ALGEBRAIC EXPRESSIONS
Rule 3. When a term with an exponent is divided by a like term with an exponent, the exponent of the divisor is subtracted from the exponent of the dividend. For
example:
x5 ÷ x3 = x5–3 = x2. Example Problem: Divide 6x2 + 9xy – 6x by 3x.
Solution: Write down the dividend and divisor in the following manner:
3x
)
6x2 + 9xy – 6x .Divide each term of the dividend by the divisor, write down each quotient, and place the appropriate sign before each quotient.
2x + 3y – 2 3x
)
6x2 + 9xy – 6x 6x29xy 9xy 6x 6x .
Example Problem: Divide x2 – 2xy + y2 by x – y.
Solution: Arrange the dividend and divisor as for long division of numbers in arithmetic.
x – y2 x – y
)
x2 – 2xy + x x2 – xy xy + y2 xy + y2.Example Problem: Divide a4 + a3 – 9a2 – 16a – 4 by a2 + 4a + 4.
Solution: Arrange the dividend and divisor as in the previous example.
a2 – 3a – 1 a2 + 4a + 4
)
a4 + a3 – 9a2 – 16a – 4a4 + 4a3 + 4a2 – 3a3 – 13a2 – 16a – 3a3 – 12a2 – 12a – a2 – 4a – 4 – a2 – 4a – 4 .
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Practice Problems
1. Find the sum of 2a + 3b and a – 6b.
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2. Add 3b – 4bc, 5a + 3bc + 2c, and 6a + 4c.
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3. Find the sum of 6a + 20b + 4ab + c, 6b – 8ab – 9c, and – 4a + 6c.
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4. Add 5x + 3xy + 4y and –8x – 2xy –3y.
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5. Find the sum of –8ab + 2a + c, 7a + 9ab – 4c, 8c – 12ab + 6a, and 8b + 4c + 3ab.
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6. Subtract 3a2 – 2b + c from 4a2 – 6b + 2c.
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7. Take 6x + 3y from 4x2 – 6xy – 4y.
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8. Subtract 6a(b + c) + 4(a + b) from 9a(b – c) – 8(b + c).
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9. Subtract 4a2 + 3ab + 6c2 from 9a2 – 8c2.
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10. Subtract 4(x + xy) – 6y from 6x2 – 6(x + 2y).
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11. Multiply x2 + 2xy + y2 by 4a.
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12. Multiply 4a + 6ab + 3c by 5ab.
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13. Multiply 3a + 4b by a – b.
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14. Multiply (a + 3b) – (a + b)2 by 4a.
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15. Simplify the expression (x + y)2 – (2x – 3y).
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128 PRINCIPLES OF ALGEBRA
16. Divide 8a2 + 4ab by –2a.
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17. Divide 27a3 + 9a2 + 12ab + 6b by 3a.
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18. Divide m4 + 57m – 70 by m2 + 3m – 5.
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19. Divide x3 – 3x2y + 3xy2 – y3 by x – y.
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20. Divide m4 + 64 by m2 + 4m + 8.
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EQUATIONS
An equation may be simple or complicated, but it is always an expression of two quantities that equal each other. For example, 2 = 2 is a simple equation. Another simple equation is 3 × 4 = 2 × 6 because 3 × 4 = 12, 2 × 6 = 12, and 12 = 12. If variables are used, an equation may read
3a = 2b where
a = 4 b = 6.
In this case, a = 4 and b = 6; so, 3 × 4 = 12, 2 × 6 = 12, and 12 = 12.
The two sides of an equation (the parts to the left and right of the equals sign) must remain equal, no matter what operation is performed to solve for the variable. If an arithmetical operation is performed on one side of the equation, the same operation must be performed on the other side of the equation.
Solving Problems with Equations
Solving algebraic equations usually involves letters or symbols that represent the numerical value of an unknown quantity. For instance, in the previous example, a = 4 and b = 6. Additional data in the problem make it possible to set up an equation. If enough data are given, the numerical value represented by the letter can be obtained. Following are the general steps for solving problems involving simple equations:
1. Let a letter or symbol represent one unknown number.
2. If a second number is given, represent it in terms of the first unknown, using the stated conditions in the problem.
3. Set up an equation, using the relations stated by the problem.
4. Solve for the unknown number.
Equations 129
5. Use the new number to solve for other unknown numbers, using the conditions given in the problem.
6. Check the numbers to see whether they meet the conditions of the problem.
Example Problem: The sum of two numbers is 92, and their difference is 4. Find the numbers.
Solution: Let x = the bigger number, and x – 4 = the smaller number. Set up the equation according to the conditions of the problem:
x + x – 4 = 92.
Solve for x:
2x – 4 = 92 2x = 92 + 4 = 96
x = 96 ÷ 2 = 48, which is the bigger number.
Then solve for the other unknown:
48 – 4 = 44, the smaller number.
Finally, check to see whether these numbers meet the conditions in the problem:
48 – 44 = 4.
This example not only shows how to set up a problem, it also shows the operations in solving an equation for an unknown. That is, to solve for x, you must move the –4 on the left side of the equation to the right side of the equation. In so doing, the sign changes from – to +, because moving a term from one side of the equation to the other changes its sign. Put another way, moving subtraction to the equation’s other side makes it addition. In the same way, moving addition to the equation’s other side makes it subtraction.
Then, to solve for 2x, which, remember, means 2 • x, you move the 2 to the right side of the equation. In so doing, you divide 96 by 2 to obtain 48 (the bigger number), because moving multiplication from one side to the other of an equation means that it becomes division. In the same way, moving division from one side to the other becomes multiplication.
Example Problem: Ten steel plates of a certain size weigh 60 pounds each. Their total weight is equal to the weight of eight plates of a different size. What is the weight of one of the different sized eight plates?
Solution: Choose y to equal the weight of one of the eight plates. Mathematically, it is stated as, “Let y = weight of one of the eight plates.”
Because ten plates each weigh 60 pounds, then 10 × 60 lb = 600 lb, the weight of the ten plates.
Thus, 8y = the weight of the other eight plates.
But, the total weights are the same: 600 lb. Therefore, 8y = 600 lb and y = 600 ÷ 8 = 75 lb, the weight of one plate.
130 PRINCIPLES OF ALGEBRA
Practice Problems
1. Let a = 12, b = 4, and c = 2. Find the value of the following:
a. 2a + 3b _______________________________________________
b. a + 2b + c______________________________________________
c. 3b – c ________________________________________________
2. In the equation a + 2b + c = 14, what is the value of a if b = 2 and c = 8?
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3. Write the following equations, using the symbols a, b, c, d, and e where a = 4, b = 1, c = 5, d = 2, and e = 3.
a. 4 + 5 + 1 = 3 + 4 + 1 + 2 __________________________________
b. 1 + 5 + 3 = 3 + 1 + 2 + 3 __________________________________
c. 5 + 4 + 2 = 5 + 3 + 1 + 2 __________________________________
4. A person receives $572 for 8 days of work. Using an algebraic equation, determine how much this person receives per day.
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5. Five pounds of copper can be purchased at $1.07 per pound. An inferior grade may be bought for 72¢ per pound. By means of an equation, deter-mine how many pounds of the inferior grade could be purchased for the same money that would be expended for 5 pounds of the better grade.
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6. Divide 35 into two parts, so that 4 times the lesser number equals 3 times the greater number.
Greater number: ________________________________________
Lesser number: _________________________________________
7. A hundred and five feet of chain link fence material is available to build a square pen. An existing wall forms one side of the pen. What is the maxi-mum area that can be enclosed by the wall and the fence material?
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8. A small ship is chartered for a weekend excursion for 100 passengers at
$150 each. The ship’s owner agrees to reduce the rate $10 per person for each additional passenger. What is the total cost per person if 130 people participate in the cruise?
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9. A student received A grades on 50% of one term’s assignments and B grades on five other assignments for the term. The remaining term assignments, or 4/9 of the total, were not handed in. How many assignments were there?
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10. A speculator bought 100 shares of stock. She sold 25 shares at twice their cost, then the company went into bankruptcy. The remaining shares were sold for 20¢ each. Her net loss was $235. What was the initial price per share?
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FORMULAS
A formula is a group of numerical symbols arranged to express briefly a single concept. A formula may be a simple term or expression, or it may be in the form of an equation. Formulas are used to save space and time when writing and working problems. For example:
Tm = D × Td where
Tm = money earned in one month D = number of days worked Td = money earned in one day.
Instead of a formula, the same information can be written: “The total amount of money earned in one month is equal to the number of days worked that month multiplied by the amount of money earned in one day.” The formula is certainly quicker to write. In all formulas, when letters and symbols are used in place of numbers, the letters or symbols must be defined.
Example Problem: An oilfield worker earns $2,485 per month and works 12 months per year. His or her average living expenses are $2,050 per month. The balance is placed in a savings account. How much does he or she save in a year?
Solution: To reduce the statement to an algebraic formula, choose a letter for each factor to be considered in solving the problem. Let D stand for the number of dollars saved in a year, S for monthly earnings in dollars, and E for monthly expenses in dollars. Now using these letters, set up the algebraic formula:
D = (12 × S) – (12 × E) or written in a simpler form,
D = 12S – 12E where
D = number of dollars saved in a year S = monthly earnings in dollars E = monthly expenses in dollars.
Then solve for D by substituting numbers for the known factors:
S = $2,485; E = $2,050.
D = (12 × 2,485) – (12 × 2,050) = 29,820 – 24,600
D = $5,220 saved in a year.
132 PRINCIPLES OF ALGEBRA
Transposition
Sometimes, to find the unknown quantity in a formula, it is necessary to transpose, or move, the terms. For example, consider the formula
Tm = D × Td where
Tm = money earned in one month D = number of days worked Td = money earned in one day.
This formula states that you can determine Tm if you know D and Td. However, suppose you know the amount of the monthly check (Tm) and the number of days worked (D), and you need the daily rate of pay (Td). In this case, the formula must be written to solve for Td, the money earned in one day. That is, you must transpose, or move, Td to one side of the equation so that it stands alone. And, you must not nullify or destroy the equation. Put another way, the goal is to move and isolate Td so that it stands by itself on one side of the equa-tion without destroying the equaequa-tion. To keep from destroying the equaequa-tion, whatever you do to one side of the equation, you must also do it to the other side. If you do not, you invalidate the equation—that is, the term or expression on one side does not equal the term or expression on the other side. In this ex-ample, to transpose, or solve for Td, first divide both sides of the equation by D.
The equation thus becomes:
Tm D × Td
––– = ———.D D
The Ds on the right side of the equation cancel each other so that the equation becomes:
Tm ––– = TD d
which is the same as
Td = Tm ÷ D.
The daily rate (Td) is equal to the monthly check (Tm)divided by the number of days worked during the month (D).
A commonly used formula that may also require transposing terms is:
a c –– = — .b d
This formula sets up proportions and allows you to solve for an unknown when three factors are known. To understand this formula, substitute numbers for the variables; in this case, let a = 12, b = 6, c = 4, and d = 2.
12 ÷ 6 = 4 ÷ 2,
or 12 4
–– = — .6 2
Since both terms are equal to 2, they are equal to each other and constitute an equation. Note, too, that if the values are cross-multiplied—that is, if the number in the denominator on one side of the equation is multiplied by the numerator
Formulas 133
on the other side—the results equal each other. In this case, 12 is multiplied by 2 and 4 is multiplied by 6 and the result is 24 = 24. Cross-multiplying is one way to use the formula.
For instance, cross-multiplying transposes the terms in the equation a/b= c/d. Thus, cross-multiplying a by d and b by c results in a formula for direct proportion:
ad = bc.
The formula for direct proportion reads as, “the product of the means equals the product of the extremes.” In a direct proportion, the first and fourth terms (a and d in the original formula, a/b = c/d, are called the extremes, and the second and third terms (b and c) are called the means. The formula ad = bc can be further transposed into four variations:
a = bc/d (both sides divided by d—remember: the ds on the left side can-cel each other)
b = ad/c (both sides divided by c—the cs on the left side cancel out) c = ad/b (both sides divided by b—the bs on the left side cancel out) d = bc/a (both sides divided by a—the as on the left side cancel out).
These transpositions can be utilized to solve proportional problems in various ways.
Example Problem: A photograph 8 inches wide by 10 inches deep is to be reduced to 3 inches wide to fit a magazine column. What is the depth of the reduced photo?
Solution: Set up the proportion by saying, “8 is to 10 as 3 is to x.” Since the fourth term d is the unknown—the x—use the formula to solve for d (the unknown x becomes d) and substitute the known numbers for the letters: a = 8, b = 10, and c = 3.
d = bc/a
d = 10 × 3 ÷ 8 = 3.75 in.
Incidentally, simply cross-multiplying the appropriate values can solve the same proportion. In the case of the 8- by 10-in. photograph’s being reduced so that the 8-in. width becomes 3 in., determining its reduced depth can be solved by setting up the proportion as:
8 3 –– = — . 10 x
In this case, cross-multiplying results in 8x = 30 = 3.75 inches.
Transposing terms and expressions of an equation can involve several steps.
The important rule to remember is that the equation remains valid as long as each side is multiplied by, divided by, added to, or subtracted from equally. That is, the same mathematical operation must be done to both sides of the equation;
if not, it is no longer an equation. Multiplication and division by the same term on the same side of the equation cancels that term, and addition and subtraction of the same term on the same side of an equation cancels the term. To illustrate, solve for b in the equation
20 a = –––––— . (b + 10)
134 PRINCIPLES OF ALGEBRA
Because you are solving for b, and the goal is to single out the b without nul-lifying the equation, the first step is to multiply each side by the expression (b + 10). This step puts an expression with b in it on the left side of the equation.
This step comes out as:
20
a(b + 10) = –––––————— . (b + 10) × (b + 10)
Since multiplying the same term on the same side of the equation cancels the terms, cancellation results in:
a(b + 10) = 20.
At this point, an a is on the left side of the equation along with a b. But, you only want a b on the left side because you are solving for b. So, to eliminate the a without destroying the equation, the next step is to divide both sides by a, which results in:
a(b + 10) 20
———–– = — .a a
The as on the left cancel to result in:
(b + 10) = — . 20 a
To eliminate the 10 on the left side without destroying the equation, the last step is to subtract 10 from each side:
20 b + 10 – 10 = — – 10, a
which yields the desired result of:
20 b = — – 10. a
Remember that when solving for an expression in an equation, the goal is to single out the expression without destroying the equation. As long as you add, subtract, multiply, or divide both sides of the equation by the same terms, you do not destroy the equation. The other factor to keep in mind is the cancellation rule:
multiplication and division by the same term on the same side of the equation cancels that term, and addition and subtraction of the same term on the same side of an equation cancels that term.
Formula Applications
Formulas are used extensively in the petroleum and other industries. Many of the commonly used formulas deal with area and volume and therefore are given in chapter six, “Practical Geometry.” Other established formulas and their ap-plications in industry are presented here and in chapter eight.
Since most calculations are carried out on a calculator, be aware that a formula may have to be restated so that it gives the desired answer if you en-ter the numbers and operation signs one afen-ter the other on the calculator. In any case, it is vital to perform the operations in the proper sequence either by properly entering the values of the formula in the calculator or by rearranging the formula.
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Well Control Formulas
Well control is a subject of concern in the petroleum industry. Briefly, well control involves recognizing and then properly handling an unwanted or unexpected entry of formation fluid into the wellbore. Such an entry is a kick. If rig crewmembers fail to recognize and take appropriate steps to control a kick, they can lose control of the well. The well can blow out. A blowout is the uncontrolled flow of formation fluids into the atmosphere or into another subterranean formation the wellbore has penetrated. Blowouts are not desirable because they can injure or kill person-nel, destroy equipment, pollute the environment, and waste valuable resources.
An important formula in well control provides rig crewmembers with a way to determine the hydrostatic pressure of the drilling fluid (usually called mud) in the wellbore. The amount of hydrostatic pressure a drilling mud devel-ops depends on its weight, or density, and on its depth, or height. The more a drilling mud weighs, the more hydrostatic pressure it develops. And, the deeper, or longer, the length of the mud column in the wellbore, the more hydrostatic pressure it develops. Usually, but not always, drilling mud completely fills the wellbore. So, often, the well’s depth and the height, or depth, of the mud in the wellbore are the same.
A well’s true vertical depth is critical to accurate calculations. Nowadays, many wells are deliberately not drilled vertically—that is, they deviate from vertical even to the point of being horizontal. (The wellbore actually runs paral-lel, rather than perpendicular, to the surface.) However, to calculate hydrostatic pressure correctly, the wellbore’s true vertical depth and not its measured depth must be used. Measured depth is the depth of the well taken along the actual path it takes through the earth. Thus, a well’s true vertical depth can be less than its measured depth, especially in horizontal wells.
In any case, the mud’s weight and its true vertical height in the wellbore govern the amount of hydrostatic pressure it develops. And hydrostatic pressure plays a large role in controlling pressures in a well. The mud column’s hydrostatic pressure must be the same as or, preferably, slightly greater than the pressure in formations the wellbore penetrates. The formula for determining hydrostatic pressure is:
HP = C × MW × TVD where
HP = hydrostatic pressure, pounds per square inch (psi) or kilopascals (kPa)
C = a constant. Its value depends on how mud weight is expressed. In most of the U.S., mud weight is in pounds per gallon (ppg). On the Pacific Coast of the U.S., however, it may be in pounds per cubic foot (pcf). Other places express mud weight in kilograms per cubic metre (kg/m3). Where the mud weight is in ppg, the constant is 0.052. Where mud weight is in pcf, the constant is 0.0069. Where the mud weight is in kg/m3, the constant is 0.0098. In this case, hydrostatic pressure is in kilopascals (kPa) and true vertical depth is in metres.
MW = mud weight, ppg, pcf, or kg/m3 TVD = true vertical depth, feet or metres.
136 PRINCIPLES OF ALGEBRA
Example Problem: What is the hydrostatic pressure in a well that is full of mud that weighs 11.2 ppg and whose true vertical depth is 4,000 feet? (The constant for ppg is 0.052.)
Solution:
HP = C × MW × TVD = 0.052 × 11.2 × 4,000 = 2,329.5999
HP = 2,330 psi. (For practical field use, round pressure to the nearest whole number.) Thus, the pressure that an 11.2-ppg mud pro-duces at the bottom of a 4,000-foot well is 2,330 psi.
Example Problem: What is the hydrostatic pressure in a well that is full of mud that weighs 83.8 pcf and whose true vertical depth is 4,000 feet? (The constant for pcf is 0.0069.)
Solution:
HP = C × MW × TVD = 0.0069 × 83.8 × 4,000 = 2,312.88
HP = 2,313 psi.
Example Problem: What is the hydrostatic pressure in a well that is full of mud that weighs 1,342 kg/m3 and whose true vertical depth is 1,219 metres? (The constant for kg/m3 is 0.0098.
Solution:
HP = C × MW × TVD = 0.0098 × 1,342 × 1,219 = 16,031.8004
HP = 16,032 kPa.
When a kick occurs, crewmembers stop pumping mud and close a large valve (a blowout preventer) and other, smaller valves at the top of the well. Shut-ting the blowout preventer and the other valves completely seals the well and prevents more formation fluids from entering. After a well is completely shut in on a kick, pressure appears on a gauge at the surface. This surface gauge is a drill pipe pressure gauge; it measures the amount of pressure the intruded fluids place on the drill string. (If the mud’s hydrostatic pressure equaled or exceeded
When a kick occurs, crewmembers stop pumping mud and close a large valve (a blowout preventer) and other, smaller valves at the top of the well. Shut-ting the blowout preventer and the other valves completely seals the well and prevents more formation fluids from entering. After a well is completely shut in on a kick, pressure appears on a gauge at the surface. This surface gauge is a drill pipe pressure gauge; it measures the amount of pressure the intruded fluids place on the drill string. (If the mud’s hydrostatic pressure equaled or exceeded