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Mat 222

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Solving Proportions
MAT 222
Roberta Bledsoe
July 21, 2013

Solving Proportions
When having to solve for proportions we have a process to go through. First we must use the Extreme-Means property. The first problem, example A., is the word problem about bear population. Putting the word problem into an equation we get b/50 = 100/2. We must solve the equation to find what b will equal. Then we have a second problem, example B., which is Y-1/x+3 = -3/4. We will solve example B. until we get to the linear equation format or an extraneous solution. Both examples of proportions will be solved in the same process though the problems themselves seem very different. A.
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?
B=bear population The variable b will be used to identify the bears in population b/50 = 100/2 I used the ratio equation to solve for b.
(2)(b)=(50)(100) I crossed multiplied the two fractions. We multiply the two extremes 2 and b. Then we multiply our means 50 and 100.
2b=5000
B=5000/2 To isolate the b I divided 2 on both sides.
B=2500 After diving the 5000 by 2 we see that b then equals 2500. B.
Y-1/x+3 = -3/4 This is the proportion will be solved into linear equation format.
-3(y-1)= 4(x+3) The equation is then cross multiplied using the extreme-means property.
-3y+3=4x+12 The 3 and 4 are then distributed into the parentheses.
-3y+3-3=4x+12-3 Subtract -3 from both sides
-3y=4x+9
-3y=4x+9 Dividing by -3 on both sides.
-3 -3 y=11/3+3 In linear equation in the form of y=mx+b
Going about solving the equations a different way we will not cross multiply.
Y-1/x+3 = -3/4 y-1=-3/4 (x+3) Multiply x+3 to both sides. y-1=3/4x-21/4 y-1=3/4x-21/4 Adding 1 to both sides.
+1 +1 y=3/4x-11/4 In these examples we can see the method of solving for proportions. The processes of using the Extreme-Means property are used on both examples. The difference, in example A., is the part where we must solve the equation to find what b will equal. Then we will put example B into linear equation format to solve the problem. Example B. could be done differently. If example B as shown. If you divide -3/4 on both sides to get the equation into quadratic format by having it now equal zero and solving from there.

References

Dugopolski, M. (2012). Elementary and intermediate algebra (4th ed.). New York, NY: McGraw-Hill Publishing.

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