3.7 Systems of Three Equations Substitution

1 x 4 y 11 3 x 4z 7 y 5 x 2z 25 2 x 4 y

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Unformatted text preview: ______________________ Solving Systems of Three Equations w/ Substitution Date________________ Period____ Solve each system by substitution. 1) x = 4 y − 11 − 3 x + 4z = − 7 y = − 5 x + 2z + 25 2) x = −4 y + 4z + 4 z = 5 x − 25 − 2 x − 5z = 17 (5, 4, 2) 3) z = −4 x + 4 y + 13 x + 2 y − 2z = 10 x = 2z + 10 (4, −5, −5) 4) z = −2 x + 5 y + 24 x = 3 y − 3z + 21 5 y − 3z = −24 (4, 0, −3) 5) −5 x − 3 y + z = −4 − 2 x − 2 y + 2z = 4 z=x+5 (3, −3, 3) 6) −4 x + 2z = 14 y = x + z + 12 − 2 x − 4z = 22 (0, 3, 5) (−5, 4, −3) -1- 7) 3 x − 3 y = − 6 z = −3 x − 3 y + 9 −4 x + 5 y + z = 8 8) x = −5 y + 4z + 1 x − 2 y + 3z = 1 2x + 3 y − z = 2 (1, 3, −3) 9) −2r − 2s + 2t = −4 −4r + 2t = −16 r + s + 6t = −12 No unique solution 10) −6 x + y − 4z = 3 5 x − 3 y = −8 − x − 5 y = −4 (3, −3, −2) (−1, 1, 1) 11) −4 y + 5z = −19 5 x − 5 y − 6z = 8 2 x + z = −5 12) −4r + 2s = 12 4r − 4s − 2t = − 4 −4r + 3t = −10 (−1, 1, −3) (−2, 2, −6) -2-...
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This note was uploaded on 07/01/2013 for the course MATH Linear Alg taught by Professor Shia during the Fall '12 term at Long Branch High.

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