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

1 Answer
Jan 15, 2018

Answer:

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.