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Saturday, October 6, 2012

Percent Mixture Problem

Category: Algebra, Chemical Engineering Math

"Published in Newark, California, USA"

A flask containing 250 mL of 80% iodine solution tips over spilling part of its contents. This was subsequently replaced by an equal amount from a mixture of 40% iodine. If a 72% solution resulted, how much was spilled?

Solution:

Let x be the amount of 40% iodine solution to replace the spilled solution of 80% iodine solution.

Let (250 - x) be the amount of remaining solution of 80% iodine after it is spilled. 

Since the given problem is about the mixture problem where the dilution process is involved, let's use the principle of chemistry in making solutions. Please remember that the

Amount of Iodine before dilution = Amount of Iodine after dilution

Therefore, 






The amount of 80% iodine solution that was spilled is 50 mL.

Friday, October 5, 2012

Integrating Rational Fractions

Category: Integral Calculus

"Published in Newark, California, USA"

Evaluate

     

Solution:

First, let's examine the numerator and denominator if they can factor or not. Since the denominator can be factored, the above item can be written as 

  


Next, split the denominator's factors into a single factor, we have



where A, B, and C are constants and we have to determine the unknown given constants. Multiply each sides by their Least Common Denominator (LCD) which is x(x - 2)(x + 2).
 



Equate each term, we have

For x2: (equation 1)

For x: (equation 2)

For x0:      (equation 3)

From equation 3,




From equation 2,




Substitute the values of A and B in equation 1, we have
 
 
 
 

Substitute the value of C in equation 2, we have




Therefore,


 

Thursday, October 4, 2012

Simplifying Algebraic Fractions

Category: Algebra

"Published in Newark, California, USA"

Simplify in lowest term for
 

Solution:

Consider the given equation
 

Group the terms in the numerator
 

You notice that the first group is the sum and the difference of two squares and the second group is having a common coefficient. Let's factor the grouped terms as follows
 

The common factor at the numerator is (x + y). We can take out (x + y) from the group
 

Simplify the grouping at the second group
 

The (x - y + 2) will be cancelled as their Greatest Common Factor (GCF)
 

Therefore, the answer is x + y.