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

1 Answer
Jan 15, 2018

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.