How to solve this algebra II problem (solving systems of linear equations in 3 variables)?

Which values should be given to a, b, and c so that the linear system shown has (-1, 2, -3) as its only solution?

x + 2y - 3z = a

- x - y + z = b

2x + 3y - 2z = c

thanks in advance! ♥

3 Answers

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  • GMT
    Lv 6
    7 years ago
    Favorite Answer

    Given that

    (-1, 2, -3) = ((x, y, z)

    x + 2y - 3z = a

    -x - y + z = b

    2x + 3y - 2z = c

    x + 2y - 3z = a =-1 + 4 + 9 = 12 = a

    -x - y + z = b = 1 - 2 - 3 = -4 = b

    2x + 3y - 2z = c = -2 + 6 + 6 = 10 = c

    a = 12

    b = -4

    c = 10

    I hope this helps!

  • 7 years ago

    Sub in the given values into the system of equations and then determine a, b and c.

    -1 + 2(2) - 3(-3) = -1 + 4 + 9 = 12 = a

    etc.

  • curren
    Lv 4
    3 years ago

    First, use the Distributive Property on both equations: .5x + .5y + 1 = 0 .5x - .5y = 2 Now, take both equation--i'm going to use the bottom one--to remedy for a variable (i will decide upon x). .5x - .5y = 2 .5x = 2 + .5y Divide all sides via half. X = 1 + y Now substitute x into the primary equation: .5x + .5y + 1 = 0 .5(1 + y) + .5y + 1 = 0 Use the Distributive Property. .5 + .5y + .5y + 1 = zero Add like terms. 1.5 + y = zero y = -1.5 Now alternative y into the primary equation: .5x + .5y + 1 = zero .5x + .5(-1.5) + 1 = 0 .5x - .75 + 1 = zero Add like phrases. .5x + .25 = 0 .5x = -.25 Divide either side through half of. X = -.5 ok! So now you've gotten y = -1.5 and x = -.5!

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