Dealing with Proportions and Basic Algebra
Chapter 4: Dealing with Proportions and Basic Algebra
3. Add the result from Step 2 to the 50 homes that were sold by March 1
First, 50 – 20 = 30 tells you that 30 homes were sold in March. Let the fraction involving the days have 20 in the numerator and 31 in the denominator.
(“Thirty days hath September . . .”) Put the xopposite the 20 and the 30 opposite the 31. So your proportion looks like this:
x 31 days 20 days
30 homes homes
=
Now cross-multiply and divide each side of the equation by the coefficient of x:
. x
x x x 31 20
30 600 31
31 600
31 31 19 354 .
=
=
=
It looks like about 19 homes were sold between March 1 and March 20, so the year-to-date number on March 20 is 20 + 19 = 39 homes. Great work!
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Chapter 4: Dealing with Proportions and Basic Algebra
When using proportions to interpolate, be careful not to assume a relatively equal rate between the consecutive entries in the table. In the practice prob- lem, if I would have tried to interpolate starting from the beginning of the year, I would have received a completely different (and incorrect) number. Only 20 homes were sold during the first two months and then 30 more during the third month. The average change is different for the different months.
Handling Basic Linear Equations
When solving proportions for an unknown value, and when performing many mathematical computations involving measures, money, or time, you’ll likely find yourself working with an equation that has a variable in it. You solve for the value of the unknown by applying basic algebraic processes. These processes keep the integrity of the equation while they change its format so you can find the value of the unknown.
When solving a basic linear equation of the form ax+ b= c, where xis an unknown variable and a, b, and care constants, you apply the following rules (not all will be used in every equation — you pick and choose per situation):
Add the same number to each side of the equation.
Subtract the same number from each side of the equation.
Multiply both sides of the equation by the same number (but don’t mul- tiply by 0).
Divide each side of the equation by the same number (but don’t divide by 0).
Distribute (multiply) each term in parentheses by the number outside the parentheses so that you can drop the parentheses.
Check your answer in the original equation to be sure it makes sense.
Here’s an example you can try: Solve this equation for the value of x: 3(2x– 7) = 4x– 13.
First distribute the 3 over the terms in the parentheses by multiplying each term by 3. You get 6x– 21 = 4x– 13.
Now subtract 4xfrom each side and add 21 to each side. Doing so allows you to get the x’s on one side of the equation and the numbers without x’s on the other side. Here’s how your math should look:
x x
x x
x x
6 21 4 13
4 4
2 21 13
21 21
2 8
- = -
- -
- = -
+ +
= +
Now divide each side of the equation by 2, and you get that x= 4. Always check your answer. If you replace the xwith 4, you get 3(2×4 – 7) = 4×4 – 13.
In the parentheses, you get 8 – 7 = 1. So the equation now reads 3(1) = 16 – 13, or 3 = 3. When you see a true statement — one in which a number equals itself — you know that you’ve done the process correctly.
Just for kicks, try out another practice problem. Solve the following propor- tion for the value of x:
x x 400 200
30
= 2 +
First reduce the fraction on the left by dividing both the numerator and denominator by 200. Then cross-multiply. Your math looks like this so far:
x x
x x
x x
x x
400 200
30 2
2 1
30 2
1 30 2 2
30 4
2 1
= +
= +
+ =
+ =
^ h ^ h
Now solve the equation for the value of x.
x x
x x
x x x 30 4 30 3 3 30
3 3 10 + =
- -
=
=
=
Check to be sure that xdoes equal 10. By checking your answer to see if it makes sense, you’re more apt to catch silly mistakes. Go back to the original proportion and replace the x’s with 10s, like so:
400 200
10 30 2 10
400 200
40 20
= +
=
^ h
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Chapter 4: Dealing with Proportions and Basic Algebra
You can stop with the check at this stage, noting that both fractions are equal to 1⁄2.Or you can cross-multiply to get 8,000 on each side of the equation. You also can reduce both fractions to get the same numbers. In any case, you’ve shown that the 10 does represent the xin the proportion.
Comparing Values with Variation
What is variation?It’s how one entity varies with relation to another. Variation comes in two varieties: direct and indirect. Consider the following statements:
The pressure of a gas is indirectly proportional to its volume.
The productivity of the team is indirectly proportional to the number of absences.
The revenue earned is directly proportional to the hours worked.
The speed of the truck is directly proportional to the acceleration.
I explain the differences between the two types of variation in the following sections.
Getting right to it with direct variation
When one quantity varies directly with another, one quantity is directly pro- portional to the other. An algebraic expression of how the two quantities a and bvary directly is: a= kb, where kis the constant of proportionality,which is the multiplier that connects the two quantities. If you divide each side of the equation by b, you get the following proportion:
b
a k
=1
And this shows you how the proportion and variation are tied together. When quantities vary directly, one quantity is a multiple of the other. If you know at least one of the quantities in the relationship, you can solve for the other (if it’s unknown). Check out the following example problem to see exactly what I mean.
The revenue earned by the sales office varies directly with the number of hours spent out in the field. Last month, the sales team earned $450,000 and spent a total of 1,200 hours meeting with customers. How much will the team earn if it spends 2,000 hours out in the field?
Using direct variation, you need to find the value of k, the constant of propor- tionality. Write the equation $450,000 = k ×1,200 for the relationship between the quantities. Divide each side of the equation by 1,200, and you get that k= 375. Now, to solve for the amount earned if the team is in the field for 2,000 hours, plug the numbers into the formula and solve: 375 ×2,000 = $750,000.
What does the k, the constant of proportionality, really represent in this problem? You can interpret kto be the number of dollars earned per hour spent with a customer — a sort of average for the business.
Going the indirect route with indirect variation
Two quantities vary inversely,or indirectly, if, as one quantity increases, the other decreases proportionately. Instead of a constant multiplier of k, like you find with direct variation, the equation tying the two quantities together is:
a b
=k
Try your hand at the following indirect variation example.
When a 2 x 4 pine board is suspended between two buildings (with just the ends of the board at the building edges), the maximum weight that the board can support varies indirectly with the distance between the buildings. If the distance between the buildings is 10 feet, the board can support 480 pounds.
What’s the maximum amount of weight that such a board can support if the buildings are 20 feet apart? How about 30 feet apart?
First write the relationship between the 10-foot distance and the 480 pounds so you can solve for k by multiplying each side of the equation by 10:
, k k 480 10
4 800
=
=
Next, solve for the amount of weight that can be supported, putting 4,800 for the value of kand the distance of 20 feet in the denominator:
, 20 4 800
240
=
The board can support half as much at that distance — or 120 fewer pounds when the distance is increased by 10 feet.
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Chapter 4: Dealing with Proportions and Basic Algebra
Now, using the equation and replacing the denominator with 30, you get:
, 30 4 800
160
=
Increase the distance by another 10 feet, and the amount of weight decreases by another 80 pounds.