How do you solve #2x + 3y - z = 1#, #3x + y + 2z = 12#, and #x + 2y - 3z = -5#?

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How do you solve #2x + 3y - z = 1#, #3x + y + 2z = 12#, and #x + 2y - 3z = -5#?

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Answer 1 · 2015-04-04T18:20:48

Have a look:

Explanation:

I would use Cramer's Rule:
enter image source here
Where the matrices #A_x,A_y and A_z# are obtained changing the columns corresponding to x, y and z.

Graphically:
enter image source here

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How do you solve #2x + 3y - z = 1#, #3x + y + 2z = 12#, and #x + 2y - 3z = -5#?

1 Answer

Explanation:

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