SOLUTION: equations in three linear variables 6x-4y+5z=31 5x+2y+2z=13 x+y+z=2 wouldn't you multiply the second equation by 2?

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: equations in three linear variables 6x-4y+5z=31 5x+2y+2z=13 x+y+z=2 wouldn't you multiply the second equation by 2?      Log On


   



Question 44704: equations in three linear variables
6x-4y+5z=31
5x+2y+2z=13
x+y+z=2
wouldn't you multiply the second equation by 2?

Answer by OHtutor(16) About Me  (Show Source):
You can put this solution on YOUR website!
no, you don't have to multiple because the equation doesn't have fractions.
You would try to move 2 of the varables to the right and work with2 of the variables.
6x=4y-5x+31
2y=-5x-2z+13
y=(-5x-2z+13)/2
Now put the above equation into each of the other 2 equations left and then work just with x and z (it will be just a 2 variable equation after that)