Question 399727
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Presuming you mean:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5}{4}z\ -\ \frac{3}{8}z]


rather than


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5}{4z}\ -\ \frac{3}{8z}]


which is an equally valid interpretation of what you wrote, proceed as follows:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5}{4}z\ -\ \frac{3}{8}z]


Find the LCD of 4 and 8.  Both 4 and 8 divide 8 evenly, so 8 is the LCD.


Convert *[tex \Large \frac{5}{4}].  You have to multiply the denominator by 2 to get from 4 to 8, so you have to multiply the numerator by 2 as well.  Result:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{10}{8}z\ -\ \frac{3}{8}z]


Now you can add the numerators.  10 plus -3 is 7, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{7}{8}z]


Now, for where I think you may have gone wrong.  You added 5 and -3 to get 2 and then simply ignored the fact that you had a different denominator in the second fraction.  Then, having erroneously decided that the sum was *[tex \Large \frac{2}{4}z], you changed *[tex \Large \frac{2}{4}] into 2.


Let's see if your arithmetic makes any sense.  I have one whole pizza and a quarter of a pizza.  The whole pizza is sliced into 8 slices, and the quarter pizza is two slices.  I give you the whole thing and tell you to eat what you want and calculate what you have when you are done.  You eat 3 of the slices and then, according to your arithmetic, you have 2 whole pizzas left.  I never believed that loaves and fishes story either.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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