How do you solve the system of equations algebraically #1.8x-z=0.7, 1.2y+z=-0.7, 1.5x-3y=3#?

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Answer 1 · 2016-12-30T14:34:24

#x=0.5#, #y=-0.75# and #z=0.2#

The three equations are

#1.8x-z=0.7# ...........................(1)

#1.2y+z=-0.7# ...........................(2)

#1.5x-3y=3# ...........................(3)

Note that adding (1) and (2) eliminates #z# and we get

#1.8x+1.2y=0# ...........................(4)

Now multiplying (3) by #6# and (4) by #5#, we get

#9x-18y=18# and #9x+6y=0# and subtracting latter from former, we get #-18y-6y=18-0# i.e. #-24y=18#

Hence #y=-18/24=-0.75#.

Putting this in (3) we get

#1.5x-3xx(-0.75)=3# or #1.5x=3-2.25=0.75#

Hence #x=0.75/1.5=0.5#

Now putting this in (1) , we get

#1.8xx0.5-z=0.7# or #0.9-z=0.7# i.e.

#z=0.9-0.7=0.2#

Hence solution is #x=0.5#, #y=-0.75# and #z=0.2#