1,-6). Answer: max= 6.25 at  x = z = 2.5    or  x = z = -2.5.

3. Write LP in standard and canonical forms:

f=  2x -  y -> max,  -1 <= x <= 9, y <= 0.
Solution.  Let u = x + 1 >= 0, w = - y >= 0., s = 10 -u = 9 - x  >= 0.
Canonical form:  -f = -2u - w +2 ->  min,  u <= 10; u, w > = 0.
Standard form: -f = -2u - w + 2 ->  min,  u + s = 10; u, w, s > = 0.

4. Solve for x,y:
 

3	x	2	y
x	-1	y	-2	= x
-1	0	x	0	= -3

 Solution. The first equation in standard form is x = 0. The second equation in standard form is x = 0.
Answer: x =0, y is arbitrary.