Question

For the following exercises, solve the systems of linear and nonlinear equations using substitution or elimination. Indicate if no solution exists.

Answer 1

Consider the following system of equations:

4x – 6y – 2z = 1/10

x – 7y + 5z = -1/4

3x + 6y – 9z = 6/5

 

Solve for x, y and z.

4x – 6y – 2z = 1/10 …… (1)

x – 7y + 5z = -1/4 …… (2)

3x + 6y – 9z = 6/5 …… (3)

 

Multiply equation (1) by 7 and multiply equation (2) by –6 and then add both equations

28  – 42y – 14z = 7/10

-6x + 42y – 30z = 6/4

          22x – 44z = 11/5

 

Hence,

22x – 44z = 11/5 …… (4)

 

Add equation (1) and equation (3) to get

4x – 6y -2z = 1/10

3x + 6y – 9z = 6/5

      7x – 11z = 13/10

 

Hence,

7x – 11z = 13/10 …… (5)

 

Multiply equation (5) by –4 and add it to equation (4)

22x – 44z = 11/5

-28x + 44z = -26/5

            -6x = -3

               x = 1/2

 

Substituting x = 1/2 into equation (4) to get

 

22(1/2) – 44z = 11/5

        11 – 44z = 11/5

               -44z = -44/5

                    z = 1/5

 

Substituting the values x = 1/2 and z = 1/5 into equation (1) to get

4(1/2) -6y – 2(1/5) = 1/10

           2 – 6y – 2/5 = 1/10

                          -6y = -3/2

                             y = 1/4

 

Therefore, the solution is the ordered triple 1/2, 1/4, 1/5.

Therefore, the solution is the ordered triple 1/2, 1/4, 1/5.