
University/College:
University of Arkansas System 
Type of paper: Thesis/Dissertation Chapter

Words: 690

Pages: 3
Solving Proportions.Elementary & Intermediate Algebra
Solving Proportions
Introduction
Problem 1
Bear population. To estimate the size of the bear population on the Keweenaw Peninsula, conservationists captured, tagged, and released 50 bears. One year later, a random sample of 100 bears included only 2 tagged bears. What is the conservationist’s estimate of the size of the bear population?
I think using a simple ratio equation would work here,
let b = bear population
=
cross multiply the expression
The equation to form will be 2b = 50×100
This will be the first solution to the above equation 2b=5000 divide 2 by 5000
Therefore, the answer will be
5000 =2b
Therefore b=2500 answer
In conclusion the conservasionists estimated the bear population to be 2500 if the whole population is assumed to remain constant.
Solution to Problem 2
In the calculation of the second problem on page 444, I am required to solve this equation for y. In order to make everything clear the first thing I decided to work on as well as the first thing I notice is that it is a single fraction (ratio) on the two sides of the equal sign. Most importantly, I basically I realized that it was a proportion which can be solved by cross multiplying the extremes and means. Therefore, I cross multiplied the both sides of the equation
The following is the solution to the algebraic expression
y1 = 3 this problem is a proportion in its own
x+3 4
y1 (x+3 = 3 (x+3) multiply both sides by x+3 – using the extreme means
Therefore the result of the multiplication is +3 4 property
Hence 1= 3x+3 add 1 to 3. A number that appears to be a solution but causes 4 0 in a denominator is called an extraneous solutions
The resulting expression follows below
y=3x+4
4 is the solution
At the end of the calculation, the appearance of equation I ended up with as the solution to problem 10 would be a linear equation. In conclusion, I noticed that the coefficient of x is different than the original problem is that x+3 and in my problem it is 3x/4.
I finally realized that I could solve the problem by cross multiplying the equation at the beginning of the problem.
References
Dugopolski, M. (2011). Elementary & Intermediate Algebra. New York : McGrawHill.
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