2x+4y+6z=0(1)4x+5y+6z=3(2)7x+8y+9z=6(3)(2)−(1)2x+y=3(2)×34x+5y+6z=312x+15y+18=9(4)(3)×214x+16y+18z=12(5)(5)−(4)2x+y=3Letx=nwheren∈Ry=3−2xy=3−2nSubstituting the value ofxandyin(1)2n+4(3−2n)+6z2n+12−8n+6z6zz=0=0=6n−12=n−2∴The general solution of the simultaneouslinear equation isx=n,y=3−2n,z=n−2,for alln∈R
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