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Find the ordered triplet (x,y,z) for the following system of equations:

x + 3y + 2z = 1

-3x + y + 5z = 10

-2x + 3y + z = 7

 May 18, 2021
 #1
avatar+115 
+3

well, I came back again to this site. 

you see the first and the last equation?

Subtract them such that the 3y is canceled out. 


$x+3y+2z-(-2x+3y+z)=1-7$
$3x+z = -6$

Neext, multiply the middle equation by $3$ to make $-9x+3y+15z=30$.

then, subtract the last equation from it to get: 
$-2x+3y+z-(-9x+3y+15z) = 7-30$
which is equal to $7x-14z=-23$

So here are the two equations we have: 

$3x+z = -6$

$7x-14z=-23$

Take the first equation and find z, which is $z=-6-3x$. Put the Z inside the second equation, $7x-14(-6-3x) = -23$, which is $7x+84+42x = -23$, or simply $49x=-107$

So, $x$ is equal to $-107/49$

Now, find z. Insert x into the equation, and get that $3(-107/49) + z = -6$, which means that $z=27/49$. 

Insert those two into the second ever equation to find y: 
$6 and 27/49 + y + 2 and 37/49= 10$, which in the end is equal to $y=9 and 15/49$

so the answer you seek is $(-107/49,27/49, 456/49)$

 May 18, 2021
 #2
avatar+3 
+2

thanks for the clear, detailed explanation

word counter

pipahaha  May 18, 2021
 #3
avatar+115 
+3

Your welcome! Have a nice night or day or afternoon! 

OofPirate  May 18, 2021
 #4
avatar+32557 
+4

I get the following:

 May 18, 2021

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