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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

1 Answer

3 votes
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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