First let us try and write this problem in an equation of unknown #x#, where:
#x= "number of windows"#
#color(red)(x_(red))= "number of windows Big red washed"=color(red)(1/2x)#
#color(green)(x_(green))= "number of windows Big Green washed"=2/3(x-color(red)(x_(red)))#
#=2/3*(x-1/2x)#
#=2/3*(2x-x)/2=cancel2/3*x/cancel2#
#=color(green)(x/3)#
#color(blue)(x_(blue))= "number of windows Big Blue washed" = 1/2*(x-color(red)(x_(red))-color(green)(x_(green)))#
#=1/2*(x-1/2x-x/3)#
#=1/2*((6x-3x-2x)/6)#
#=1/2*x/6#
#color(blue)(x/12)#
Total numbers of windows washed:
#x=color(red)(x_(red))+color(green)(x_(green))+color(blue)(x_(blue))+color(yellow)(x_(yellow))#
#x=color(red)(1/2x)+color(green)(x/3)+color(blue)(x/12)+color(yellow)(10)#
#x=color(red)((1*6)/(2*6)x)+color(green)((x*4)/(3*4))+color(blue)(x/12)+color(yellow)(10)#
#x=(color(red)(6x)+color(green)(4x)+color(blue)(x))/12+color(yellow)(10)#
#x-11/12x=cancel(11/12x)cancel(-11/12x)+10#
#12*(x-11/12x=10)#
#rarr 12x-11x=120#
#x=120 windows in total#
Now substituting the value of x in #color(red)(x_(red)),color(green)(x_(green)),color(blue)(x_(blue))#:
#color(red)(x_(red))=1/2(120)=color(red)(60#
#color(green)(x_(green))=120/3=color(green)(40)#
#color(blue)(x_(blue))=(120)/12=color(blue)(10#
