SOLUTION: what is the solution set of this linear system, in detailes Y - 4Z = 8 2X - 3Y + 2Z = 1 5X - 8Y + 7Z = 1

Algebra ->  Linear-equations -> SOLUTION: what is the solution set of this linear system, in detailes Y - 4Z = 8 2X - 3Y + 2Z = 1 5X - 8Y + 7Z = 1       Log On


   

Question 1150840: what is the solution set of this linear system, in detailes
Y - 4Z = 8
2X - 3Y + 2Z = 1
5X - 8Y + 7Z = 1

Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Y+-+4Z+=+8........eq.1
2X+-+3Y+%2B+2Z+=+1........eq.2
5X+-+8Y+%2B+7Z+=+1........eq.3
start with
Y+-+4Z+=+8........eq.1, solve for Y+
Y+=4Z%2B+8...........eq.1a
go to
2X+-+3Y+%2B+2Z+=+1........eq.2, substitute Y+
2X+-+3%284Z%2B+8%29+%2B+2Z+=+1
2X+-+12Z-24+%2B+2Z+=+1
2X+-+10Z+=+1%2B24
2X+-+10Z+=+25......solve for X
2X++=+10Z%2B25
X++=+5Z%2B25%2F2..........eq.2a
go to
5X+-+8Y+%2B+7Z+=+1........eq.3, substitute Y+
5X+-+8%284Z%2B+8%29+%2B+7Z+=+1
5X+-+32Z-+64+%2B+7Z+=+1
5X+-+25Z=64%2B+1
5X+-+25Z=65
5X+=25Z%2B65............both sides divide by 5
X+=5Z%2B13........eq.3a
from eq.2a and eq.3a we have
+5Z%2B25%2F2=5Z%2B13......as you can see, we have 5Z on both side which means it will cancel each other
therefore, highlight%28no%29 highlight%28solutions%29 highlight%28exist+%29


Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


If you were to try to solve this system using Cramer's rule, the denominator would be the determinant of the coefficient matrix. The coefficient matrix is
   | 0  1 -4 |
   | 2 -3  2 |
   | 5 -8  7 |

The determinant is

0%28-21%2B16%29-1%2814-10%29-4%28-16%2B15%29+=+0-4%2B4+=+0

The determinant of the coefficient matrix is 0; that means the denominator of the fraction used in Cramer's rule is 0; since division by 0 is not allowed, that means the system has no solutions.