First, multiply both sides of the equation by #color(red)(35)# to eliminate the fractions while keeping the equation balanced. #color(red)(35)# was selected because it is the common denominator for the two fractions (#7 xx 5)#:
#color(red)(35)(x/5 - (3 + x)/7) = color(red)(35) xx 0#
#(color(red)(35) xx x/5) - (color(red)(35) xx (3 + x)/7) = 0#
#(cancel(color(red)(35))7 xx x/color(red)(cancel(color(black)(5)))) - (cancel(color(red)(35))5 xx (3 + x)/color(red)(cancel(color(black)(7)))) = 0#
#7x - color(red)(5)(3 + x) = 0#
Next, expand the terms within parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis:
#7x - (color(red)(5) xx 3) + (color(red)(5) xx x) = 0#
#7x - 15 - 5x = 0#
#7x - 5x - 15 = 0#
#(7 - 5)x - 15 = 0#
#2x - 15 = 0#
Then, add #color(red)(15)# to each side of the equation to isolate the #x# term while keeping the equation balanced:
#2x - 15 + color(red)(15) = 0 + color(red)(15)#
#2x - 0 = 15#
#2x = 15#
Now, divide each side of the equation by #color(red)(2)# to solve for #x# while keeping the equation balanced:
#(2x)/color(red)(2) = 15/color(red)(2)#
#(color(red)(cancel(color(black)(12)))x)/cancel(color(red)(12)) = 15/2#
#x = 15/2#
