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 conservationists 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

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 : McGraw-Hill.

Source document

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 conservationists 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

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 : McGraw-Hill.

Source document