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Topic: **Solving systems of equations in three variables homework****Question:**
I got part of it but I need help with the rest.
Solve this system in three variables:
x + 4y - z = 20
3x + 2y + z = 8
2x - 3y + 2x = -16
I got "y", which was 4, but I need help finding x and z... Can someone help,and explain the process to me as well, because I'm a little stuck. I just need to find the two other variables, x and z.
Thank you so so so much.

May 19, 2019 / By Dindrane

if u found y and ur answer is correct u can plug y=4 into the first and second equations and get... x+16-z=20 x-z=4 3x+8+z=8 3x+z=0 so the 2 equations are x-z=4 3x+z=0 add these two equations 4x=4 x=1 plug this into x-z=4 1-z=4 -z=3 z=-3 so.... x=1 and z=-3 and y=4

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I shouldn't do this but hey i cheat on my homework sometimes lol. 5x-5=3x+9 5x-3x-5-9=0 2x-14=0 then x=14/2 then x=7.

put the y=4 in any two equations , from first two,you will get x+16-z=20 3x+8+z=8 or x-z=4 3x+z=0 adding both, get 4x=4 x=1 and put this valve in one of the above equation. 1-z=4 z=-3

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in case you opt for to element it right into a appropriate sq., it (would be of the variety (ax+b)^2, because is a quadratic. look on the co-useful of x^2. considering that (ax+b)^2=(a^2)(x^2)+2abx+b^2=25x^2+20x+o... by using equating the co-efficients of x^2 on the two components we detect that a^2=25, so a=5. Now, equate the co-efficients of the x words. we detect that 2ab=20, and plugging in a=5 we get 10b=20, b=2. Now, equating the consistent words, we get that ok=b^2, and substituting b=2 we get that ok=2^2=4, this is the respond. wish this helps!

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Parameter=2x+2y=44, x+y=22, x=22-y...(1) y+2=3x, 3x-y=2, 66-3y-y=2, y=16, x=6, area=96

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