Question

How do you solve the system `4x+4y+z=24`, `2x-4y+z=0`, and `5x-4y-5z=12`?

Answer 1

Add the equations to eliminate variables and solve.

Explanation:

`4x+4y+z=24`
`2x-4y+z=0`
`5x-4y-5z=12`

Choose two equations that will easily cancel out.
`4x+4y+z=24`
`2x-4y+z=0`
The y's cancel out, leaving you with `6x+2z=24`. This can be simplified to `3x+z=12`.

Choose another two equations to get another binomial.
`4x+4y+z=24`
`5x-4y-5z=12`
Again, the y's cancel out, leaving you with `9x-4z=36`.

Use substitution, making `3x+z=12` into `z=12-3x`. Plug this into `9x-4z=36` to get `9x-4(12-3x)=36`. Distribute to get the equation `9x-48+12x=36`. Combine like terms, making the equation `21x=84`. Divide both sides by `21` to get `x=4`.

You can then use this value in the equation `z=12-3x`. Plug it in to get `z=12-3(4)`, which simplifies to `z=0`.

With these two values you can use any of the original equations to find the remaining value. `2(4)-4y+(0)=0`. Simplify to get `8-4y=0`. Add `4y` to both sides to get `8=4y`, and divide both sides by `4` to get `y=2`.

The final values are `x=4`, `y=2` and `z=0`.
This can be checked by plugging these values into the two remaining original equations.