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11 May, 11:47

2. Solve the following system. Show all your work. - 3x + 2y + z = - 6 x + 3y + 2z = 5 4x + 4y + 3z = 13 Is there a step I'm skipping because every time I add the first 2 equations, I keep getting - 2x + 5y + 3z with no variables eliminated.

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  1. 11 May, 14:26
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    -3x + 2y + z = - 6

    (3) x + 3y + 2z = 5 (3) We want to get rid of any variable within 2 equations. The easiest one would be by multiplying both sides of eq#2 by 3 and add it to eq#1.

    -3x+2y+z=-6

    3x+9y+6z=15

    Add them together to get 11y+7z=9

    (-4) x+3y+2z=5 (-4) Now, multiply the original eq#2 by - 4 to cancel out x in the last equation

    -4x-12y-8z=-20

    4x + 4y + 3z = 13

    add them together to get - 8y-5z=-7

    Now multiply the eq 11y+7z=9 by 5 and the equation - 8y-5z=-7 by 7 to cancel out the z variable when we add them together.

    Once we multiply them, we get

    -56y-35z=-49

    55y+35z=45

    add them together to get - y=-4 and we know y=4

    substitute y into any equation here that has y and any other 1 variable like - 8y-5z=-7 and get z=-5

    Finally, we know y=4 and z=-5, we can substitue that into any of the original eqs like - 3x + 2y + z = - 6 and we get x=-3

    To check, just substitute these values into any of the equations above
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