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**HallsofIvy** Or, more "unsophisticatedly" (my favorite kind- although it usually involves more work) solve the equations directly.

For example, if we add twice the first equation, 2x+ 2y+ 4z= 0, to the second equation, -2x- ay+ z= 4, we eliminate x: (2-a)y+ 5z= 4.

We can also eliminate x by adding -2a times the first equation, -2ax- 2ay- 4az= 0, to the third equation, $\displaystyle 2ax+ 3a^2y+ 9z= 4: (3a^2- 2a)y+ (9- 4a)z= 4$.

Now that you have two equations in y and z only, you could combine them to eliminate, say, z, leaving an equation for y only (and a, of course). Solving that equation for y will involve dividing one expression in a by another. There will be no solution if and only if the denominator is 0 but the numerator is not.