Question 346673
Algebra- solving sytems of equations in three varaiables: 
( Answer is needed within 1.5 hours, by 11"30 AM Pacific time - Thanks!) 
3x+5y+4z =13
5z+2y+3z=-9  (--> assuming this is 5x + 2y + 3z = -9)
6x+3y+4z=-8


using method of elimination, choosing z first since I see two 4z's


use 1st equation and multiply 3rd equation by -1


 3x + 5y + 4z = 13
-6x - 3y - 4z =  8
add these to get -3x + 2y = 21


use 1st equation multiplied by 3 and 2nd equation multipled by -4


  9x + 15y + 12z = 39
-20x -  8y - 12z = 36
add these to get -11x + 7y = 75


we now have:
 -3x + 2y = 21 (multiply this one by 7)
-11x + 7y = 75 (multiply this one by -2)


to get:
-21x + 14y =  147
 22x - 14y = -150
add these to get x = -3


plug x = -3 into -3x + 2y = 21
-3(-3) + 2y = 21
9 + 2y = 21
2y = 12
y = 6


now going back to the original equations, plug x = -3 and y = 6
into 3x + 5y + 4z = 13
3(-3) + 5(6) + 4z = 13
-9 + 30 + 4z = 13
21 + 4z = 13
4z = -8
z = -2


solution is (-3,6,-2)


hope this helps you