Algebra

Running head: SOLVING PROPROTIONS 1 Running header should use a shortened version of the title if the title is long. Page number is

located at right margin.

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Solving Proportions

John Q. Student

MAT 222 Week 1 Assignment

Instructor’s Name

Date

SOLVING PROPORTIONS 2

Solving Proportions (title required on first line)

Proportions exist in many real-world applications, such as finding unit price or estimating an

animal population. By comparing data from at least two experiments, conservationists are able to

predict patterns of animal increase or decrease. In this situation, 300 bluegill in Spearhead Lake were

tagged and released to estimate the size of the bluegill population. A month later, after capturing 120

bluegill from the same lake, proportions were used to determine the lake’s population.

In reading over the “Capture-Recapture” method in #55 on page 437 (Dugolpolski, 2012), the

concept of proportions allows the assumption the ratio of originally tagged fish to the whole

population is equal to the ratio of recaptured tagged fish to the size of the sample. To determine the

estimated solution, variables will be defined and rules for solving proportions used.

The ratio of originally tagged bluegill to the whole population is 300 /x.

The ratio of recaptured tagged bluegill to the sample size is 45/120.

300 = 45 This is the proportion set up and ready to solve. Cross multiplication is required

X 120 at this point. The extremes are 120 and 300. The means are x and 45.

120(300) = 45x

36000 = 45x Divide both sides by 45.

45 45

x = 800 The bluegill population of Spearhead Lake is estimated to be around 800 fish.

For the second problem in this assignment, the equation must be solved for y. Continuing the

discussion of proportions, a single fraction (ratio) exists on both sides of the equal sign so basically it

is a proportion, which can be solved by cross multiplying the extremes and means.

y + 5 = -1

x-2 2

2(y + 5) = -1(x – 2) The result of the cross multiplying.

SOLVING PROPORTIONS 3

2y + 10 = -x + 2 Distribute 2 on the left and -1 on the right.

2y + 10 – 10 = -x + 2 – 10 Subtract 10 from both sides.

2y = -x – 8

2y = -x – 8 Divide both sides by 2

2 2

y = -½ x - 4 This is a linear equation in the form of y = mx + b.

After comparing the solution to the original problem, it is noticed that the slope, -½ ,is the

same number on the right side of the equation. This indicates another method exists for solving

the equation.

y + 5 = -1

x-2 2

y + 5 = -1/2 (x-2) I would multiply both sides of the equation by (x – 2), which cancels the

y+ 5 -5= -1/2x + 1 -5 denominator on the left, subtract 5 from both sides to get y alone, and

y = -12/x -4 then simplify the right side. It would save a couple of steps from solving it

the other way.

Conclusion paragraph would go here. Remember to include 4-5 sentences to make a

complete paragraph.

SOLVING PROPORTIONS 4

Reference Dugopolski, M. (2012). Elementary and intermediate algebra (4th ed.). New York, NY:

McGraw-Hill Publishing.

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