Pharm help *Math*

what is the question asking?
I would think that you would cross multiply the inner #'s and then the outer #'s and get rid of X. Example:
4:5/6=X:51
0.83x=204 and divide both sides by 0.83...
I could be totally way off...is that 51 or 5I that you posted..?

good luck!

Ok for ratio/proportion problems you would

1. Multiply the inner numbers (means)
2. Multiply the outer numbers (extremes)
3. Put the number with the X on the left
4. Solve for X

Example
2:10::5:X
2X = 50
X = 25

Hope it helps

Ratios are nothing but fractions. The example you posted can also be written as an equation of fractions. If I re-write the ratio as fractions in an equation, it becomes an algebra problem:

4 (numerator)/5/6 (denominator) = X (numerator)/5 (denominator)

Note that you have a fraction that has a fraction in the denominator. To find the answer to this problem you cross multiply, so you then have:

(4)(5) = (5X/6)

Clear the "6" out of the numerator of the right side of this equation. Remember that what you do to one side of the equation you must also do to the other side:

(4)(5)(6) = (5X/6)(6)

Which leaves you with:

(4)(5)(6) = (5X)

Isolate the X by dividing both sides of the equation by "5" to remove the "5" from the X variable and isolate it"

(4)(5)(6)/5 = (5X)/5

This results in:

(4)(6) = X, or 120/5 = X

X = 24

Here are webpages and websites that contain all the rules you need to know to work problems with fractions:

• http://www.mathleague.com/help/fractions/fractions.htm#simplifyingcomplexfractions - this web page has everything you want to know about fractions
• http://www.visualfractions.com/ - Visual Fractions Tutorials
• http://www.aaamath.com/fra.html#topic26 - Tutorials on working with Fractions