Solving a proportion as we learned this week, means that you are missing an import number in your equation or fraction, and you need to solve for that missing value. As in my example, I did not know what percentage of bills we each should pay. We knew each other’s salaries; but we really had to sit down, crunch numbers, and figure it out. For this week’s assignment we were asked to work through two proportions. For the first proportion, number 56 on page 437 of Elementary and Intermediate Algebra by Dugopolski: 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? Beginning to solve this equation first, we needed to make a proportion with the number of tagged bears in the sample and in the population:

Number of the tagged bears in the sample compared to the sample size equals the number of tagged bears in the population: Population size. The population size is “x”, we need to solve for “x” as such: 2/100 = 50/x

Cross multiply or use the extremes means property
2 * x = 100 * 50
2x = 5000
Because we want to solve for “x” we must isolate it by dividing both sides by two. x = 5000/2 = 2500
Answer: x= 2500 bears
For the second problem, number 10 on page 444: we must solve the equation for y: Y-1 = - 3
X+3 4

Cross multiplying or use the extremes means property. According to Dugopolski, “We use the extremes-means property to solve proportions” (Dugopolski, 2011).

4*(y-1) = -3*(x+3)

[Distribute by 4 on the left and -3 on the right.

4y - 4 = -3x - 9

Add 4 to both sides and reduce to lowest terms.

4y - 4 + 4 = - 3x - 9 + 4
4y = -3x – 5

Divide each side by 4
y = (-3/4)x - (5/4) or y= -3x-5
4

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Chapter 1, Problems 2, 4, 12, 14, 20, 22
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