1. solve the system by elimination. 2.solve the system by substitution.

Step One

Add the first and third

-2x + 2y + 3z = 0

2x + 3y + 3z = 5

        5y + 6z = 5

Step Two

Add the second and third together.

- 2x - y + z = -3

2x + 3y + 3z = 5

       2y + 4z = 2            Divide this result by 2

         y + 2z = 1             multiply this result by 5

        5y + 10z = 5

Step three

Subtract the result of step 2 from the result of step one

         5y + 10z = 5

         5y +  6z = 5        Subtract

                  4z = 0

                    z = 0

Step Four

Solve for y

5y + 6z = 5     But z = 0

5y = 5

y = 1

Step Four

Solve for x . Use equation 3 in the givens.

2x + 3y + 3z = 5

2x + 3*1) + 3*0 = 5

2x = 5 - 3

2x = 2

x = 1

Answer

x = 1

y = 1

z =0       This has been checked and found to be correct.

Problem 2

Substitution is not the best way to do this. Divide the last equation by 2 and add it to the first one.

Step One

divide equation 3 by 2.

x + z = 2

Step Two

Add step one to equation 1

-x - y - z = - 8

x    +    z =  2

    - y = - 6

      y =   6

Step Three

Put this result into equation 1.

-x - 6 - z = - 8

-x - z = - 2

x + z = 2     Multiply this equation by 4

4x + 4z = 8

Step four

Put y = 7 into the second equation.

-4x + 4y + 5z =  7

-4x + 4*6 + 5z = 7

-4x + 24 + 5z = 7

-4x          +5z = -17

Step 5

Add the results from Step 3 and Step 4 together.

-4x + 5z = - 17

4x + 4z = 8

9z = - 9

z = - 1

Step 6

Find x

-x - y - z = - 8

-x -6 + 1 =- 8

-x - 5 = - 8

-x = - 3

x = 3      

Answers

x = 3

y = 6

z = -1    verified to be correct.