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19 March, 23:12

Solve the system, using substitution. 3a-b = 11, 2a+3b=0.

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Answers (2)
  1. 20 March, 00:19
    0
    3a-b=11; 2a+3b=0

    Solve 3a-b=11 for b

    3a-b+-3a=11+-3a (Add - 3a)

    -b=-3a+11

    -b - 1 = - 3a+11 / - 1 (Divide - 1)

    b=3a-11

    Substitute (3a-11) for b in 2a+3b=0

    2a+3b=0

    2a+3 (3a-11) = 0

    11a-33=0 (Simplify)

    11a-33+33=0+33 (Add 33)

    11a=33

    11a/11 = 33 / 11 (Divide by 11)

    a=3

    Substitute (3) for a in b=3a-11

    b=3a-11

    b = (3) (3) - 11

    b=-2 (Simplify)

    b=-2 and a=3
  2. 20 March, 01:05
    0
    3a-b=11 = > b=3a-11

    2a+3b=0

    2a+3 (3a-11) = 0

    2a+9a-33=0

    11a=33

    =>a=3

    => b=3*3-11=9-11=-2
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