- #1

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2x-2y+z=0

-2x+y+z=0

- Thread starter TonyC
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- #1

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2x-2y+z=0

-2x+y+z=0

- #2

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I guess I should have also put:

How can I break this down?

How can I break this down?

- #3

FredGarvin

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- #4

TD

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You'll get an infinite number of solutions, for every t (so z), you have a couple (x,y).

- #5

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The answers to choose from are:

x=3/4t, y=-t,z=1/2t

x=-3/4t,y=-t,z=1/2t

z=-3/4t,y=t,z=1/2t

z=3/4t,y=t,z=1/2t

This is why I am stumped.

- #6

TD

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In your case, y was substituted for t. Then solve it as if t was a parameter for x and z.

- #7

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? I am still confused.

- #8

TD

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[tex]\left\{ \begin{gathered}

2x - 2y + z = 0 \hfill \\

- 2x + y + z = 0 \hfill \\

\end{gathered} \right[/tex]

Substitute y = t, t is now a parameter, and solve the following (2x2)-system for x and z

[tex]\left\{ \begin{gathered}

2x + z = 2t \hfill \\

- 2x + z = - t \hfill \\

\end{gathered} \right[/tex]

- #9

HallsofIvy

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(Am I the only person who

- #10

TD

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(No :yuck:)HallsofIvy said:(Am I the only person whohatesmultiple choice questions in mathematics?)

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