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solve the system of equations in three variables

  • math -

    We need two more equations.

  • math -



  • math -

    Your goal is to take these 3 equations in 3 variables down to 2 equations in 2 variables then the problem becomes much easier.

    you have to decide which variable to eliminate first and you have to do this twice.

    Looking at the problem. I might try to eliminate the y's

    I will take -2 times the first equation and add it to the second equation.

    6x -2y + 2z = 20
    -4x +2y +3z = -1

    This reduces to 2x + 5z = 19 Save this equation..

    Now, I have to use the 3rd equation

    2x + 3y -2z = 5

    It is easier for me to eliminate the 3y if I use the first equation again.

    I multiply the first equation by -3

    2x + 3y -2z = -5
    9x -3y+3z =30

    add these two: 11x + z = 25

    Now, go to the problem you saved

    2x + 5z = 19
    11x + z = 25

    Multiply the second equation by -5 and that will help you eliminat the z's. solve for x.

    Take your answer for x and subsitute into one of these two equations to find z.

    Take your answers for x and z and substitute these values into one of the original equations to find y.

    Once you have x, y, z... check all 3 in each of the three equations to make sure you haven't made a mistake.

    It is really, really easy to make a mistake with these equations. If you get lost. Just start over try eliminating a different variable. Sometimes that helps.

  • math -

    y = 3x+z-10
    substituting values from the first into the others,

    -4x+2(3x+z-10)+3z = -1
    2x+3(3x+z-10)-2z = -5
    2x+5z = 19
    11x+z = 25

    Now, z=25-11x, so

    2x + 5(25-11x) = 19
    -53x = -106
    x = 2
    z = 25-11x = 3
    y = 3x+z-10 = -1

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