Question 388168
Let numerator = x
Problem statment: demoninator = (x+2)
So the fraction looks like:
x/(x+2)

Second part of problem statement: 
Numerator (x-1)
Denominator (x+2)-1

So the new fraction (forget about the old one!) looks like:
{{{(x-1)/(x+2-1)}}}

For subtraction simplifies to:
{{{(x-1)/(x+1)}}}

Set the equation. Problem statement says this fraction thing is equal to 1/2.
{{{(x-1)/(x+1) = 1/2}}}

The 1/2 fraction doesn't look too nice so let's just make it 1. We can do that by multiplying both sides of the equation by 2, i.e. 2/1.

{{{(2/1)((x-1)/(x+1)) = (2/1)(1/2)}}}

The right side just becomes 1 (the 2s cancel out.)
The left side is tricky. You must multiply the numerator by 2 only.
{{{(2)*(x-1)}}} becomes {{{2x-2}}}
 
The denominator is multiplied by 1, so it does not change at all. 
The new left side and right side of the equation is now:
{{{(2x-2)/(x+1) = 1}}}

Much nicer to look at. And when we have a fraction = 1, we can always multiply both sides by the denominator (x+1). This cancels the denominator on the left, and moves the (x+1) to the right side. This gets rid of the fraction, which is what we wanted.
{{{(x+1)((2x-2)/(x+1)) = (x+1)(1)}}}
{{{(2x-2) = (x+1)}}}
{{{2x-2 = x+1}}}
Combine like terms. We can put x terms on the left, and the constant (non-x) terms on the right. First subtract x from both sides.
{{{2x-2-x = x-x+1}}}
{{{x-2 = 1}}}
Now add 2 to both sides.
{{{x-2+2 = 1+2}}}
{{{x = 3}}}

Check your answer by putting x into the original equation.
{{{(x-1)/(x+1) = 1/2}}}
{{{(3-1)/(3+1) = 1/2}}}
{{{(2)/(4) = 1/2}}}