SOLUTION: Solve the system by back substitution. 2x+4y+z+3w=1 y+6z+2w=8 z+w= 2 2w= 6 The solution set is: _, _, _, _

Algebra ->  Finance -> SOLUTION: Solve the system by back substitution. 2x+4y+z+3w=1 y+6z+2w=8 z+w= 2 2w= 6 The solution set is: _, _, _, _       Log On


   

Question 1117716: Solve the system by back substitution.
2x+4y+z+3w=1
y+6z+2w=8
z+w= 2
2w= 6

The solution set is: _, _, _, _
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


1. Solve the last equation for .

2. Substitute the value of you discovered in step 1 into the third equation and then solve for .

3. Substitute the value of from step 1 and the value of from step 2 into the second equation and then solve for .

4. Substitute the values of , , and from the first three steps into the first equation and solve for
The way you are asking for the answer is incorrect. A list of the values of the four variables is NOT the solution set of the 4X4 system of equations. The solution set of this system has a single element, namely the ordered quadruple of the form . The solution set would then need to be written: .

John

My calculator said it, I believe it, that settles it