Haven't found the Essay You Want?
For Only $12.90/page

Solving Proportions.Elementary & Intermediate Algebra Essay

Solving Proportions


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

            y-1 = -3 this problem is a proportion in its own

            x+3 4

            y-1 (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


            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.


Dugopolski, M. (2011). Elementary & Intermediate Algebra. New York : McGraw-Hill.

Source document

Essay Topics:

Sorry, but copying text is forbidden on this website. If you need this or any other sample, we can send it to you via email. Please, specify your valid email address

We can't stand spam as much as you do No, thanks. I prefer suffering on my own