systems of linear equations: w-2z+y+4z=0,4w-8x+5y+20z=1,6w-

[Angela123]

Solve the following system of linear equations:

w- 2x+ y+ 4z=0
4w- 8x+ 5y+20z=1
6w-12x+15y+60z=9

I solved it by doing:
-4(w-2x+y+4z=0)-->-4w+8x-4y-16z=0
4w-8x+5y+20z=1-->4w-8x+5y+20z=1
=y+4z=1
-6(w-2x+y+4z=0) -->-6w+12x-6y-24z=0
6w-12x+15y+60z=0-->6w-12x+15y-60z=0
=9y+36z=0

-9(y+4z=1)-->-9y+36z=-9
9y+36z=0--> -9y-36z=0

X and Z are arbitrary solutions and W and Y are infinite. This was wrong but I thought: w=2x-y-4z and y=-w+2x-4z. Where did I go wrong?

 

[Deleted member 4993]

Re: linear equations

Solve the following system of linear equations:

w- 2x+ y+ 4z=0
4w- 8x+ 5y+20z=1
6w-12x+15y+60z=9

I solved it by doing:
-4(w-2x+y+4z=0)-->-4w+8x-4y-16z=0
4w-8x+5y+20z=1-->4w-8x+5y+20z=1
=y+4z=1
-6(w-2x+y+4z=0) -->-6w+12x-6y-24z=0
6w-12x+15y+60z=0-->6w-12x+15y-60z=0 <<<< That should be 6w-12x+15y+60z=9
=9y+36z=0

-9(y+4z=1)-->-9y+36z=-9 <<<<< That should be -9y - 36z = -9
9y+36z=0--> -9y-36z=0



X and Z are arbitrary solutions and W and Y are infinite. This was wrong but I thought: w=2x-y-4z and y=-w+2x-4z. Where did I go wrong?

 

[Angela123]

Re: linear equations

Ok, I see I made an error, but:
9y+36z=9
-9y-36z=-9 so it all cancels out and I am still left with the question as to what w=? and y=?

 

[Deleted member 4993]

Re: linear equations

I am still left with the question as to what w=? and y=?

Since you need to find w and y - start by eliminating 'x' or 'z'.

You started with eliminating 'w' - in general you would not be solve 'w' by eliminating 'w'.