Guitarman said:

So the equation is as follows:

. . . . .\(\displaystyle \L \frac{1}{1\,-\,\frac{3}{2\,+\,w}}\,= \,60\)

Since the numerator is equivalent to "1/1", since 1 - 3/(2 + w) = [(2 + w) - 3]/(2 + w) = (w - 1)/(2 + w), and since, when dividing by a fraction, one inverts and multiplies, this simplifies as:

. . . . .\(\displaystyle \L \frac{2\,+\,w}{w\,-\,1}\, =\,60\)

"Cross-multiplying", we get:

. . . . .\(\displaystyle \L 2\,+\,w\,= \,60w\,-\,60\)

Solve by the usual methods.

Note: The solution to any "solving" exercise may be checked by plugging it back into the original problem. You have proposed "w = 120" as a solution.

Checking:

. . . . .\(\displaystyle \L \frac{1}{1\,-\,\frac{3}{2\,+\,(120)}}\)

. . . . . . .\(\displaystyle \L =\,\frac{1}{1\,-\,\frac{3}{122}}\)

. . . . . . .\(\displaystyle \L =\,\frac{1}{\left(\frac{122\,-\,3}{122}\right)}\)

. . . . . . .\(\displaystyle \L =\,\frac{1}{\left(\frac{119}{122}\right)}\)

. . . . . . .\(\displaystyle \L =\,\frac{122}{119}\)

. . . . . . .\(\displaystyle \L \neq\,60\)

So the solution is likely some other value.

Eliz.