[i] … … … x + 3y - z = 9
[ii] … … … 2x + 9y + 4z = 12
[iii] … … … x + 4y + z = 7
Eliminate z by combining …
• … 4 times equation [i] and equation [ii] :
4 (x + 3y - z) + (2x + 9y + 4z) = 4•9 + 12
(4x + 12y - 4z) + (2x + 9y + 4z) = 36 + 12
6x + 21y = 48
2x + 7y = 16
• … equation [i] and equation [iii] :
(x + 3y - z) + (x + 4y + z) = 9 + 7
2x + 7y = 16
Since we ended up with 2 copies of the same equation, we have infinitely many solutions for x, y, and z. That is, we have infinitely many choices for x and y that satisfy 2x + 7y = 16, and consequently infinitely many choices for z to satisfy any of the 3 original equations.
We can parameterize the solution by letting, for instance, x = t; then the solution set is
x = t
2x + 7y = 16 ⇒ y = (16 - 2t)/7
x + 3y - z = 9 ⇒ z = t + 3 (16 - 2t)/7 - 9 ⇒ z = (t - 15)/7
where t is any real number.
