SOLUTION: I need to solve the following system of linear equations with three variables.
Here is the system:
2x-4y+3z=17
x+2y-z =0
4x -y -z=6
O.
Algebra.Com
Question 378503: I need to solve the following system of linear equations with three variables.
Here is the system:
2x-4y+3z=17
x+2y-z =0
4x -y -z=6
O.K. I know i need to reduce it to a system of 2 equations with 2 variables. This is what i have so far. I'm not certain it's right. I did this:
2x-4y+3z.....2x-4y+3z=17
3(x+2y-z=0)...3x+6y-3z=0
That got rid of my "z". now i THINK i"m supposed to add? That gives me:
5x+2y=17. O>K> There's my 1st equation with 2 variables. Now for the 2nd one.
I did: x+2y-z (that's equation #2 in the original)
-1(4x-y-z=6)...I multiplied the original 3rd equation by -1 to get rid of ITS "z". I got:
-3x+3y=-6. There's my 2nd 2-variable equation. Now I did:
5x+2y=17
-3x+3y=-6...Now to solve: I chose to get rid of "y" and solve for x like this:
-3(5x+2y=17)...-15x-6y=-57
2(-3x+3y=-6)..-6x+6y=-12
Adding...............................
21x =-69
69/21
x=3.29 or 3
My two questions are: A) is this right up til now? and B)What do i do now? I'm lost. Please help!:(
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
YOU ARE DOING GREEEEAT!!!!!!!
Down toward the end you have:
-3(5x+2y=17)...-15x-6y=-57
---and you made a simple error. It should be.....-15x-6y=-51 (not 57)
Now lets see how this changes the outcome
Instead of -- -21x=-69 you have -21x=-63
and x=3
Now the next step is to substitute x=3 into either of the equations with two unknowns and solve for y. Then the next step would be to substitute the value you got for y plus the value you have for x (3) into either of the equations with 3 unknowns and then solve for z.
You are doing good---keep it up.
Hope this helps ---ptaylor
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