Solving Proportions: Methods and Real-World Applications

Introduction to Proportions

Proportions are found in many different areas of our everyday lives. Proportions are used in baking, shopping, driving, and even estimating the population of animal species. Conservationists can use data from two or more experiments to estimate the size of an animal population and to determine if it is increasing or decreasing. In this real-world example, fifty bears were captured, tagged, and released on the Keweenaw Peninsula to estimate the size of the bear population. One year later, two tagged bears were found in a random sample of one hundred bears. Proportions will be used to determine the population of the bears.

The bear population example can be solved by applying the concept of proportions, such as the one used in problem 55 on page 437 of the textbook Elementary and Intermediate Algebra (Dugopolski, 2012). When using the concept of proportions, the ratio of originally tagged bears to the whole population of bears is equal to the ratio of recaptured bears to the size of the random sample. To determine the estimated population, variables will be used, as well as rules for solving proportions.

Setting up the proportion requires identifying two equal ratios. The ratio of originally tagged bears to the whole bear population is 50/x. The ratio of recaptured tagged bears to the random sample size is 2/100. These ratios form the following proportion:

50/x = 2/100

This is the set-up proportion and is now ready to solve. In this proportion, the extremes are fifty and one hundred, and the means are x and two. Cross multiplication will be used at this point.

50(100) = 2x
5000 = 2x

The Bear Population Problem

Now we divide both sides by two:

5000/2 = 2x/2
x = 2500

The bear population of the Keweenaw Peninsula is estimated to be 2,500 bears. This method, known as the capture-recapture technique, is widely used in wildlife management and conservation to estimate population sizes without counting every individual.

For the second problem in this assignment, the equation needs to be solved for y. Since the equation has a single fraction on both sides of the equal sign, it can be considered a proportion. This proportion can be solved by cross multiplying the extremes and means, as we did in the first problem.

(y − 1)/(x + 3) = −3/4

This is the original equation. The first step is to cross multiply:

4(y − 1) = −3(x + 3)

Solving Algebraic Proportions

This is the result from cross multiplying. Next, distribute the four and the negative three:

4y − 4 = −3x − 9

Add four to both sides:

4y − 4 + 4 = −3x − 9 + 4
4y = −3x − 5

In this step, divide both sides by four:

4y/4 = (−3x − 5)/4
y = (−3x − 5)/4

This can also be written as y = (−3/4)x − 5/4, which is a linear equation in slope-intercept form, y = mx + b. The slope −3/4 is the same as the coefficient on the right side of the original equation. This solution does not contain an extraneous solution, as all steps are reversible and the variable is not restricted.

Conclusion

In conclusion, you can see that proportions are used in many different situations. In the first problem, proportions are used to determine the estimated bear population of the Keweenaw Peninsula. In the second problem, we used the rules of proportions and cross multiplied the extremes and means in order to solve the equation. It is very important to understand proportions, as it can help in solving a variety of practical and theoretical problems.