Solve the following system

Question

2x-3y+z=5

 

  x+y+z=0

 

4x+2y+4z=4

Answer 1

      Given

        2x-3y+z =5    (say equation 1)

         x+y+z=0       (say equation 2)

       4x+2y+4z=4   (say equation 3)

Multiply equation 2 by 4 on both sides

      4(x+y+z)= 4*0=0

      4x+4y+4z = 0    ( say equation 4)

  Let us deduct equation 3 from equation 4

      4x+4y+4z=0

     -(4x+2y+4z)=-4

       0x+2y+0z  =-4

               y         =-2

Let us put the value of y in equation 1 

      2x-3(-2)+z=5 

     2x+6+z=5

     2x+z=-1 ( say equation 5)

Let us put the value of y in equation 2

      x+(-2)+z=0

     x+z=2    (say equation 6)

 Let us deduct equation 6 from equation 5

     2x+z=-1

    -(x+z)=-2

     x +0z=-3

    x=-3

Let us put value of x and y in equation 1

  2*(-3)-3*(-2)+z=5

   -6+6+z=5

   z= 5+6-6= 5

Thus we find that     x=-3

                               y=-2

                               z =5

 

 

 

Answer 2

2x - 3y + z = 5                   eq1

x + y + z = 0                      eq2

4x + 2y + 4z = 4                 eq3

 

 

We have three equations in which each one has 3 different variables.  We cannot solve a single equation with 3 different variables, so we must substitute equations into another and at the same time, maybe some elimination methods.  Since eq2 has variables where their coefficients are 1, we can substitute that equation into both eq1 and eq3.

 

From eq2,

 

x = - y - z

 

 

Now, we substitute this value of x from eq2 into the other equations.

 

2(-y - z) - 3y + z = 5

-2y - 2z - 3y + z = 5

-5y - z = 5                    new eq1

 

 

4(-y - z) + 2y + 4z = 4

-4y - 4z + 2y + 4z = 4

-2y = 4                           new eq3

 

 

We now only have 2 equations to work with.

 

-5y - z = 5                  new eq1

-2y = 4                       new eq3

 

 

We can solve for y from new eq3.

 

-2y = 4

   y = -2

 

Substitute this value of y into new eq1 to solve for z.

 

-5y - z = 5

-5(-2) - z = 5

10 - z = 5

      - z = -5

        z = 5

 

Finally, we can substitute these values of y and z that we found into the original eq2.

 

x = -y - z

x = -(-2) - (5)

x = 2 - 5

x = -3

 

 

Your solutions are

 

x = -3

y = -2

z = 5