How do you solve the system of equations algebraically #2x-2y+3z=6, 2x-3y+7z=-1, 4x-3y+2z=0#?

1 Answer
Sep 6, 2017

There are no solutions


The system is presented as thus:

(A) #2x-2y+3z=6#
(B) #2x-3y+7z =-1#
(C) #4x-3y+2z =0#

We will pick two pairs of equatwins in the system, and use addition and subtraction to eliminate the same variable from both pairs. We will use A and B, and B and C, for this, eliminating x:

#A-B-> (2x-2x) + (-2y+3y) + (3z-7z) = 6+1 -> y-4z=7# A*

#c - 2b -> x (4-4) +y (-3+6) +z (2-14) = (0+2) -> 3y -12z =2# B*

Note that the variable portion of B is a multiple of the same section in A, i.e. #3y-12z = 3(y-4z)#. This means the system is linearly independent, so there cannot be just one solution. If multiplying both sides of A by 3 makes the right hand side of the equathin the same as that of B, then there are infinitely many solutions. If they are unequal, there is no solution. To demonstrate, we try to solve the same as above, using 3A - B

3A-B#-> (3Y-3Y) - (-12Z + 12Z) = 21-2 -> 0 = 19#

Solutions will,only be present if 0 = 19. Because this is not the case, there are no solutions.