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15 January, 12:51

In the solution to this system, what is the value of y?

x+y+z=1 x+y+z=1

2x+y-z=8 2x+y-z=8

x-y+z=-5

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  1. 15 January, 13:22
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    If you have multiple equations with multiple variables, you can either do clever substitutions, or turn it into a matrix on which you can perform linear combinations or multiplications (Gauss elimination)

    1 1 1 1

    2 1 - 1 8

    1 - 1 1 - 5

    (note how the above 3 rows represent the 3 equations, just got rid of the variables, plus sign and equals sign)

    subtract row1 from row3, that eliminates x and z from row 3.

    1 1 1 1

    2 1 - 1 8

    0 - 2 0 - 6

    divide row3 by - 2, that will give y a factor of 1

    1 1 1 1

    2 1 - 1 8

    0 1 0 3

    The last row now says y=3
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