HP50g or other: U. of Houston calc exams

[Hlib]

Thanks, CR Haeger and Gilles, you rescued me from a failure at exam :-)
On a statement of the problem we have to find parameter "a" without solving system of the linear equations. You simply solved ordinary system, having replaced "a" with "z":
2x -3y = 6 - 7z
-3x + 5y = 7+ 2z
7x +11y = 4 - 6z
I have the same result. It seems to me, it is not a proved trick. We must have:
rank[[2 -3][-3 5][7 11]]=rank[[2 -3 6-7a][-3 5 7+2a][7 11 4-6a]]<=2,
therefore det[[2 -3 6-7a][-3 5 7+2a][7 11 4-6a]]=0.

Code:

[[2 -3 ′6 - 7a′ ] [-3 5 ′7+ 2a′] [7 11 ′4 - 6a′]] DET 0 = ′a′ ZEROS

a=235/128.
Initially I was going to solve quite so.
††††What would you do, if:
2x -3y = 6 - 7*SQ(a)
-3x + 5y = 7+ 2a
7x +11y = 4 - 6a ?