How do we solve $\displaystyle 4x-y+2z=1$? What does it even mean to "solve" this? (I cansolve4x=1). Thanks.

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- Aug 1st 2012, 05:40 PMmathDadSolve 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. - Aug 1st 2012, 08:18 PMrichard1234Re: Solve 4x-y+2z=1
You can't solve for a unique solution. All solutions (x,y,z) lie on a plane.

- Aug 2nd 2012, 06:39 AMmathDadRe: 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. :D