First, add #color(red)(7)# to each side of the equation to isolate the #q# term and keep the equation balanced:
#3/4q - 7 + color(red)(7) = 8 + color(red)(7)#
#3/4q - 0 = 15#
#3/4q = 15#
Next, multiply each side of the equation by #color(red)(4)/color(blue)(3)# to solve for #q# while keeping the equation balanced:
#color(red)(4)/color(blue)(3) xx 3/4q = color(red)(4)/color(blue)(3) xx 15#
#cancel(color(red)(4))/cancel(color(blue)(3)) xx color(blue)(cancel(color(black)(3)))/color(red)(cancel(color(black)(4)))q = color(red)(4)/color(blue)(3) xx (3 xx 5)#
#q = color(red)(4)/cancel(color(blue)(3)) xx (color(blue)(cancel(color(black)(3))) xx 5)#
#q = 4 xx 5#
#q = 20#