SOLUTION: Please help me solve this system of equations: 36x-15y+50z=-10 2x+25y=40 54x-5y+30z=-160 Thank you so much!

Question 265448: Please help me solve this system of equations:
36x-15y+50z=-10
2x+25y=40
54x-5y+30z=-160
Thank you so much!
Found 2 solutions by vksarvepalli, palanisamy:
Answer by vksarvepalli(154) (Show Source): You can put this solution on YOUR website!
36x-15y+50z=-10 ----------------------- 1
2x+25y=40 ----------------------- 2
54x-5y+30z=-160 ----------------------- 3
multiply eq. 1 by 3 and eq. 3 by 5 and then subtract both (to eliminate z)
so we get 108x-45y+150z=-30
270x-25y+150z=-800
-162x-20y=770
now multiply eq.2 by 81 and add to the above eq.

we get 2005y=4010
so y=2

from eq. 2
2x+50=40 => 2x=-10
=> x=-5
from eq.1 -180-30+50z=-10
so 50z=200
=> z=4
so x=-5, y=2, z=4
Answer by palanisamy(496) (Show Source): You can put this solution on YOUR website!
The given equations are,
36x-15y+50z=-10 ...(1)
2x+25y=40 ...(2)
54x-5y+30z=-160 ...(3)
(1)*3=> 108x-45y+150z = -30 ...(4)
(2)*5=> 270x-25y+150z = -800 ...(5)
(4)-(5)=> -162x-20y = 770 ...(6)
(6)*5=> -810x-100y = 3850 ...(7)
(2)*4=> 8x+100y = 160 ...(8)
(7)+(8)=> -802x = 4010
x = 4010/(-802)
x = -5
(2)=> 2*(-5)+25y = 40
-10+25y = 40
25y = 40+10
y = 50/25
y = 2
(1)=> 36*(-5)-15*2+50z = -10
-180-30+50z = -10
50z = 180+30-10
50z = 200
z = 200/50
z = 4
So the solution is, x= -5, y= 2 and z= 4
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