For this particular problem, we do not have any parenthesis or exponents. Therefore, we must go off of just #"MDAS"# for this particular problem. So, the first step would be #"M"#, or Multiplication.
Here is the challenging part. We need to find the LCM, or Least Common Multiple, or the two denominators (on the bottom of a fraction) in this particular question. In this case, we have #9# and #5# as our two denominators.
If you prime-factorize #9#, you get #3*3#. If you prime factorize #5#, you simply get #5#. Since the prime factors of the denominators do not share any factors in common, then the LCM is just the product (multiply) of the denominators. In less complicated words, the LCM is the product of #5# and #9#, or #5*9=45#.
Now, let's get back to #"M"#. We still have to get rid of those denominators, so let's multiply both sides of the equation by the LCM that we found, #45#. We get this:
#3+1/9x=1/5x#
#45(3+1/9x)=45(1/5x)#
#135+5x=9x#
Now, let's continue on with the order of operations. We have to go to #"D"# next, but we don't have anything to divide yet. So what about #"A"# and #"S"#? Yeah, we do have something to do there!
Now, we can subtract #5x# from both sides of the equation:
#135+5x=9x#
#135=4x#
Lastly, we have one more step, which is to divide both sides by #4# in order to isolate #x#.
#x=135/4#
#x=33.75#
That's all, I hope it helps!
