Elimination Method of 3 Variables

[LocaChica16]

Wow guys, I really need your help. I've been working on this problem and a substition problem for the past 2 hours atleast and I can't for the life of me figure it out... it's driving me absolutly insane and if I don't see how it's done it'll bug me forever.. so if you can show me how it's done, I'd appreciate it very much.. thank you. :x

2x-5y-3z=7
-4x+10y+2z=6
6x-15y-z=-19

 

[stapel]

What have you done? How far have you gotten?

There is no one "right" way to go with this, and different people prefer different methods and paths, so, if you asked four different people, you'd probably see four different solutions, and they'd all be different from what you'd tried. :shock: That's just how this works.

My first step would be to take twice the first row and add it to the second row, and take three times the first row and subtract it from the third row. This would give me:

. . . . .2x - 5y - 3z = 7
. . . . . . . . .. . .-4z = 20
. . . . . . . . . . . .8z = -40

This lets you solve for the value of z, but it looks like x is going to be in terms of y. That is to say, you can't get a unique numerical solution; you can only get a parameterized one.

Hope that helps a bit.

Eliz.

 

[tkhunny]

You say "this problem and a substitutino problem". Does this mean you are to solve this one by methods other than substitution?

#1) 2x-5y-3z=7
#2) -4x+10y+2z=6
#3) 6x-15y-z=-19

Take twice #1 and add to #2, giving a new #2

4x-10y-6z=14
-4x+10y+2z=6
0*x + 0*y -4*z = 20
Simplify
-4*z = 20
z = -5 <== New #2

Take -3 multiplied by #1 and add to #3, giving a new #3

-6x+15y+9z=-21
6x-15y-z=-19
0*x + 0*y + 8*z = -40
Simplify
8*z = -40
z = -5 <== New #3

That leaves:

#1) 2x-5y-3z=7
#2) z = -5
#3) z = -5

That is fine. Where do we go from here?

 

[LocaChica16]

well, for this problem we're supposed to use specifically the elimanation method, and thats what runs me into problem.s

 

[stapel]

Well, you've gotten two replies showing the solution by elimination, so work from them.

Eliz.