Question

Please someone help me with this....What is the solution of the system of equations?
3x+2y+z=7
5x+5y+4z=3
3x+2y+3z=1

Answer 1

Try subtracting the third equation from the first.

Answer 2

I actually figured out already but thanks for trying to help :)

Answer 3

Great! I'm glad you didn't need me. :) Keep up the good work!

Answer 4

Given 3x+2y+z=7.............(1)
5x+5y+4z=3............(2)
3x+2y+3z=1 …..........(3)
subtract equation (1) and (3)
=>3x+2y+z=7
=>3x+2y+3z=1
….-.....-....-.....-
..------------------
...0+0-2z=6
=>z=-3
substitute z value in equation (1) and (2)
=>3x+2y=10.....(1a)
=>5x+5y=15.....(2a)
divide equation (2a) by 5
=>x+y=3.......(2b)*2
=>2x+2y=6.....(2b)
subtract (2b) and (1a)
=>3x+2y=10
=>2x+2y=6
….-....-......-
.----------------
...x=4
and 4+y=3
=>y=-1
hence x=4,y=-1 and z=-3



CBSE Previous Year Question Papers Class 10 http://goo.gl/HzqHts