First, multiply the fraction #11/(2x)# by the appropriate form of #1# to have a common denominator for the two fractions on the left side of the equation:
#(2/2 xx 11/(2x)) + 5/(4x) = 3/4#
#(2 xx 11)/(2 xx 2x) + 5/(4x) = 3/4#
#22/(4x) + 5/(4x) = 3/4#
Next, add the two fractions on the left side of the equation:
#(22 + 5)/(4x) = 3/4#
#27/(4x) = 3/4#
Then, multiply each side of the equation by #color(red)(4)/color(blue)(3)# to eliminate the fraction on the right side of the equation while keeping the equation balanced:
#color(red)(4)/color(blue)(3) xx 27/(4x) = color(red)(4)/color(blue)(3) xx 3/4#
#cancel(color(red)(4))/cancel(color(blue)(3)) xx color(blue)(cancel(color(black)(27))9)/(color(red)(cancel(color(black)(4)))x) = cancel(color(red)(4))/cancel(color(blue)(3)) xx color(blue)(cancel(color(black)(3)))/color(red)(cancel(color(black)(4)))#
#9/x = 1#
Now, multiply each side of the equation by #color(red)(x)# to solve for #x# while keeping the equation balanced:
#color(red)(x) xx 9/x = color(red)(x) xx 1#
#cancel(color(red)(x)) xx 9/(color(red)(cancel(color(black)(x)))) = x#
#9 = x#
#x = 9#