How do you solve #x+2y+z=34#, #x+2y+2z=38#, #2x+y+3z=54#?

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Answer 1 · 2016-07-02T04:44:28

#x=18,y=6,z=4#

#color(green)("Three given equations are") #

#x+2y+z=34.......(1)#,

#x+2y+2z=38.......(2)#,

#2x+y+3z=54......(3)#

#color(blue)("Subtracting (1) from(2) we have") #

#x+2y+2z-(x+2y+z)=38-34#

#=>cancelx+cancel(2y)+2z-cancelx-cancel(2y)-z=4#

#=>z=4#

#color(blue)("Inserting the value of z in (1)") #

#x+2y+4=34#,

#x+2y=30.....(4)#

#color(blue)("Inserting the value of z in (3)") #

#2x+y+3xx4=54#

#2x+y =42.......(5)#

#color(blue)("Multiplying (4) by 2 and subtracting the resulting equation from (5)") #

#2x+y -2xx(x+2y)=42-2xx30#

#=>2x+y -2x-4y=42-60#

#=>-3y=-18#

#=>y=6#

#color(blue)("Inserting the value of y in (4) "#

#x+2xx6=30#

#=>x=18#