Solving Systems

Question

I'm trying to finish my homework but I keep doing this wrong :/. Can anyone help?

 

6x-y+3y=7

x+3y-z=1

3x+3y-4z=11

Answer 1

6x - y + 3z = 7                   eq1
x + 3y - z = 1                      eq2
3x + 3y - 4z = 11                eq3

 

 

I am not sure where you went wrong, but follow these steps to solve these types of problems.  Well, the first equation has 2 terms with y variables.  Each equation should have an x, y, and z variable.  Change 3y to 3z in the first equation.

 

Since we have a system of equations, we want to use the substitution/elimination methods to reduce the number of equations and variables.

 

We can substitute eq2 into the other equations.  From eq2,

 

x = 1 - 3y + z

 

 

6(1 - 3y + z) - y + 3z = 7

6 - 18y + 6z - y + 3z = 7

-19y + 9z = 1                        new eq1

 

 

3(1 - 3y + z) + 3y - 4z = 11

3 - 9y + 3z + 3y - 4z = 11

-6y - z = 8                               new eq3

 

 

The bolded equations become the new equations to work with.  Multiply new eq3 by 9.  Keep new eq1.

 

-19y + 9z = 9                 new eq1

-54y - 9z = 72                new eq3

 

 

Now add the equations together to eliminate the z terms.

 

-73y = 81

 

     y = -81 / 73

 

   

Then you will substitute this value of y into new eq1 to solve for z.  Once you have the values of y and z, substitute those values into eq2 to solve for x.