SOLUTION: We have been studying the system Linear Equations in Three Variables by elimination method of x,y,z and I'm lost. PLEASE help! x+y+z=4 x-y+2z=1 x-y-3z=-4

Algebra ->  Linear-equations -> SOLUTION: We have been studying the system Linear Equations in Three Variables by elimination method of x,y,z and I'm lost. PLEASE help! x+y+z=4 x-y+2z=1 x-y-3z=-4      Log On


   



Question 155120: We have been studying the system Linear Equations in Three Variables by elimination method of x,y,z and I'm lost. PLEASE help!
x+y+z=4
x-y+2z=1
x-y-3z=-4

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
x+y+z=4
x-y+2z=1
x-y-3z=-4

  x + y +  z =  4
  x - y + 2z =  1
  x - y - 3z = -4


1. If there happens to be one or more equations which contains 
   only two of the 3 unknowns, then write it down and skip to 
   step 3.

2. Choose two of the equations and an unknown which they both 
   contain to eliminate. Then eliminate that unknown from them.
   Then write the resulting equation down.

3. Pick a pair of equations which both contain the SAME letter
   that is missing in the equation you have written down, either
   from step 1 or 2.  Then use multiplication to eliminate that
   SAME unknown from them. Write that resulting equation under
   the other equation.

4. Now you have a system of two equations in two unknowns. Solve
   that for both unknowns.

5. Find the third unknown by substituting the values found in step
   4 into any one of the original equations which contains it.
    
Now we'll go through those rules:

  x + y +  z =  4
  x - y + 2z =  1
  x - y - 3z = -4

1. If there happens to be one equation which contains only 
   two of the unknowns, then write it down and skip to step 3.

There is no equation with fewer than 3 unknowns so we cannot
skip step 2.

2. Choose two of the equations and an unknown which they both 
   contain to eliminate. 

Let's choose the first and second equations and choose y to
eliminate because that's the easiest to eliminate:

  x + y +  z =  4
  x - y + 2z =  1

We just add them as they are and eliminate the y's

 2x     + 3z =  5

Then we write the resulting equation down.

     2x + 3z = 5

3. Pick a pair of equations which both contain the SAME letter
   that is missing in the equation you have written down, either
   from step 1 or 2.  Then eliminate that SAME unknown from them. 
  
Let's pick the first and third original equations, and we must eliminate
the SAME variable, y, from them

  x + y +  z =  4
  x - y - 3z = -4

We just add them as they are and eliminate the y's



 2x    -  2z =  0

Then write the resulting equation under
   the other equation.

    2x + 3z = 5
    2x - 2z = 0

4. Now you have a system of two equations in two unknowns.
   Solve that system for both unknowns.

I assume you know how to solve that system either by 
elimination or substitution. 

The solution to that is x=1 and z=1

5. Find the third unknown, y, by substituting the values 
   found in step 4 into any one of the original equations 
   which contains it.  Solve for that letter.

Let's choose the first original equation:

      x + y + z = 4
  (1) + y + (1) = 4
      1 + y + 1 = 4
          2 + y = 4
              y = 2
                
So the solution is (x, y, z) = (1, 2, 1)

Edwin