Answer:
(x, y, z) = (1-z, z, z) . . . . . . . an infinite number of solutions
Step-by-step explanation:
Use the first equation to substitute for x in the remaining two equations.
(-5y +4z +1) -2y +3z = 1 . . . . substitute for x in the second equation
-7y +7z = 0 . . . . . . . . . . . . . . simplify, subtract 1
y = z . . . . . . . . . . . . . . . . . . . . divide by -7; add z
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2(-5y +4z +1) +3y -z = 2 . . . . substitute for x in the third equation
-7y +7z = 0 . . . . . . . . . . . . . . subtract 2; collect terms
y = z . . . . . . . . . . . . . . . . . . . . divide by -7; add z
<em>This is a dependent set of equations</em>, so has an infinite number of solutions. Effectively, they are ...
x = 1 -z
y = z
z is a "free variable"
