Solve the system of triple ordered pairs

[Jackiesuavey]

Help anyone please. I need a solid structure on how to eliminate Z variable twice to solve for [Y] and [X] in two variable form.

-x - 2y - z = 9

2x + y + z = 1

x - y + 2z = -8

 

[HallsofIvy]

Help anyone please. I need a solid structure on how to eliminate Z variable twice to solve for [Y] and [X] in two variable form.

-x - 2y - z = 9

2x + y + z = 1

x - y + 2z = -8

It looks straight forward to me. Seeing "-z" in the first equation and "+z" in the second, you should immediately think "add the first two equations. Seeing "+2z" in the third, multiply the first equation by 2 to get "-2z" in it, then add to the third equation.

 

[Steven G]

Help anyone please. I need a solid structure on how to eliminate Z variable twice to solve for [Y] and [X] in two variable form.

-x - 2y - z = 9

2x + y + z = 1

x - y + 2z = -8

Here are the steps
1) pick any two equations and pick any variable.
2) using the two equations from 1) eliminate that chosen variable.
3) Box that new equation obtained from 2)
4) using the equation you did NOT use in 1) along with one of the equations you did use in 1) eliminate that same variable you chose in 1)
5) box this equation
6) now you have two equations (the boxed ones) with 2 unknowns. Solve this for the two unknowns.

With experience you will get better at deciding how to determine which 2 equations and variables you should chose in 1) as well as which equation from 1) to use in step 4.

HallsofIvy gave you great hints for step 1 and 4. Look at this advice very closely to 'see' why those choices were made.

Take your time, follow the steps closely and you will get the correct answer.

Tell us what you get.

 

[lookagain]

I'm seconding this advice. It gets to the point.

It looks straight forward to me. Seeing "-z" in the first equation and "+z" in the second, you should immediately think
"add the first two equations. Seeing "+2z" in the third, multiply the first equation by 2 to get "-2z" in it, then add to the third equation.