1. Evaluating Algebraic Expressions
Using Technical Applications
Evaluating Expressions

2. Technical Application
Scientists, engineers and technicians need, develop, and use mathematics to explain, describe, and predict what nature, processes, and equipment do.
Many times the math they use is the math that is taught in ALGEBRA 1!

3. There are 3 Objectives of this presentation:1)
To evaluate algebraic expressions involving multiplication and division of real numbers. 2)
To simplify algebraic expressions by using the rules for order operations to evaluate algebraic expressions. 3)
To use algebra 1 to help understand a technical application.

4. 1 3 -9 -27 3 = = -27 3 (-27) ( ) = d = 1 2 ) ( b d 1 2 ) ( b d 1 2 ) (
Evaluating Expressions
The rules for dividing real numbers involves the mathematical concept of RECIPROCALS.
The division by a number is defined as multiplying by the reciprocal or multiplicative inverse of the number.
A common symbol technicians, scientists and engineers use for multiplication.
TWO EXAMPLES
The fraction one third is the reciprocal of 3
(a) Specific Situation 1 3 -9
-27 3 = =
-27 3
(-27) ( ) =
Divide –27 by 3 is the same as multiply –27 by the reciprocal of 3.
the fraction 1 ) ( b
is the reciprocal of b
(b) General Situation d = 1 2 ) ( b d 1 2 ) ( b d 1 2 ) ( b 2 b 1 ) ( d 2 b 1 ) ( d d 2 ) ( b = ( ( =

5. 4 EASY PRACTICE PROBLEMS
Evaluating Expressions
The division by a number is defined as multiplying by the reciprocal or multiplicative inverse of the number.
4 EASY PRACTICE PROBLEMS
(a) Divide the number 1 by the number 4
The fraction one fourth 4 = 1 ? 1 1 = 4
0.25 1
is the reciprocal of the number 4 1 4
Dividing 1 by 4 is the same as multiplying 1 by the fraction
Dividing 1 by 4 is the same as multiplying 1 by the reciprocal of the number 4 1 4
is the reciprocal of the number 4

6. 4 EASY PRACTICE PROBLEMS (continued)
Evaluating Expressions
The division by a number is defined as multiplying by the reciprocal or multiplicative inverse of the number.
4 EASY PRACTICE PROBLEMS (continued)
(b) 9 = 1 ?
0.11 1 = 9

7. 4 EASY PRACTICE PROBLEMS (continued)
Evaluating Expressions
The division by a number is defined as multiplying by the reciprocal or multiplicative inverse of the number.
4 EASY PRACTICE PROBLEMS (continued)
(c)
If the values of d and b are 5 and 36 respectively, what is the value of the following algebraic expression? d = 1 10 ) ( b ( d 1 10 ) b = = 5 1 10 ) ( 36 ? 1 10 ) ( 5 36 = 1 b
The variable b
and the variable
are reciprocals of each other. =
(0.10)
(0.14) =
0.014

8. l l l ) ( 4 EASY PRACTICE PROBLEMS (continued) 10 b d 1 = = ? (d)
Evaluating Expressions
The division by a number is defined as multiplying by the reciprocal or multiplicative inverse of the number.
4 EASY PRACTICE PROBLEMS (continued)
(d)
Sometimes it is fun in Algebra to use a letter from the Greek alphabet as well as letters like “d” and “b”. Try the following problem using the Greek letter LAMBDA l
the Greek letter Lambda 10 b ) ( d 1 l = = ? l
when b equals 36 and d equals 5.

9. l l l l l l ) ( ) ( ) ( ) ( 10 b 1 = 10 b d 1 = ? 10 b d 1 10 b 1 = (
Evaluating Expressions
The rules for dividing real numbers involve the mathematical concept of reciprocals. The division by a number is defined as multiplying by the reciprocal or multiplicative inverse of the number.
What is the reciprocal of ? 10 b ) ( 1
4 Easy Practice Problems (continued) l
the Greek letter Lambda = l 10 b ) ( d 1
(d) l =
when b equals 36 and d equals 5. ? 10 b ) ( d 1 10 ) ( b 1 = ( d ) ? = l 5 10 ( ) 36 = 5
360 ( ) l = l
1800

10. Evaluating Expressions
Technicians, scientists and engineers use Algebra all the time. When using algebra, they also like to use combinations of letters and numbers as algebra symbols.
The previous example problem used the Greek letter lambda as well as the letters “d” and “b”. = l 10 b ) ( d 1 = l 10 b ) ( d 1
A technician might use this algebraic expression with lambda and the letters “d” and “b” replaced by symbols that are combinations of letters and numbers. 10 b ) ( 1 l 1 = d 1

11. l l l l ) ( ( ) ( ) ( ) 10 d 1 1 1 = d 1 10 b 1 1 ( 10 ) 1 b d = = 5
Evaluating Expressions
What is the reciprocal of ? 10 d ) ( 1
Technicians, scientists and engineers use Algebra all the time. However, they also like to use combinations of letters and numbers as algebra symbols. 1 l 1 ( ) = d 1 10 b 1
when b1 equals 36 and d1 equals 5. l 1 ( 10 ) 1 b d = = 5 10 ( ) 36 =
1800 ?
PRACTICE PROBLEM
when b1 equals 36 and d1 equals 5. d1 10 1 ( ) b1 l 2 = d1 10 ) b1 ( = ? 5 10 1 ( ) 36 50 =
0.72 l 2 = = 36

12. Scientists, technicians and engineers also use algebraic symbols that are combinations of letters and numbers because they often work with the same algebraic expression but substitute different numbers.
substitute different numbers.
TWO QUICK EXAMPLE PROBLEMS
These examples use the following algebraic expression; b 1 = 10 ) ( d l
(1)
Let b1 equal 36 and d1 equal 5. = ? ( 10 ) 1 b d 5 36 l
1800
(2)
This time, let b1 equal 35 and d1 equal 5. l 2 = ( 10 ) 1 b d = 5 10 ( ) 35 =
1750 ?

13. l l l l l ( ) ( ) ( ) ( ) ) ( ) ( ) ( ) ( ) ( 1 = 50 35 = 10 7 = 1 10
Sometimes an engineer, scientist or technician may select symbols that are similar when the algebraic expressions are different.
This often happens when there is a connection between the answers after the expressions have been evaluated.
2 EASY EXAMPLE PROBLEMS
In both problems b1 equals 35 and d1 equals 5.
(1) 1 = ( 50 ) 35 = ( 10 ) 7 = l 1 10 ) ( d b = l 1 10 ) ( d b
= ?
1.43 1 = ( 35 ) 50 = ( 7 ) 10
(2) = l 2 10 ) ( d 1 b = l 2 10 ) ( d 1 b = l 2 10 ) ( d 1 b
= ?
0.7

14. Evaluating Expressions
= ?
(1)
(2)
0.7 = l 1 10 ) ( d b
1.43 7 50 35 2
What is the connection between and ? l 1 2 l 1 2
is the reciprocal of l 2 1
is the reciprocal of ?
Or, if you wish l 1 2 ( )
= (1.43) (0.7) = 1.00 1

15. THREE QUICK REVIEW QUESTIONS to see what we rememberl 1 l 2 If
= 1.43
and
= 0.7 l 2 1
What is the connection between
and ? 1)
One is the reciprocal of the other. 2)
What is the answer if you multiply reciprocals together?
You always get the number 1 as the answer.
Try this with a calculator. Is there a problem?
What is the reciprocal of ? 10 d ) ( 1 3) d 1 10 ) (

16. What do you think? 1)
Is one half the reciprocal of 2? Why/why not? 2)
Do all fractions have reciprocals? Why/Why not? 3)
Two of the most popular manufacturers of calculators (TI and HP) have a different style (way to do calculation) for getting answers to multiplication and division problems. One of them was developed with a knowledge (use) of reciprocals in mind. Which one is it? Why? 4)
Use the Web (if you have to) and examine the arrangement of the scales on a slide rule. One of those scales is know as the reciprocal scale. Which one is it and why is it named so?

17. 2)
Simplifying Expressions
Evaluating Expressions
Objective:
To simplify algebraic expressions by using the rules for order operations to evaluate algebraic expressions

18. ) ( ) ( ) ( ) ( TWO EXAMPLES 14- 10 3 +10 = 14 -30+10 = -16 +10 = -4 =
Simplifying Expressions
Rules
Perform operations within parenthesis first.
Multiply (divide) in order from left to right.
Add (subtract) in order from left to right.
TWO EXAMPLES
(a)
14- 10 3
+10 = =
= -4
(b) = ) ( - 9 1 4 10 ) ( -
0.11
0.25 10 1 ) (
0.14
0.1 ) (
0.14 10 1 =
0.014 = =

19. 14 - +10 = (10 3) = (10 3) 14- +10 = = 14 -30+10 = = -16 +10 = -4
Simplifying Expressions
Rules that help the Simplification of Expressions
Perform operations within parenthesis first.
Perform operations within parenthesis first.
Multiply (divide) in order from left to right.
Add (subtract) in order from left to right.
Add (subtract) in order from left to right.
TWO SIMPLE EXAMPLES
(a)
14 -
+10 =
(10 3) ? =
(10 3)
14-
+10 = = = =
= -4

20. Simplifying Expressions= ) ( - 9 1 4 10
0.11
0.25
0.14
0.1
0.014
(b)
0.014
Rules used?
Perform operations within parenthesis first.
Multiply (divide) in order from left to right.
Multiply (divide) in order from left to right.
Another way that technicians, scientists and engineers often simplify this type of algebraic expression.
(9-4)
(4 9) ) ( - 9 1 4 10 10 1 ( ) ) ( 5 36 10 1 = = =
Rule to use first?
0.014 ) (
0.14 10 1 =
0.014
Perform operations within parenthesis. =

21. ) ( ( ) ) ( = - d 1 b 10 10 1 (d – b) (b d) - 9 1 4 10
Simplifying Expressions
Rules that help the Simplification of Expressions
Perform operations within parenthesis first.
Multiply (divide) in order from left to right.
Add (subtract) in order from left to right.
TWO GENERALIZED EXAMPLES = ) ( - d 1 b 10 10 1 (
(d – b)
(b d) )
(a)
For the previous problem,
b was equal to 4 and
d was equal to 9 ) ( - 9 1 4 10

22. ) ( ) ( = - n2 1 n1 10 (n2– n1) (n1 n2) = - d 1 b 10 (d – b) (b d)
Simplifying Expressions = ) ( - d 1 b 10
(d – b)
(b d)
Technical workers often use different symbol combinations for the letters b and d. = ) ( - n2 1 n1 10
(n2– n1)
(n1 n2)
(b)
This time, the symbol n1 replaces the letter b, and the symbol n2 replaces the letter d.

23. Simplifying Expressions
Simplification of a new expression: ) ( - n2 1 n1 10 2
= ?
This time, let
n1 equal 2 and n2 equal 3 = ) ( - 3 1 2 10
(9 -4)
(4 9) ) ( 5 36 10 1 = ) (
0.14 10 1 =
= ?
0.014
Note:
The calculations inside the parenthesis were completed before multiplying by one tenth.

24. [ ] [ ] [ ] [ ] [ ] [ ] Reciprocal Expressions 10 10 1 = = 0.10
Simplifying Expressions
Reciprocal Expressions
Rules
Perform operations within parenthesis first.
Perform operations within parenthesis first.
Multiply (divide) in order from left to right.
Add (subtract) in order from left to right.
THREE EASY EXAMPLE PROBLEMS a) -1 10 [ ] 10 [ ] 1 = =
0.10
There is nothing to do inside this parenthesis. b)
2 + 3(2) +2 [ ] 1 = [ ] 1 = 10 [ ] 1 = -1 10 [ ] =
0.10
There is something to do inside this parenthesis.

25. [ ] [ ] [ ] [ ] [ ] [ ] ) ( ) ( ) ( ) ( Reciprocal Expressions 10 1 10
Simplifying Expressions
Reciprocal Expressions
Rules
Perform operations within parenthesis first. c) 10 [ ] 1 -1 10 [ ] ) ( - 4 1 3 20 [ ] = [ ] 1 ) ( 4 2 20 =
This version is popular in technical applications because it takes up less space on a piece of paper and is easier to type on a computer. =
A typical reciprocal (inverse) expression used in technology. ) ( - n2 1 n1 2 10 [ ] -1 ) ( - n2 1 n1 2 10 [ ] =
These two expressions are same.

26. [ ] ) ( Reciprocal Expressions - n2 1 n1 10 = PRACTICE PROBLEM
Simplifying Expressions
Reciprocal Expressions
Rules
Perform operations within parenthesis first.
PRACTICE PROBLEM ) ( - n2 1 n1 2 10 [ ] = -1
What is the value of this expression when n1 equals 2 and n2 equals 3?

27. [ ] [ ] [ ] [ ] [ ] [ ] ) ( ) ( ) ( ) ( ( ) ( ) 0.014 = - n2 1 n1 10 =
Simplifying Expressions
Note:
Perform operations within parenthesis first
0.014 [ ] -1 1
0.014 ( ) = 1) ) ( - n2 1 n1 2 10 [ ] -1
Note: 2 1
0.014 ( )
= 2 times 2 = 4
is the number 71.4 2) 3 2
n1 equals 2 and n2 equals 3
= 3 times 3 = 9 = ) ( - 3 1 2 10 [ ] -1
(9 -4)
(4 9) = ) ( 5 36 10 1 [ ] -1 ) (
0.14 10 1 [ ] -1 =
0.014 [ ] -1
The calculation of the inverse is the last thing done =
71.4 =

28. ) ( ) ( ( ) ( ) - n2 1 n1 (n2– n1) (n1 n2)
Simplifying Expressions
THREE QUICK REVIEW QUESTIONS. 1)
What are, in the correct order of use, the rules for simplifying
algebraic expressions?
Perform operations within parenthesis first.
Multiply (divide) in order from left to right.
Add (subtract) in order from left to right. 2)
What is another way to write the following algebraic expression? ) ( - n2 1 n1 ) (
(n2– n1)
(n1 n2) = 3)
What is
(a) the inverse of ? 1
0.014 ( )
71.4 1
0.014 ( )
(b)
the reciprocal of the the number 71.4?

29. [ ] [ ] ) ( ) ( - n2 1 n1 10 - n2 1 n1 10 What do you think? 1)
Simplifying Expressions
What do you think? 1)
Are the two algebraic expressions show below equal? Why/why not?
(a)
(b) ) ( - n2 1 n1 2 10 [ ] -2 ) ( - n2 1 n1 2 10 [ ] 2)
Is the inverse of a number always the same as the reciprocal of that number? Why/Why not?

30. Simplifying Expressions3)
Technology Application
Objective:
To use algebraic expressions to describe and understand a technical application.

31. Using Algebra in a Technical Application
Consider the feasibility of using a hydrogen laser beam for application as a welding tool outside the space station.
A laser beam is made when the an electron from many different atoms moves back from the same outer orbit to the same orbit closer to the atom’s nucleus.
SIMPLIFIED LASER
EXCITATION SOURCE
Laser welding tool
Atoms emit light as photons after excitation
Some travel parallel to the laser tube and bounce off the mirrored ends
LASER TUBE
Monochromatic light (LASER LIGHT) leaves the partially mirrored end

32. Using Algebra in a Technical Application
Consider the feasibility of using a hydrogen laser beam for application as a welding tool outside the space station.
A laser beam is made when the an electron from many different atoms moves back from the same outer orbit to the same orbit closer to the atom’s nucleus.
2st orbit out from nucleus (n2)
3nd orbit out from nucleus (n3)
A photon of light is emitted
Atom’s
nucleus
If the same electrons from
enough of the same type
of atoms go through this
process, we may see a collection
of light waves as a beam of
a specific colored light.
If, for example, electrons move from 3rd orbit to second orbit, a single color light will be emitted.

33. In 1875 Balmer determined a mathematical relationship that is used today to predict the wavelength of laser light generated by atomic gases.
Technology Example:
4.5l l ) ( - n3 1 n2 2
91.0 [ ] -1 =
Courtesy of NASA
Balmer’s wavelength value in nanometers
This wave is 4½ wavelengths long.
Shortest distance to same point on the wave is one wavelength distance
A technician, engineer, or scientist will know what color the light is if the length of the wave is known. The length of one cycle of a wave is its wavelength.

34. [ ] [ ] [ ] [ ] [ ] [ ] ) ( ) ( ) ( ) ( ) ( l l l l - 3 1 2 91 - n3 1
If the light beam for a hydrogen gas laser is to be generated when one electron from many individual hydrogen atoms in the gas move from the 3rd orbit to the 2nd obit, what color is the light that is observed.
Technology Application Problem Statement ) ( - 3 1 2 91 [ ] -1 ) ( - n3 1 n2 2 91 [ ] -1 l = = ) ( 91 1
(9 -4)
(4 9) [ ] -1
Symbol people use to represent the wavelength value in nanometers l
= ? = ) ( 5 36 91 1 [ ] -1 ) (
0.14 91 1 [ ] -1 l =
0.0015 [ ] -1 l =
670 nanometers =

35. l [ ] ) ( = 5 36 91 1 -1 0.14 - n2 n1 0.0015 670 nanometers
If the light beam for a hydrogen gas laser is to be generated when an electron moves from the 3rd orbit to the 2nd obit of many individual hydrogen atoms in the gas, what color is the light that is observed?
Technology Application Problem Statement = ) ( 5 36 91 1 [ ] -1
0.14 l - n2 n1 2
wavelength value
0.0015
670 nanometers
The hydrogen laser will emit light of 670 nanometers
What color do you thing the light wave is?

36. The hydrogen laser will emit light of 670 nanometers
Technology Application Problem Statement
If the light beam for a hydrogen gas laser is to be generated when an electron moves from the 3rd orbit to the 2nd obit of many individual hydrogen atoms in the gas, what color is the light that is observed?
The hydrogen laser will emit light of 670 nanometers
What color do you thing the light wave is?
657 nm
486 nm
434 nm
410 nm
The 670 nm light emitted from the hydrogen laser will be slightly to the right of the color at 657 nm

37. l l l 1 2 = .0015 = 670 2 1 Technology Example:
THREE QUICK REVIEW QUESTIONS l 1 l 2 If
= .0015
and
= 670 l 2 1
What is the relationship between
and ? 1)
They are complements. 2)
What is the answer if you multiply complements together?
You always get the number 1 as the answer.
Try this with a calculator. What’s the problem? 3)
What was the last name of the person who determined the mathematical model that is used to predict the color of light caused by electrons moving from one orbit to another?
His last name was Balmer and he worked out his formula in 1875.

38. Technology Example:
What do you think?
Do you think Balmer figured out his mathematical relationship so that he could understand the Bohr atomic model? Why/Why not? 1) 2)
Can you make a laser beam by focusing sun light through a prism? Why/ Why not? 3)
Does the Balmer model predict that a laser beam will occur if electrons go from the 2nd orbital to the first orbital of an atom? Why/why not.