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
+115 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 $(-107/49,27/49, 456/49)$
