# math

Solve the following linear system of equations.
3x-2y+z=2
x-y+z=2
5x+10y-5Z=10

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1. What have you done at school, substitution, elimination or echelon form?

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2. substitution

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posted by Nick
3. Take the last equation and divide all by 5 to simplify it, so:

3x-2y+z=2 ...(1)
x-y+z=2....(2)
x+2y-Z=2...(3)

By inspection, if we add (1) and (3), we eliminate y and z at the same time:
3x-2y+z+x+2y-z = 2+2
4x=4
x=1

Substitute x=1 into (2) and (3) to get:
-y+z=1 ...(2a)
2y-z=1 ....(3a)

y +0 = 2, so
y=2
Substitute x=1, y=2 in (2) to get
1-(2)+z = 2
so
z=3

Now substitute x=1, y=2, z=3 into equations (1), (2) and (3) as a check.

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