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Equations

Module by: Wade Ellis, Denny Burzynski. E-mail the authors

Summary: This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. Operations with algebraic expressions and numerical evaluations are introduced in this chapter. Coefficients are described rather than merely defined. Special binomial products have both literal and symbolic explanations and since they occur so frequently in mathematics, we have been careful to help the student remember them. In each example problem, the student is "talked" through the symbolic form. Objectives of this module: understand the meaning of an equation, be able to perform numerical evaluations.

Overview

  • Equations
  • Numerical Evaluation

Equations

Equation

An equation is a statement that two algebraic expressions are equal.

An equation is composed of three parts.

     =           =     

Each of the boxes represents an algebraic expression. An equation consists of two expressions separated by an equal sign. The equal sign makes the statement that the two expressions are equivalent, that is, they represent the same value. For example:

Example 1

f=32a f=32a .
The equation expresses the relationship between the variables f f and a a . It states that the value of f f is always 32 times that of a a .

Example 2

y=6x+8 y=6x+8 .
The equation expresses the relationship between the variables x x and y y . It states that the value of y y is always 8 more than 6 times the value of x x .

Numerical Evaluation

Numerical evaluation

Numerical evaluation is the process of determining a value by substituting numbers for letters.

Formulas

In various areas (business, statistics, physics, chemistry, astronomy, sociology, psychology, etc.), particular equations occur quite frequently. Such equations are called formulas. Numerical evaluation is used frequently with formulas.

Sample Set A

Example 3

f = 32a. Determinethevalueoffifa=2. f = 32(2) Replaceaby2. = 64 f = 32a. Determinethevalueoffifa=2. f = 32(2) Replaceaby2. = 64

Example 4

p= 10,000 v p= 10,000 v .

This chemistry equation expresses the relationship between the pressure p p of a gas and the volume v v of the gas. Determine the value of p p if v=500 v=500 .

p = 10,000 500 Replacevby500. = 20 p = 10,000 500 Replacevby500. = 20

On the Calculator
Type 10000 Press ÷ Type 500 Press = Displayreads: 20 Type 10000 Press ÷ Type 500 Press = Displayreads: 20

Example 5

z= xu s z= xu s .

This statistics equation expresses the relationship between the variables z,x,uands z,x,uands . Determine the value of zifx=41,u=45,ands=1.3 zifx=41,u=45,ands=1.3 . Round to two decimal places.

z = 4145 1.3 = 4 1.3 = 3.08 z = 4145 1.3 = 4 1.3 = 3.08

On the Calculator
Type 41 Press Type 45 Press = Press ÷ Type 1.3 Press = Displayreads: 3.076923 We'llroundto3.08 Type 41 Press Type 45 Press = Press ÷ Type 1.3 Press = Displayreads: 3.076923 We'llroundto3.08

Example 6

p=5 w 3 + w 2 w1 p=5 w 3 + w 2 w1 .

This equation expresses the relationship between p p and w w . Determine the value of p p if w=5 w=5 .

p = 5 (5) 3 + (5) 2 (5)1 = 5(125)+25(5)1 = 625+2551 = 644 p = 5 (5) 3 + (5) 2 (5)1 = 5(125)+25(5)1 = 625+2551 = 644

On the Calculator
Type 5 Press y x Type 3 Press = Press × Type 5 Press = Press + Type 5 Press x 2 Press Type 5 Press Type 1 Press = Displayreads: 644 Type 5 Press y x Type 3 Press = Press × Type 5 Press = Press + Type 5 Press x 2 Press Type 5 Press Type 1 Press = Displayreads: 644

Practice Set A

Exercise 1

f=32a. Determinethevalueoffifa=6. f=32a. Determinethevalueoffifa=6.

Solution

192

Exercise 2

p= 10,000 v . Determinethevalueofpifv=250. p= 10,000 v . Determinethevalueofpifv=250.

Solution

40

Exercise 3

F= 9 5 C+32. DeterminethevalueofFifC=10. F= 9 5 C+32. DeterminethevalueofFifC=10.

Solution

50

Exercise 4

y=9x14. Determinethevalueofyifx=3. y=9x14. Determinethevalueofyifx=3.

Solution

13

Exercise 5

m=5 p 3 2p+7. Determinethevalueofmifp=2. m=5 p 3 2p+7. Determinethevalueofmifp=2.

Solution

29 29

Exercises

For the following problems, observe the equations and state the relationship being expressed.

Exercise 6

x=6y x=6y

Solution

Thevalueofxis equal to six times the value of y. Thevalueofxis equal to six times the value of y.

Exercise 7

y=x+4 y=x+4

Exercise 8

e=g9 e=g9

Solution

e is equal to 9 less then the value of g. e is equal to 9 less then the value of g.

Exercise 9

y=x7 y=x7

Exercise 10

3t=6s 3t=6s

Solution

Thevalueof three times t is equal to six times s. Thevalueof three times t is equal to six times s.

Exercise 11

u= v 5 u= v 5

Exercise 12

r= 2 9 s r= 2 9 s

Solution

Thevalueof ris equal to two ninth times the value of s. Thevalueof ris equal to two ninth times the value of s.

Exercise 13

b= 3 4 a b= 3 4 a

Exercise 14

f=0.97k+55 f=0.97k+55

Solution

Thevalueof f is equal to 55 more then  97 100 times the value of k. Thevalueof f is equal to 55 more then  97 100 times the value of k.

Exercise 15

w=4 z 3 21 w=4 z 3 21

Exercise 16

q 2 =9 x 8 +2y q 2 =9 x 8 +2y

Solution

Thevalueof  q 2 is equal to nine times the value of  x 8 plus two times the value of y. Thevalueof  q 2 is equal to nine times the value of  x 8 plus two times the value of y.

Exercise 17

I= m 2 q b 5 +3.115p I= m 2 q b 5 +3.115p

Use numerical evaluation on the equations for the following problems.

Exercise 18

Geometry (circumference of a circle)
C=2πr. FindCifπisapproximatedby3.14andr=5. C=2πr. FindCifπisapproximatedby3.14andr=5.

Solution

31.4 31.4

Exercise 19

Geometry (area of a rectangle)
A=lw. FindAifl=15andw=9. A=lw. FindAifl=15andw=9.

Exercise 20

Electricity (current in a circuit)
I= E R . FindIifE=21andR=7. I= E R . FindIifE=21andR=7.

Solution

3

Exercise 21

Electricity (current in a circuit)
I= E R . FindIifE=106andR=8. I= E R . FindIifE=106andR=8.

Exercise 22

Business (simple interest)
I=prt. FindIifp=3000,r=.12andt=1. I=prt. FindIifp=3000,r=.12andt=1.

Solution

360

Exercise 23

Business (simple interest)
I=prt. FindIifp=250,r=0.07andt=6. I=prt. FindIifp=250,r=0.07andt=6.

Exercise 24

Geometry (area of a parallelogram)
A= bh. FindAifb=16andh=6. A= bh. FindAifb=16andh=6.

Solution

96

Exercise 25

Geometry (area of a triangle)
A= 1 2 bh. FindAifb=25andh=10. A= 1 2 bh. FindAifb=25andh=10.

Exercise 26

Geometry (perimeter of a rectangle)
P=2l+2w. FindPifl=3andw=1. P=2l+2w. FindPifl=3andw=1.

Solution

8

Exercise 27

Geometry (perimeter of a rectangle)
P=2l+2w. FindPifl=74andw=16. P=2l+2w. FindPifl=74andw=16.

Exercise 28

Geometry (perimeter of a rectangle)
P=2l+2w. FindPifl=8 1 4 andw=12 8 9 . P=2l+2w. FindPifl=8 1 4 andw=12 8 9 .

Solution

42 5 18 42 5 18

Exercise 29

Physics (force)
F=32m. FindFifm=6. F=32m. FindFifm=6.

Exercise 30

Physics (force)
F=32m. FindFifm=14. F=32m. FindFifm=14.

Solution

448

Exercise 31

Physics (force)
F=32m. FindFifm= 1 16 . F=32m. FindFifm= 1 16 .

Exercise 32

Physics (force)
F=32m. FindFifm=6.42. F=32m. FindFifm=6.42.

Solution

205.44 205.44

Exercise 33

Physics (momentum)
p=mv. Findpifm=18andv=5. p=mv. Findpifm=18andv=5.

Exercise 34

Physics (momentum)
p=mv. Findpifm=44andv=9. p=mv. Findpifm=44andv=9.

Solution

396

Exercise 35

Physics (momentum)
p=mv. Findpifm=9.18andv=16.5. p=mv. Findpifm=9.18andv=16.5.

Exercise 36

Physics (energy)
E= 1 2 m v 2 . FindEifm=12andv=5. E= 1 2 m v 2 . FindEifm=12andv=5.

Solution

150

Exercise 37

Physics (energy)
E= 1 2 m v 2 . FindEifm=8andv=15. E= 1 2 m v 2 . FindEifm=8andv=15.

Exercise 38

Physics (energy)
E= 1 2 m v 2 . FindEifm=24.02andv=7. E= 1 2 m v 2 . FindEifm=24.02andv=7.

Solution

588.49 588.49

Exercise 39

Astronomy (Kepler’s law of planetary motion)
P 2 =k a 3 . Find P 2 ifk=1anda=4. P 2 =k a 3 . Find P 2 ifk=1anda=4.

Exercise 40

Use a calculator. Astronomy (Kepler’s law of planetary motion)
P 2 =k a 3 . Find P 2 ifk=8anda=31. P 2 =k a 3 . Find P 2 ifk=8anda=31.

Solution

238,328 238,328

Exercise 41

Use a calculator. Astronomy (Kepler’s law of planetary motion)
P 2 =k a 3 . Find P 2 ifk=4anda=5.1. P 2 =k a 3 . Find P 2 ifk=4anda=5.1.
(Hint: On the calculator, Type 5.1, Press y x y x , Type 3, Press = = , Press × × , Type 4, Press = = .)

Exercise 42

Use a calculator. Astronomy (Kepler’s law of planetary motion)
P 2 =k a 3 . Find P 2 ifk=53.7anda=0.7. P 2 =k a 3 . Find P 2 ifk=53.7anda=0.7.

Solution

18.4191 18.4191

Exercise 43

Business (profit, revenue, and cost)
P=RC. FindPifR=3100andC=2500. P=RC. FindPifR=3100andC=2500.

Exercise 44

Business (profit, revenue, and cost)
P=RC. FindPifR=4240andC=3590. P=RC. FindPifR=4240andC=3590.

Solution

650

Exercise 45

Geometry (area of a circle)
A=π r 2 . FindAifπisapproximately3.14andr=3. A=π r 2 . FindAifπisapproximately3.14andr=3.

Exercise 46

Geometry (area of a circle)
A=π r 2 . FindAifπisapproximately3.14andr=11. A=π r 2 . FindAifπisapproximately3.14andr=11.

Solution

379.94 379.94

Exercise 47

t=21x+6. Findtifx=3. t=21x+6. Findtifx=3.

Exercise 48

t=21x+6. Findtifx=97. t=21x+6. Findtifx=97.

Solution

2,043 2,043

Exercise 49

Use a calculator. E=m c 2 . FindEifm=2andc=186,000. E=m c 2 . FindEifm=2andc=186,000.
(Hint: The number 10 that occurs on the display a few spaces away from the other number on the display is the exponent of 10 in the scientific notation form of the number.)

Exercise 50

Use a calculator. E=m c 2 . FindEifm=5andc=186,000. E=m c 2 . FindEifm=5andc=186,000.

Solution

1.7298× 10 11 1.7298× 10 11

Exercise 51

An object travels on a horizontal line. The distance it travels is represented by d d and is measured in meters. The equation relating time of travel, t t , and distance of travel, d d , is
d= t 2 4t+20 d= t 2 4t+20
Determine the distance traveled by the object if it has been in motion for 6 seconds.

Exercise 52

In medicine, there are several rules of thumb used by physicians to determine a child’s dose, D c D c , of a particular drug. One such rule, Young’s Rule, relates a child’s dose of a drug to an adult’s dose of that drug, D a D a . Young’s Rule is
D c = t t+12 · D a D c = t t+12 · D a
where t t is the child’s age in years. What dose should be given to a child 8 years old if the corresponding adult dosage is 15 units?

Solution

6 units

Exercise 53

A hemispherical water tank of radius 6 feet has water dripping into it. The equation relating the volume, V V , of water in the tank at any time is V=6π h 2 π 3 h 3 V=6π h 2 π 3 h 3 ,where h h represents the depth of the water. Using 3.14 3.14 to approximate the irrational number π π , determine the volume of water in the tank when the depth of the water is 3 feet.
A water tank in the shape of a hemisphere with a radius of six feet. The depth of the water in the tank is labeled as h.

Exercise 54

The equation W=3.51L192 W=3.51L192 has been established by the International Whaling Commission to relate the weight, W W (in long tons), of a mature blue whale to its length, L L (in feet). The equation is only used when L70 L70 . When
0<L<70 0<L<70
blue whales are considered immature. At birth, a blue whale is approximately 24 feet long. Determine the weight of a blue whale that measures 83 feet in length.

Solution

99.33 tons 99.33 tons

Exercise 55

A relationship exists between the length of a cantilever beam and the amount it is deflected when a weight is attached to its end. If a cantilever beam 20 feet long has a 600 pound weight attached to its end, the equation relating beam length and amount of deflection is
d= 60 x 2 x 3 16,000 d= 60 x 2 x 3 16,000
where d d is the amount of deflection measured in inches and x x is the length from the supported part of the beam to some point on the beam at which the amount of deflection is measured. Find the amount of deflection of the beam 17 feet from the supported end.

Deflection of a twenty feet long cantilever beam. A weight of six hundred pound is attached to its end. The amount of deflection of the beam is labeled as d. The length between the supported part of the beam, and a point on the beam at which the amount of deflection is being measured, is labeled as x.

Exercise 56

There is a relationship between the length of a suspension bridge cable that is secured between two vertical supports and the amount of sag of the cable. If we represent the length of the cable by c c , the horizontal distance between the vertical supports by d d , and the amount of sag by s s , the equation is c=d+ 8 s 2 3d 32 s 4 5 d 3 c=d+ 8 s 2 3d 32 s 4 5 d 3 . If the horizontal distance between the two vertical supports is 190 feet and the amount of sag in a cable that is suspended between the two supports is 20 feet, what is the length of the cable?

A suspension bridge with its suspesion cables secured between two vertical supports. The horizontal distance between the vertical supports is labeled as d. The amount of sag of the cable is labeled as s.

Solution

195.46474 feet 195.46474 feet

Exercises for Review

Exercise 57

((Reference)) simplify (4 x 3 y 8 )(3 x 2 y) (4 x 3 y 8 )(3 x 2 y) .

Exercise 58

((Reference)) simplify |8| |8| .

Solution

8 8

Exercise 59

((Reference)) Find the value of 4 2 · 8 2 3 2 4 2 · 8 2 3 2 .

Exercise 60

((Reference)) For the expression 5(a+b)+2 x 2 5(a+b)+2 x 2 , write the number of terms that appear and then write the terms themselves.

Solution

2;5( a+b ),2 x 2 2;5( a+b ),2 x 2

Exercise 61

((Reference)) How many x y 3 's x y 3 's are there in 5 x 2 y 5 5 x 2 y 5 ?

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