When an equation has fractions, you can get rid of the fractions by multiplying through by the LCD. (in this case, it is 2)
#color(blue)(2xx)15-(color(blue)(cancel2xx)5x)/cancel2= color(blue)(2xx)9+(color(blue)(cancel2xx)1x)/cancel2#
#color(white)(xxxxxxx)30-5x= 18+x " "larr "no fractions!"#
#color(white)(xxxxxxx)30-18 = x+5x " "larr# re-arrange terms
#color(white)(xxxxxlxxxxx)12 =6x#
#color(white)(xxxxxxxxxxx)x=2#
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Background notes.
If you want to solve #4x= 28" "# you divide by #4# to isolate #x#
#(cancel4x)/cancel4 =28/4" "# which gives #x =7#
If you have #x/5 =8" "# you multiply by #5# to cancel the #5#
#(cancel5xxx)/cancel5 = 5xx8" # which gives #x=40#
The question above uses the same process, except there are two denominators.
Multiply every term in the equation by #2# so that you can cancel the denominators.
To re-arrange:
#color(white)(xxxxxxx)30-5x= 18+x" "larr# add #5x# to both sides
#color(white)(xxxxxxx)30-5xcolor(red)(+5x)= 18+x color(red)(+5x)#
#color(white)(xxxxxxxxxxxxxx)30= 18+6x#
Now subtract #18# from each side.
#color(white)(xxxxxxxxxxx)30color(blue)(-18)= 18color(blue)(-18)+6x#
#color(white)(xxxxxxxxxxxxxx)12= 6x#
Divide both sides by #6#
#color(white)(xxxxxxxxxxxxxx)12/6= (6x)/6#
#color(white)(xxxxxxxxxxxxxxx)2= x#
