If you were to use the elimination method to solve the following system, choose the new system of equations that would result after the variable z is eliminated in the first and third equations, then the second and third equations.

x – y + 2z = –2
2x + 2y + z = 7
3x + 3y – z = 3

Question

If you were to use the elimination method to solve the following system, choose the new system of equations that would result after the variable z is eliminated in the first and third equations, then the second and third equations. 

x – y + 2z = –2 
2x + 2y + z = 7 
3x + 3y – z = 3

hartnn · Answer

add 2nd and 3rd equation, what u get ?

anonymous · Answer

x - y + 2z = - 2
2x + 2y + z = 7
3x + 3y - z = 3
--------------
Step 1:
add 1st and 3rd equation multiplying if necessary to eliminate z
x - y + 2z = -2
3x + 3y -z = 3 --->(2)3x +  3y - z = 3
---------------
x - y + 2z = -2
6x + 6y - 2z = 6 (result of multiplying 2nd equation by 2)
--------------
7x + 5y = 4 ====> new equation resulting from 1st and 3rd equation

Now for the 2nd and 3rd equation.....
2x + 2y + z = 7
3x + 3y - z = 3 
-------------no multiplying is needed because the z's cancel out
5x + 5y = 10 ===> new equation resulting from 2nd and 3rd equation

easy enough :)

anonymous · Answer

thanks That helped a lot.

anonymous · Answer

no problem...anytime :)