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anonymous

  • 5 years ago

How do you solve x+y+2=6 x-y+2z=7 2x-y-4z=-9 step by step please

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  1. anonymous
    • 5 years ago
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    you can start off by writing the system in augmented matrix form. Then, solve it by gaussian elimination and back substitution method : )

  2. anonymous
    • 5 years ago
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    i dont know how to do that lol can you explain further please i am in algebra 2 and this is very confusing to me

  3. anonymous
    • 5 years ago
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    x+y+2-2=6-2 xy=4

  4. anonymous
    • 5 years ago
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    An augmented matrix is associated with each linear system like {█(x+y+2=6@x-y+2z=7@2x-y-4z=-9)┤ . Then the system can be written as {█(x+y =4@x-y+2z=7@2x-y-4z=-9)┤ The matrix associated to this system can be written like that ■(1&1&0@1&-1&2@2&-1&-4) = ■(4@7@-9) to the left of the equal is called the coefficient matrix. First multiply row 2 by (-1). Then add row 1 to row 2 and replace row 2 with the result. ■(1&1&0@0&2&-2@2&-1&-4) = ■(4@-3@-9) .Multiply row 3 by – ½. Then add row 1 to row 3 and replace row 3 with the result. ■(1&1&0@0&2&-2@0&3/2&2) = ■(4@-3@9/2) . Multiply row 3 by – 4/3. Then add row 2 to row 3 and replace row 3 with the result. ■(1&1&0@0&2&-2@0&0&-14/3) = ■(4@-3@-14) . Multiply row 3 by – 3/14.Then multiply row 2 by ½. Then add row 3 to row 2 and replace row 2 with the result. ■(1&1&0@0&1&0@0&0&1) = ■(4@3/2@3) . Multiply row 1 by –1. Then add row 2 to row 1 and replace row 1 with the result. ■(-1&0&0@0&1&0@0&0&1) = ■(-5/2@3/2@3) . Multiply row 1 by and get the final result. The solution is (5/2, 3/2, 3).

  5. anonymous
    • 5 years ago
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    I hope you can 'decode' what I send to you.:) Everything is written as matrix.

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