# Thread: Solve 4x-y+2z=1

1. ## Solve 4x-y+2z=1

How do we solve $\displaystyle 4x-y+2z=1$? What does it even mean to "solve" this? (I can solve 4x=1). Thanks.

2. ## Re: Solve 4x-y+2z=1

You can't solve for a unique solution. All solutions (x,y,z) lie on a plane.

3. ## Re: Solve 4x-y+2z=1

Oh, hehe. I misunderstood the question on an honors precalculus study guide. I thought the question was:
Solve $4x-y+2z=1$
Solve $x+3y-z=2$, and
Solve $7x+4y+5z=3$.

The question is actually:
Solve the system
$4x-y+2z=1$
$x+3y-z=2$
$7x+4y+5z=3$.

Which is easy. Stupid mistake.