Word Problem - Age Problem

Category: Algebra

"Published in Newark, California, USA"

Six years ago, Rosario was 4 times as old as her daughter. Ten years from now, she will be only twice as old as her 
daughter. How old are they now?

Solution:

The given word problem is about age problem where you   need to solve for the age of a person or people. Let's analyze the given word problem above as follows:

Let         x = be the present age of Rosario
              y = be the present age of Rosario's daughter
         x - 6 = be the age of Rosario six years ago
         y - 6 = be the age of Rosario's daughter six years ago
      x + 10 = be the age of Rosario ten years from now
      y + 10 = be the age of Rosario's daughter ten years from now.

If the first statement says, "Six years ago, Rosario was 4 times as old as her daughter." then the first working equation will be

If the second statement says, "Ten years from now, she will be only twice as old as her daughter." then the second working equation will be

Next, equate the two working equations in order to solve the age of Rosario's daughter as follows

Substitute the value of y to either of the two working equations in order to solve for the age of Rosario as follows

Therefore, in present time, Rosario is 38 years old while her daughter is 14 years old.

More Integration Procedures, 7

Category: Integral Calculus

"Published in Suisun City, California, USA"

Evaluate

Solution:

Consider the give equation above

If

then

Therefore, the given equation above can be integrated by simple integration using integration by power as follows

Solving Trigonometric Equations, 4

Category: Trigonometry, Algebra

"Published in Suisun City, California, USA"

Find the value of x for 

Solution:

Consider the given equation above

By looking at the equation above, the factors of 4sin x cos x are 2sin x and 2cos x while the factors of -1 are 1 and -1. Since there are 2sin x, -2cos x, and -1 in the equation above, then we can rewrite the above equation as follows

Equate each factor to zero and solve for the value of x.

If

then

Since the value of sine is positive at 1st and 2nd Quadrant, then 

where n is the number of revolutions.

If

then

Since the value of cosine is negative at 2nd and 3rd Quadrant, then 

where n is the number of revolutions.