SOLUTION: Hi would you please help me with this question and could you show me the steps on how to solve it please. Solve the system of equations: x + y+ z=9 2x+4y+2z=14 -x+8y-3z= -39

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: Hi would you please help me with this question and could you show me the steps on how to solve it please. Solve the system of equations: x + y+ z=9 2x+4y+2z=14 -x+8y-3z= -39       Log On


   

Question 904087: Hi would you please help me with this question and could you show me the steps on how to solve it please.
Solve the system of equations:
x + y+ z=9
2x+4y+2z=14
-x+8y-3z= -39
My professor said the solution set is {(5,-2,6)}
Can you kindly show me how he got those answers please. Thanks!
Found 3 solutions by richwmiller, josgarithmetic, Edwin McCravy:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
He most likely opened his book to the answer page
BTW
Don't submit similar problems, that is, same type problem with different numbers.
Most of the numbers were the same.
I did check.The professor did give you the right answer.

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
How he solved to get the answer he would know better than anyone else. More than one way to do it.



Simplify the second equation with the "14";



Add R1 to R3 for new R3;



Revise R2 as R2-R1;



The y variable is now solved as found in the second row, that y=-2.
Re-order the rows:



Revise R2 as R2-9*R3;



Revise R2 as -(1/2)*R2;



That row 2 now shows highlight%28z=6%29.
And you also know highlight%28y=-2%29.

Avoiding further matrix operations, the Row 1 is x%2By%2Bz=9
x=9-y-z
x=9%2B2-6
x=9-4
highlight%28x=5%29;
Solved for x, y, and z to the same values as your professor found.

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
You may not have studied the augmented matrix method yet. Here's how to
solve it by elimination:

1.  Pick a letter to eliminate.
2.  Eliminate that letter from any two of the 3 equations that contain 
    that letter.
3.  Eliminate the SAME letter for one of those equations you just used but
    this time using the third equation.
4.  Now you have a new system of only two equations in two unknowns.
5.  Solve this system for those two unknowns.
6.  Substitute those values in one of the original equations to find the
    remaining letter.  [It will be the first letter eliminated.    


system%28x+%2B+y%2B+z=9%2C%0D%0A2x%2B4y%2B2z=14%2C%0D%0A-x%2B8y-3z=+-39%29

Add the first and third equations term by term to eliminate x

 x +  y +  z =   9
-x + 8y - 3z = -39
------------------
     9y - 2z = -30

Now eliminate x from the 2nd and 3rd equations

 2x + 4y + 2z =  14
 -x + 8y - 3z = -39

To make the x's cancel we must multiply the second one
above through by 2


 2x +  4y + 2z =  14
-2x + 16y - 6z = -78
--------------------
      20y - 4z = -64

We can simplify that by dividing through by 4

      5y -  z = -16

Now we have this system which is 2 equations and 2 unknowns.

system%289y+-+2z+=+-30%2C5y+-++z+=+-16%29

To make the z's cancel we must multiply the second one
above through by -2

  9y - 2z = -30
-10y + 2z =  32
---------------
  -y      =   2

Divide both sides by -1

        y = -2

Substitute -2 for y in

      5y -  z = -16
    5(-2) - z = -16
      -10 - z = -16
           -z = -6
            z = 6

Substitute both y = -2 and z = 6 in one
of the original equations:

    x + y + z = 9 
 x + (-2) + 6 = 9
    x - 2 + 6 = 9
        x + 4 = 9
            x = 5

Solution (x,y,z) = (5,-2,6)

Edwin