First, divide each side of the equation by #color(red)(5)# to eliminate the need for parenthesis while keeping the equation balanced:
#(5(2x + 1))/color(red)(5) = 50/color(red)(5)#
#(color(red)(cancel(color(black)(5)))(2x + 1))/cancel(color(red)(5)) = 10#
#2x + 1 = 10#
Next, subtract #color(red)(1)# from each side of the equation to isolate the #x# term while keeping the equation balanced:
#2x + 1 - color(red)(1) = 10 - color(red)(1)#
#2x + 0 = 9#
#2x = 9#
Now, divide each side of the equation by #color(red)(2)# to solve for #x# while keeping the equation balanced:
#(2x)/color(red)(2) = 9/color(red)(2)#
#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = 9/2#
#x = 9/2#
