SOLUTION: x + y + z =6 x+2y+3z=10 x+2y+az=b)

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Question 1191180: x + y + z =6
x+2y+3z=10
x+2y+az=b)
Found 3 solutions by MathLover1, Edwin McCravy, ikleyn:
Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!

........1)
.......2)
.........3)
___________________________


........1)
.......2)
____________________________subtract

.......solve for
............1.1)

........1).........substitute
.....solve for


...............1.2)


.........3)........substitute , ,
.....solve for




...........1.3)


then
...............1.2)
and
............1.1)


The solutions to the system of equations are:



where


Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!


Case 1: a=3.





-1R1+R2->R2



-1R2+R1->R1









(x,y,z) = (2+z,4-2z,z) where z is arbitrary.

Some solutions from case 1 are (x,y,z)=(2,4,0), (x,y,z)=(3,2,1),
(x,y,z)=(4,0,2), (x,y,z)=(5,-2,3), (x,y,z)=(1,6,-1), etc., etc.,...

Case 2. x ≠ 3    



-1R2+R1->R1



-1R1+R2->R2



-1R1+R3->R3



-1R2+R1->R1



-1R2+R3->R1



-1R2+R3->R3



1/(a-3)*R3->R3



R3+R1->R1



-2R3+R2->R2



Simplifying:







Edwin
Answer by ikleyn(52803)   (Show Source): You can put this solution on YOUR website!
.

            This post came with no question, which makes me very sad
            (as usual as I see such an inaccuracy and disrespect to the tutors).

            The short resume to this problem is as follows. 


(1)  The case a= 3 is special.


     If  a= 3 and b= 10, then there are infinitely many solutions.

                         In this case, the system is dependent.


     If a= 3 and b =/= 10, then there is NO solutions and the system is inconsistent.



(2)  If a =/= 3, then we have three independent equations;

                 the system is consistent and has a unique solution.

Answered.



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