Question 897529
Hello. I hope I put this question in the correct topic. I am norwegian, and I don't know the English equivalent of my question. Anyway, here goes:

x+2y+3z=14           I
4x-y-z=-1            II
7x+5y+z=20           III

What is x, y and z? (If you can show it from where I last learned, and figured out, that would be great):
x = 14-2y-3z
-> 4(14-2y-3z)-y-z=-1
-> 7(14-2y-3z)+5y+z=20


PS: I got
-9y-13z=-57
-9y-20z=-78         What then? (...if correct up to here)

Thank you so much in advance!! :)
<pre>
 x + 2y + 3z =  14 ------- eq (i)      
4x -  y -  z = - 1 ------- eq (ii)
7x + 5y +  z =  20 ------- eq (iii)

PS: I got
-9y-13z=-57
-9y-20z=-78         What then? (...if correct up to here)

Yes, you are CORRECT up to here. Good job!!

- 9y - 13z = - 57 -------- eq (iv)
- 9y - 20z = - 78 -------- eq (v)
        7z =   21 -------- Subtracting eq (v) from eq (iv)
z = {{{21/7}}}, or {{{z = 3}}}
<u>**Notice that when eq (v) was subtracted from eq (iv), the y variable was ELIMINATED</u>

Next, you substitute 3 for z in either eq (iv) or eq (v)
- 9y - 13(3) = - 57 -------- Substituting 3 for z in eq (iv)
- 9y -    39 = - 57
- 9y         = - 57 + 39
- 9y         = - 18
y = {{{(- 18)/(- 9)}}}, or {{{y = 2}}}

x + 2(2) + 3(3) = 14 ------- Substituting 3 for z, and 2 for y in eq (i)
x + 4 + 9 = 14
x + 13 = 14
x = 14 - 13
{{{x  = 1}}}
Solution: {{{highlight_green(highlight_green(system(x = 1, y = 2, z = 3)))}}}
You can do the check!! 

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