Question 236724
If I understand correctly, you're trying to find the value of Y, right?
{{{Y=(5x/3x)+9}}}
Since you have an x to the power of 1 in the numerator and in the denominator, we can cancel the x from the equation.  So:
{{{Y=(5/3)+9}}}
This is the same as to say:
{{{Y=(5/3)+(9/1)}}}
So now we look for a common denominator, which would be 3.
{{{Y=(5/3)(1/1)+(9/1)(3/3)}}}
{{{Y=(5/3)+(18/3)}}}
Now you can add the fractions as they have same denominator.
{{{Y=23/3}}}