Posted by rayne on Monday, October 1, 2012 at 10:26pm.
i need help with this problem
2x+2y+3z=1
3x5y2z=21
7x+3y+5z=10

math  Reiny, Monday, October 1, 2012 at 10:35pm
What method have you learned?

math  rayne, Monday, October 1, 2012 at 10:37pm
EITHER SUBSTITUTION OR ELIMINATION

math  Reiny, Monday, October 1, 2012 at 10:55pm
Ok, elimination then
work on the z's
1st times 2 > 4x + 4y + 6z = 2
2nd times 3 > 9x  15y  6z = 63
add them
13x  11y = 61 (#4)
2nd times 5 > 15x  25y  10z = 105
3rd times 2 > 14x + 6y + 10z = 20
add them
29x  19y = 125 (#5)
going to be messy .....
#4 times 19 > 247x  209y = 1159
#5 times 11 > 319x  209y = 1375
subtract them
72x = 216
x = 216/72 = 3 , Yeahhh!
back in #4
13(3)  11y = 61
11y = 6139
11y = 22
y = 2
back in #1
2(3)+2(2) + 3z = 1
3z = 1 2
z = 1
x=3 , y=2 , z=1 
math  rayne, Monday, October 1, 2012 at 11:26pm
CAN ANYONE JUST GET ME HEADED IN THE RIGHT
DIRECTION ON THIS PROBLEM
2x+2y+3z=1
3x5y2z=21
7x+3y+5z=10 
math  Reiny, Tuesday, October 2, 2012 at 12:01am
And exactly what did you not like about the solution which is staring at you right above you ?????