8 May, 21:19

# 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. 8 May, 21:26
0
-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