How would you go about finding the solution to this system of three equations in three variables? Be specific. For example, you
might want to add the first two equations first, or multiply the third one by two first and use elimination, etc. (No right answer exists.) 2x – y + z = 7
I:2x – y + z = 7 II:x + 2y – 5z = -1 III:x – y = 6
you can first use III and substitute x or y to eliminate it in I and II (in this case x): III: x=6+y -> substitute x in I and II: I': 2*(6+y)-y+z=7 12+2y-y+z=7 y+z=-5
II':(6+y)+2y-5z=-1 3y+6-5z=-1 3y-5z=-7
then you can subtract II' from 3*I' to eliminate y: 3*I'=3y+3z=-15
3*I'-II': 3y+3z-(3y-5z)=-15-(-7) 8z=-8 z=-1
insert z in II' to calculate y: 3y-5z=-7 3y+5=-7 3y=-12 y=-4
insert y into III to calculate x: x-(-4)=6 x+4=6 x=2
Ok so if u are shipping boxes in a right rectangular prism u will need to either divide or multiply. if he wants to make each box to be exactly 98 cubic feet then he will need to divide so when u do the u will need to find the first part of the answer then u will divide it by 98 cubic feet :> hope u get what i meant