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15 September, 18:03

What is the solution to the following system? x+2y+=9; x-y+3z=13; 2z=10

x = 10, y = 2, z = 5

x = - 4, y = 2, z = 5

x = 0, y = 2, z = 5

x = 4, y = 6, z = 5

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Answers (2)
  1. 15 September, 18:13
    0
    2z=10

    z=10/2=5 so that is easy

    substitute z=5 in x-y+3z=13

    x-y+3*5=13

    x-y=-2

    subtract x-y=-2 from x+2y + (?) = 9

    3y + (?) = 11

    if (?) is nothing, then y=11/3 or 3.67 and x=5/3 or 1.67

    but if (?) is 1z, then 3y+5=11

    3y=6

    y=2

    x=-2+y=-2+2=0
  2. 15 September, 18:24
    0
    x + 2y + z = 9

    x - y + 3z = 13

    2z = 10

    Solve the 3rd equation for z.

    z = 5

    Now substitute z = 5 into both the first and seconds equations.

    x + 2y + 5 = 9

    x - y + 3 (5) = 13

    x + 2y = 4

    x - y = - 2

    Subtract the first equation from the second equation above.

    -3y = - 6

    y = 2

    Now substitute z = 5 and y = - 2 into the first original equation, and solve for x.

    x + 2 (2) + 5 = 9

    x + 9 = 9

    x = 0

    Answer: x = 0; y = 2; z = 5
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