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# Mathematics and Logic - Skill and Concept Development

with lessons and lesson ideas at many levels. If one site element is not to your liking, try another. Each one is different.

30 pages en Francais || Parents - Help Your Child or Teen Learn
Online Volumes: 1 Elements of Reason || 2 Three Skills For Algebra || 3 Why Slopes Light Calculus Preview or Intro plus Hard Calculus Proofs, decimal-based.
More Lessons &Lesson Ideas: Arithmetic & No. Theory || Time & Date Matters || Algebra Starter Lessons || Geometry - maps, plans, diagrams, complex numbers, trig., & vectors || More Algebra || More Calculus || DC Electric Circuits || 1995-2011 Site Title: Appetizers and Lessons for Mathematics and Reason

Mathematics Concept & Skill Development Lecture Series: Webvideo consolidation of site lessons and lesson ideas in preparation. Price to be determined.

Bright Students: Top universities want you. While many have high fees: many will lower them, many will provide funds, many have more scholarships than students. Postage is cheap. Apply and ask how much help is available. Caution: some programs are rewarding. Others lead nowhere. After acceptance, it may be easy or not to switch.

Are you a careful reader, writer and thinker? Five logic chapters lead to greater precision and comprehension in reading and writing at home, in school, at work and in mathematics.
- 1 versus 2-way implication rules - A different starting point - Writing or introducting the 1-way implication rule IF B THEN A as A IF B may emphasize the difference between it or the latter, and the 2-way implication A IF and ONLY IF B.
- Deductive Chains of Reason - See which implications can and cannot be used together to arrive at more implications or conclusions,
- Mathematical Induction - a light romantic view that becomes serious.
- Responsibility Arguments - his, hers or no one's
- Islands and Divisions of Knowledge - a model for many arts and disciplines including mathematics course design: Different entry points may make learning and teaching easier. Are you ready for them?

#### Early High School Arithmetic

Deciml Place Value - funny ways to read multidigit decimals forwards and backwards in groups of 3 or 6.
- Decimals for Tutors - lean how to explain or justify operations. Long division of polynomials is easier for student who master long division with decimals.
- Primes Factors - Efficient fraction skills and later studies of polynomials depend on this.
- Fractions + Ratios - See how raising terms to obtain equivalent fractions leads to methods for addition, comparison, subtraction, multiplication and division of fractions.
- Arithmetic with units - Skills of value in daily life and in the further study of rates, proportionality constants and computations in science & technology.

#### Early High School Algebra

What is a Variable? - this entertaining oral & geometric view may be before and besides more formal definitions - is the view mathematically correct?
- Formula Evaluation - Seeing and showing how to do and record steps or intermediate results of multistep methods allows the steps or results to be seen and checked as done or later; and will improve both marks and skill. The format here allows the domino effects of care and the domino effects of mistakes to be seen. It also emphasizes a proper use of the equal sign.
- Solve Linear Eqns with & then without fractional operations on line segments - meet an visual introduction and learn how to present do and record steps in a way that demonstrate skill; learn how to check answers, set the stage for solving word problems by by learning how to solve systems of equations in essentially one unknown, set the stage for solving triangular and general systems of equations algebraically.
- Function notation for Computation Rules - another way of looking at formulas. Does a computation rule, and any rule equivalent to it, define a function?
- Axioms [some] as equivalent Computation Rule view - another way for understanding and explaining axioms.
- Using Formulas Backwards - Most rules, formulas and relations may be used forwards and backwards. Talking about it should lead everyone to expect a backward use alone or plural, after mastery of forward use. Proportionality relations may be use backward first to find a proportionality constant before being used forwards and backwards to solve a problem.

#### Early High School Geometry

Maps + Plans Use - Measurement use maps, plans and diagrams drawn to scale.
-
- Coordinates - Use them not only for locating points but also for rotating and translating in the plane.
- What is Similarity - another view of using maps, plans and diagrams drawn to scale in the plane and space. Many human-made objects are similar by design.
- 7 Complex Numbers Appetizer. What is or where is the square root of -1. With rectangular and polar coordinates, see how to add, multiply and reflect points or arrows in the plane. The visual or geometric approach here known in various forms since the 1840s, demystifies the square root of -1 and the associated concept of "imaginary" numbers. Here complex number multiplication illustrates rotation and dilation operations in the plane.
- Geometric Notions with Ruler & Compass Constructions :
1 Initial Concepts & Terms
2 Angle, Vertex & Side Correspondence in Triangles
3 Triangle Isometry/Congruence
4 Side Side Side Method
5 Side Angle Side Method
6 Angle Bisection
7 Angle Side Angle Method
8 Isoceles Triangles
9 Line Segment Bisection
10 From point to line, Drop Perpendicular
11 How Side Side Side Fails
12 How Side Angle Side Fails
13 How Angle Side Angle Fails

www.whyslopes.com >> Volume 2 Three Skills For Algebra >> Chapter 15. Solving Linear Equations Next: [Chapter 16. Painless Theorem Proving.] Previous: [ Chapter 14. Forward and Backward Use of a Formula.]   [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19][20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32] [33] [34] [35] [36] [37] [38] [39] [40] [41] [42]

# Chapter 15. Solving Linear Equations

Volume 2, Three Skills for Algebra

Here are some more examples in which we solve equations. Our aim is to become familiar or at ease with handling and manipulating equations. So we look at the algebraic solution of equations containing one or more unknown numbers.

### 1  One Unknown

#### 1.1  First Example

When we let x = 5, we have 2x = 10 and 4x ¹ 15. Suppose now we forgot the value of x which made 2x = 10, could we find the value of x from the equation 2x = 10? The answer is yes. We can solve for the unknown or forgotten value of x as follows:

 x = 2x 2 = 10 2 = 5
In this solution, we used the property [(ab)/(b)] = b with the role of a played by x and the role of b played by 2. This gives the first equality. The second equality follows from assumption that 2x = 10. The latter allows 2x to be replaced by its value 10. Another way to look at this solution is to say
 2x = 10
Therefore
 2x 2 = 10 2
Hence
 x = 5
The manipulation process here creates new equalities from previous ones until an expression
 x = a numerical value
appears. How we get find the value of x from an equation involving x or other unknowns is a matter of taste.1

#### 1.2  Second Example

Problem:   Find the value of x which satisfies the equation 7x+9 = 65.

Solution:   The aim is to manipulate (or change or massage) the given equation

 7x+9 = 65
to get a new one of the form
 x = a numerical value.
The first step is to subtract 9 from both sides. This gives
 7x = 65-9
Some of you may know that 65-9 = 56. We could write 56 instead of 65-9. A next step to further isolate x is to divide by 7 (or multiply by [1/7]) since [(7x)/7] = x. This manipulation gives
 7x 7 = (65-9) 7
Therefore
 x = (65-9) 7 = 56 7 = 8
The isolation of x is complete. The solution is x = 8. To check this, just in case we made a mistake, observe when x = 8, we have 7x+9 = 7·8+9 = 56+9 = 65. So the original the equation 7x+9 = 65 holds (is satisfied, is true) when x = 8.

#### 1.3  Third Example

Problem:   Find the value of x which satisfies 5x+6 = 117.

Solution:   The aim is to manipulate the given equation

 5x+6 = 117
to get a new one of the form
 x = a numerical value
A first step is to subtract 6 from both sides. This gives
 5x = 117-6
A next step to further isolate x is to divide by 5 (or multiply by [1/5]) since [(5x)/5] = x. This manipulation gives
 x = 5x 5 = (117-6) 5
Therefore
 x = 111 5 = 22+ 1 5 = 22 1 5
The isolation of x is done. The solution is x = [111/5]. To check this, just in case we made a mistake, observe when x = [111/5], we get 5x+6 = 111+6 = 117.

The solution

 x = 111 5 = 111 ×2 5 ×2 = 222 10 = 22.2 = 22 1 5
can be written in several ways. Which way we prefer is a matter of taste.

Here are some more examples in which we solve equations. Our aim is to become familiar or at ease with handling and manipulating equations. So we look at the algebraic solution of equations containing one or more unknown numbers.

### 2  Algebraic Shorthand Solution

In a play or movie, the roles are more important than the actors, stars excepted. That is, any role can be played by any actor. But after the cast is selected, each role is usually played by only one actor, and each actor usually plays only one role. Once a play (or scene) is finished, the actors can take roles in another play (or scene). Likewise, in algebra, we have choice in the selection of the shorthand notation in which a problem or its solution is posed. But after the selection, the choice should be fixed at least temporarily. Once the problem and solution have been treated, the shorthand in it can be recycled in another plot.

#### 2.1  Third Example Revisited

The role of x in the third example can be played by any other letter, for instance y. We will repeat the third example problem with y in place of x. (This is mathematics ad nauseum.)

Problem:   Find the value of y which satisfies 5y+6 = 117. (This problem is identical to the previous one, except the shorthand symbol for the forgotten or unknown number is now the letter y instead of the letter x. The solution is identical. It is given or repeated next. Excuse the repetition, but you must see that it is a repetition.)

Solution:   The aim is to manipulate the given equation

 5y+6 = 117
to get a new one of the form
 y = a numerical value
A first step is to subtract 6 from both sides. This gives
 5y = 117-6
A next step to further isolate y is to divide by 5 (or multiply by [1/5]) since [(5y)/5] = y. This manipulation gives
 y = 5y 5 = (117-6) 5
Therefore
 y = 111 5 = 22+ 1 5 = 22 1 5
The isolation of y is done. The solution is y = [111/5]. To check this, just in case we made a mistake, observe when y = [111/5], we get 5y+6 = 111+6 = 117.

#### 2.2  An Algebraic Pattern

Each of the above examples has the form ax+b = c in which the numbers a, b and c are given, and x is initially unknown. In the first example, the roles of a, b and c were played or given by 7, 9 and 65. That gave the equation 7x+9 = 65. In the second example 5x+6 = 117, the number 5 is used in place of a, the number 6 plays the role of b and the number 117 is given by c.

General Problem:   Find x if ax+b = c.

ALGEBRAIC SHORTHAND SOLUTION. We follow the pattern set in the previous examples. First we subtract b from both sides of the equation ax+b = c. This gives

 ax = c-b
Next, we observe if a is nonzero,
 x = ax a = (c-b) a
Thus the formula for x is
 x = (c-b) a
This gives a recipe for x no matter what values of a, b and c are given in the problem: find x if ax+b = c. The formula can be used when a ¹ 0. Division by zero is not permitted or done in arithmetic. It is not possible.

Check: When x = [(c-b)/(a)], we see ax+b = a·[(c-b)/(a)] = (c-b)+b = c as hoped.

The recipe
 x = (c-b) a
describes and gives the solution to many problems of the form ax+b = c.

 Problem Solution ax+b = c x = [(c-b)/(a)] 5x+6 = 65 x = [(65-6)/5] 7x+9 = 117 x = [(117-9)/7] 7y+9 = 117 y = [(117-9)/7] 123x+456 = 12067 x = [(12067-456)/123] 100x+(-20) = 800 x = [(800-(-20))/100] = [(800+20)/100] = 8.2 100x-20 = 800 x = [(800-(-20))/100] = [(800+20)/100] = 8.2 [4/5]x+4 = 10.2 x = [(10.2-4)/([4/5])] 3z+7 = 19 z = [(19-7)/3]

The formula x = [(c-b)/(a)] describes and gives a solution to many problems of the form ax+b = c. We can further use this recipe without repeating each time, the reasoning that led to it.

EXTRA. The above formula for x can be used to solve the equation ax-d = c by putting b = -d. The equation ax-d = c can be rewritten as ax+(-d) = c since subtraction of d can be replaced by the addition of the number -d.

Here are some more examples in which we solve equations. Our aim is to become familiar or at ease with handling and manipulating equations. So we look at the algebraic solution of equations containing one or more unknown numbers.

### 3  Systems with More Unknowns

Equations with more than one unknown can be solved if they are manipulated or massaged into a simpler form. For equations with more than two or three unknowns, we can obtain complicated formulas for their solution, but the equations can be solved more efficiently without these formulas. In this case the method of solution is easier to remember than the formula. This contrasts with the expectation that I have (accidentally) built earlier. The shorthand description of calculations that could be done, that is, formulas for obtaining numbers and quantities, are useful tools. But we should not insist on using them all the time. Your own experience or that of others is needed to say when a purely algebraic approach appears best.

Problem:   Find the unknown or forgotten values of x,y and z when

 50
 =
 10y
 21
 =
 2x+3y
 14
 =
 x+y+2z
Solution:   Don't panic. From the first equation we see y = [50/10] = 5. We use this information in the second equation 21 = 2x+3y to get
 21 = 2x+15
So 6 = 21-15 = 2x. This gives 2x = 6 or x = 3. Finally, we use the values x = 3 and y = 5 in the third and last equation14 = x+y+2z to get 14 = 3+5+2z. Therefore 14 = 8+2z. So 14-8 = 2z. The value of z = [(14-8)/2] = [6/2] = 3. The unknown or forgotten values should be x = 3, y = 5 and z = 3. Are they correct?

Check:
 10y =10·5
 =
 50
 2x+3y = 2·3 + 3·5 = 6+15
 =
 21
 x+y+2z= 3+5+2·3 = 8+6
 =
 14

### 4  Simplified Problems

#### Addition-Multiplication Method for non-triangular systems

In the previous problem, we could find the unknowns one at a time. One method for solving equations is to change, massage and manipulate them into a form where we can find the unknowns one at a time. We can do this by adding multiples of equations together.

Example:   Solve

 7 =
 2x+y
 5 =
 2x-y

Solution:   Keep the first equation as is, and add it to the second. This gives 7 = 2x+y and

 7+5 = 2x+y+(2x-y)
The left-hand side 7+5 = 12 and the right-hand side simplifies to 2x+y+2x-y = 4x since adding y and then subtracting it gives the same result as not doing anything. Therefore the second becomes or is replaced by
 12 = 4x
The new second equation suggests x = [12/4] = 3. We use this value in the first equation to find 7 = 2·3+y. Therefore x = 3 and 7 = 6+y. So the solution should be x = 3 and y = 7-6 = 1. It is easy to check that this proposed solution satisfies the two equations.

Example:   Solve the following system (set) of equations
 12 =
 2x+3y
 19 =
 5x+2y
Solution:   To get two equations with equal coefficients for x, (i) multiply the both sides of the first equation by the number 5 and (ii) multiply both sides of the second equation by 2. This gives a system of two new equations
 60 =
 10x+15y
 38 =
 10x+4y
Now keep the second equation as is, and subtract it from the first. This gives a third system
 60-38 =
 11y
 38 =
 10x+4y
This system has a simple form. We can find y and then x. Here y = [((60-38))/11] = [22/11] = 2. We can now use the known value 2 of y to find x. From 38 = 10x+4y, we may get the value of x in two ways:
• An arithmetic way: the equation 38 = 10x+4y gives 38 = 10x+4 ·2. So 38 = 10x+8. Therefore 10x = 38-8 = 10. Thus x = [30/10] = 3.
• An algebraic way: the equation 38 = 10x+4y implies or gives 10x = 38-4y. So x = [(38-4y)/10] = [(38-4·2)/10] = [30/10] = 3.
Either way the solution is given by x = 3 and y = 2. Exercise: check this.

### 5  Examples with Three Unknowns

Example: Solve the following equations (a1), (a2) and (a3).

 (a1 ):         5x +2y +2z = +27 (a2 ):       11x +8y +2z = +57 (a3 ):       -3x +1y -2z = -16

A Solution:   First subtract equation (a1) from (a2). This implies yields equation (b2) below.

 (b1 :)         5x +2y +2z = +27 (b2 :)         6x +6y = +30 (b3 :)       -3x +1y -2z = -16
Equations (b1) and (b3) are identical to equations (a1) and (a3), respectively. We have just changed the labels on them.

Second, add equation (b1) to equation (b3). This gives

 (c1 ):       5x +2y +2z = +27 (c2 ):       6x +6y = +30 (c3 ):       2x +3y = +11

The above steps have eliminated z from the last two equations.

Third, from equation (c2) subtract two times equation (c3). This implies

 (d1 ):       5x +2y +2z = +27 (d2 ):       2x = +8 (d3 ):       2x +3y = +11

Finally, we may change the order of equations. This yields the more suggestive system of equations:

 (d2 ):    2x = +8 (d3 ):    2x +3y = +11 (d1 ):    5x +2y +2z = +27

This last step was optional. Now we can do the following.

•

• Solve equation (d2) for x. This gives
 x = 8/2 = 4
• Solve equation (d3) for y. This yields
 y = 1 3 (11-2x) = 1 3 ·(11-8) = 1
• Solve equation (d1) for z. This implies
 z = 1 2 (27-5x-2y) = 1 2 (27-20-2) = 1 2 (5) = 5 2
This in summary yields the solution
 (x,y,z) = (4,1, 5 2 )
respectively.

Exercise:   Solve

 (a1 :)         x +y + 3z = +10 (a2 :)         x -y +2z = +5 (a3 :)       2x +4y -5z = +9

Note that you can and should check your answer (values for x, y and z) satisfy each equation. If one is not satisfied then there is an error somewhere in your work, the solution or the check.

Exercise:   Solve

 (a1 :)         x +y + 3z = +10 (a2 :)         x -y +2z = +5 (a3 :)       2x +4y +5z = +9

See the difference a "small" change in the problem makes.

### 6  Useful Arithmetic Rules & Advice

#### 6.1 Useful Arithmetic Rules

In the previous examples we have used rules or properties of arithmetic to help us in our algebraic manipulations. A short description of them follows. These rules or properties are all related to the idea that if two expression are supposed to give the same number when computed, then using one of them in place of the other in larger expressions should not change the values given by the larger expressions.

2Talking about the preservation of equality can be a wordy exercise - to be done at least once, but not more than thrice.

When a, b and c are shorthand symbols or expressions representing real numbers, the following properties are useful in solving equations.

• If a = b, that is if a and b are two expressions which when evaluated should give the same number (or quantity) then we should have
ac = bc
a+c = b+c, and
a
-c = b-c.
whenever we multiply, add or subtract by another number or expression c. In brief, equality should be preserved (kept) if what we do to one side, we also do to the other.2
• If a+b = c+b then a = c. This follows by subtracting b from both sides.
• If ab = cb and b is nonzero then a = c.
• If ab = cb and b is nonzero then a = c. This follows by dividing both sides by b or by multiplying both sides by [1/(b)] = 1¸b.
• If a = b and c is nonzero then  a/c = b/c

www.whyslopes.com >> Volume 2 Three Skills For Algebra >> Chapter 15. Solving Linear Equations Next: [Chapter 16. Painless Theorem Proving.] Previous: [ Chapter 14. Forward and Backward Use of a Formula.]   [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19][20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32] [33] [34] [35] [36] [37] [38] [39] [40] [41] [42]

Road Safety Messages for All: When walking on a road, when is it safer to be on the side allowing one to see oncoming traffic?

Play with this [unsigned] Complex Number Java Applet to visually do complex number arithmetic with polar and Cartesian coordinates and with the head-to-tail addition of arrows in the plane. Click and drag complex numbers A and B to change their locations.

#### Pattern Based Reason

Online Volume 1A, Pattern Based Reason, describes origins, benefits and limits of rule- and pattern-based reason and decisions in society, science, technology, engineering and mathematics. Not all is certain. We may strive for objectivity, but not reach it. Online postscripts offer a story-telling view of learning: [ A ] [ B ] [ C ] [ D ] to suggest how we share theory and practice in many fields of knowledge.

#### Site Reviews

1996 - Magellan, the McKinley Internet Directory:

Mathphobics, this site may ease your fears of the subject, perhaps even help you enjoy it. The tone of the little lessons and "appetizers" on math and logic is unintimidating, sometimes funny and very clear. There are a number of different angles offered, and you do not need to follow any linear lesson plan. Just pick and peck. The site also offers some reflections on teaching, so that teachers can not only use the site as part of their lesson, but also learn from it.

2000 - Waterboro Public Library, home schooling section:

CRITICAL THINKING AND LOGIC ... Articles and sections on topics such as how (and why) to learn mathematics in school; pattern-based reason; finding a number; solving linear equations; painless theorem proving; algebra and beyond; and complex numbers, trigonometry, and vectors. Also section on helping your child learn ... . Lots more!

2001 - Math Forum News Letter 14,

... new sections on Complex Numbers and the Distributive Law for Complex Numbers offer a short way to reach and explain: trigonometry, the Pythagorean theorem,trig formulas for dot- and cross-products, the cosine law,a converse to the Pythagorean Theorem

2002 - NSDL Scout Report for Mathematics, Engineering, Technology -- Volume 1, Number 8

Math resources for both students and teachers are given on this site, spanning the general topics of arithmetic, logic, algebra, calculus, complex numbers, and Euclidean geometry. Lessons and how-tos with clear descriptions of many important concepts provide a good foundation for high school and college level mathematics. There are sample problems that can help students prepare for exams, or teachers can make their own assignments based on the problems. Everything presented on the site is not only educational, but interesting as well. There is certainly plenty of material; however, it is somewhat poorly organized. This does not take away from the quality of the information, though.
... section Solving Linear Equations ... offers lesson ideas for teaching linear equations in high school or college. The approach uses stick diagrams to solve linear equations because they "provide a concrete or visual context for many of the rules or patterns for solving equations, a context that may develop equation solving skills and confidence." The idea is to build up student confidence in problem solving before presenting any formal algebraic statement of the rule and patterns for solving equations. ...

#### Senior High School Geometry

- Euclidean Geometry - See how chains of reason appears in and besides geometric constructions.
- Complex Numbers - Learn how rectangular and polar coordinates may be used for adding, multiplying and reflecting points in the plane, in a manner known since the 1840s for representing and demystifying "imaginary" numbers, and in a manner that provides a quicker, mathematically correct, path for defining "circular" trigonometric functions for all angles, not just acute ones, and easily obtaining their properties. Students of vectors in the plane may appreciate the complex number development of trig-formulas for dot- and cross-products.
Lines-Slopes [I] - Take I & take II respectively assume no knowledge and some knowledge of the tangent function in trigonometry.

#### Calculus Starter Lessons

Why study slopes - this fall 1983 calculus appetizer shone in many classes at the start of calculus. It could also be given after the intro of slopes to introduce function maxima and minima at the ends of closed intervals.
- Why Factor Polynomials - Online Chapter 2 to 7 offer a light introduction function maxima and minima while indicating why we calculate derivatives or slopes to linear and nonlinear curves y =f(x)
- Arithmetic Exercises with hints of algebra. - Answers are given. If there are many differences between your answers and those online, hire a tutor, one has done very well in a full year of calculus to correct your work. You may be worse than you think.