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# Numerical Solution of Algebraic Equations

By Frances Reyes,2014-06-17 01:48
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Numerical Solution of Algebraic Equations ...

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Visual Algebra for the Early Grades

George Mondras

There are two ways to solve an algebraic equation: 1) the axiomatic method that will be

taught in the later grades and 2) the visual method that can be taught in the early grades.

Visual Algebra is a method that was prevalent at the dawn of the mathematical sciences. At

the University of Alexandria, 310 B.C.E., geometric methods were used in the solution of ?algebraic equations. For the equation: 2 x = 8, each term in the equation was depicted as an

area and a geometric diagram was constructed. In the first textbook on mathematics, The

Elements, by Euclid, the fourth axiom, “Things which coincide with one another are equal to one another,” is compelling evidence that algebra was geometric in that era. A student would ??conclude that equality requires that the two rectangular areas are equal; therefore length x must

coincide with length 4.

?The geometric solution: 2 x = 8

?? 2 x ?? 2 8 2

x4

?????? Symmetry in algebra is achieved when an equation is reformulated such that the form of

the equation is the same on both sides of the equality. Visual algebra does not solve for the

unknown value of the variable, but applies the axioms to reformulate and render an equation

symmetrical. The solution is then found implicitly by a one-to-one matching of terms.

The reflective symmetry of the human form is a natural concept that is intuitively understood, is

ingrained in everyone and was once utilized in the early geometric development of algebra. The

visual method is geometric algebra, with symmetry substituted for geometry. Reformulating an

equation to obtain symmetry utilizes the student’s most primitive ingrained concepts.

The solution using symmetry:

?2 x = 8

??2 x = 2 4 x = 4, from the one-to-one matching of terms

??The concept of one-to-one matching is most likely deduced from the result of a simple

experiment that everyone does very early in life: match thumb to thumb, index finger to index ????finger, and so on to obtain a perfect left to right hand match. A classroom demonstration that

one-to-one matching is within the purview of the students; ask anyone to respond if they think

the students in the room is greater than the number of chairs. No one will respond. The students

need only observe one empty seat or that everyone has a seat and they can say with certainty that

the number of chairs is greater or equal to the number of students. There is a more advanced way

to determine the facts, count all the chairs and count all the students then compare the two

numbers. Counting is advanced and requires prior knowledge whereas one-to-one matching is

primitive.

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Definition of Algebra:

Algebra is the language of higher mathematics.

1) The numbers are the nouns of the language.

?, ?, ?, ?, ?, , and so on, tell us what action to take, so they play the 2) The operators,

roll of verbs in the language.

3) Assumptions that are postulated for algebra are the grammar and syntax of the language;

five assumptions were postulated to develop the early geometric algebra: 1) Things that are ??equal to the same thing or to equal things are equal to each other. 2) If equals are added to equals, then the

results are equal. 3) If equals are subtracted from equals, then the remainders are equal. 4) Things that

coincide with one another are equal to one another. 5) The whole is greater than the part.

Algebra is the first level of abstraction in mathematics and requires the use of literals to

represent numbers. In an equation, the literals initially have unknown or variable values that can

be determined by algebraic means. The literals are generally the letters of an alphabet: A, a, B, b,

C, c, X, x, and so on; using capitals letters first and lower case later in parallel with student early

learning. Avoid the use of the literals I, i, O, o, and lower case ell l as they are too easily

confused with zero and one. Do not overuse the literal x and, if the numbers have a tangible meaning, try to choose a literal that acts as a memory aid: N for number, M for Mary’s age, V for

velocity, T for temperature, etc.

One possible way to introduce visual algebra is to initially have the students conduct the

finger-matching experiment (see page 1). The pedagogy can then proceed through equations that

are already written in translational symmetry form. The K-1 students would only be asked to

identify and state the unknown value of the variable by the one-to-one matching of terms. As

skills develop, the equations can require arithmetic operations to achieve translational symmetry.

Define Visual Algebra:

Apply the axioms to one side of an equation to reformulate and achieve symmetry. Solve for

the variable implicitly using a one-to-one matching of terms. Check any modified expression.

Visual Mantra: Rewrite the equation until both sides look the same.

Apply the axioms to one side of the equation to achieve final symmetry a?x?b?c

c?bc?b Translational symmetry; x = by one-to-one matching of terms a?x?b?a?()+baa??c?b Reflective symmetry is a possible reformulation but is not likely a?x?b?b?()?aa

????Examples: Acronym: TS = Translational symmetry

3?x?5?3?2?5a) ? Translational symmetry; x = 2 by the one-to-one matching of terms ??

3?A?5?3?(1?2)?5b) ? TS requires 1 operation; note that A is now 3 3?B?5?3?(2?2)?2?3c) ? TS requires 2 operations; note that B is now 4 ??

3?D?5?3?(3?4?2)?7?2d) ? TS requires 3 operations; note that D is now 6

??3?d?5?3?(6?2?4?2)?6?1e) ? TS requires 4 operations; note that d is now 5

??3?X?5?3?(3?6?2?10)?4?1f) ? TS requires 4 operations; note that X is now 1

??3?a?5?3?a?5g) ? Equations, written in translational symmetry form, can be reformulated

?? in many ways to enable the students to acquire both arithmetic and algebraic skills.

??

??

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Visual Solution of Equations:

Acronyms: LHS/RHS = Left/Right Hand Side.

English sentence: Three times the quantity x, in addition to five is eleven.

x + 5 = 11 Algebraic sentence: 3 ?

3?x?5?11Algebraic equation: ? Asymmetric equation

??

3?x?5?11 ? RHS: Reformulate to achieve translational symmetry

3?x?5?(11?5)?5?? ? Step 1, RHS: add 0 = 5 + 5; group the unmatched terms (11 5)

11?51?? ? Step 2, RHS: multiply by to match LHS coefficient 3 3?x?5?3?()?51?3?33

??3?x?5?3?2?5? ? Step 3, RHS: evaluate: (11 5) 3 = 2

x ? 2 ? Visual solution: x = 2, from the one-to-one matching of terms

????3?x?5?3?2?5 ? Translational symmetry is the most likely reformulation. ????

3?x?5?5?2?3 ? Reflective symmetry is a possible reformulation, but not likely. ??

x?2 ? Reflective symmetry is the solution via the axiomatic method ?? ???? Reflective symmetry, the Geometric Solution: ?? th 3x + 5 = 5 + 6, Euclid’s 4 Axiom, area equality requires 3 +5 = 5 + 3 3x 6 that x must coincide with 2. Geometric algebra has been ?? absent from the pedagogy for approximately 2000 years

2 x

The example above demonstrates all aspects of the Visual Method. Step 1, the equivalent of

0 or 5 + 5 is added to obtain a matching term +5 = +5. Step 2, the unmatched terms, 11 5, are

?grouped, (11 5), and multiplied by the equivalent of 1 or 3 ? 3 or 3 (1/3) to obtain the matching x coefficient 3 = 3. That completes the reformulation procedure; the x-value is found visually by the one-to-one matching of terms. The steps on the right-hand side of the equation

throughout are only a reformulation of the number 11: ??

11?53?2?5?1111 5 + 5 = 11, = 11, and . 3?()?53

The subtraction 11 5, can be done by integer addition as shown on the next page.

??

?? 11 5 = 1 + 1 + 4 = 6

4 Special Case of Subtraction By 9’s Complement Addition:

Develop an algorithm such that the operation of subtraction can be accomplished by addition.

Acronym: 9C = 9’s complement of the subtrahend, that is, subtrahend + 9’s complement = 9

Subtraction Table: subtrahend digit greater than the corresponding minuend digit.

10 11 12 13 14 15 16 17 18

1 9

2 8 9

3 7 8 9

4 6 7 8 9

5 5 6 7 8 9

6 4 5 6 7 8 9

7 3 4 5 6 7 8 9

8 2 3 4 5 6 7 8 9

9 1 2 3 4 5 6 7 8 9

To accomplish subtraction by addition, start with the equation: 15 7 = 8 and note that 1 + 5 is the sum of the minuend digits and 2 is the 9’s complement of subtrahend 7, that is, 7 + 2 = 9.

Algorithm ( by +): 15 7 = 1 + 5 + 2 ? 15 7 = 6 + 2 ? 15 7 = 8 Algorithm verified: 15 7 = (9 + 6) (9 2) ? 15 7 = 9 + 6 9 + 2 ? 15 7 = 8

The 9’s complement table for the digits 0 to 9 are: (9 + 0), (8 + 1), (7 + 2), (6 + 3), (5 + 4) = 9 Examples of subtraction by addition using the above algorithm:

18 = 9 + 9 1 + 8 = + 9 ? Add the minuend digits 1 + 8

9 = 0 9 + 0 ? 0 is the number that added to subtrahend 9 equals 9, or 9 + 0 = 9

9 = 9 + 0 + 9 ? Difference verified by: 18 9 = (9 + 9) (9 0) = 9 + 0 = 9

16 = 7 + 9 1 + 6 = + 7 ? Add the minuend digits 1 + 6

8 = 1 9 + 1 ? 1 is the number that added to subtrahend 8 equals 9, or 8 + 1 = 9

8 = 8 + 0 + 8 ? Difference verified by: 16 8 = (9 + 7) (9 1) = 7 + 1 = 8

13 = 4 + 9 1 + 3 = + 4 ? Add the minuend digits 1 + 3

7 = 2 9 + 2 ? 2 is the number that added to subtrahend 7 equals 9, or 7 + 2 = 9

6 = 6 + 0 + 6 ? Difference verified by: 13 7 = (9 + 4) (9 2) = 4 + 2 = 6

11 = 2 + 9 1 + 1 = + 2 ? Add the minuend digits 1 + 1

6 = 3 9 + 3 ? 3 is the number that added to subtrahend 6 equals 9, or 6 + 3 = 9

5 = 5 + 0 + 5 ? Difference verified by: 11 6 = (9 + 2) (9 3) = 2 + 3 = 5

12 = 3 + 9 1 + 2 = + 3 ? Add the minuend digits 1 + 2

5 = 4 9 + 4 ? 4 is the number that added to subtrahend 5 equals 9, or 5 + 4 = 9

7 = 7 + 0 + 7 ? Difference verified by: 12 5 = (9 + 3) (9 4) = 3 + 4 = 7

Note-1: The Algorithm is valid in the ranges, Minuend = 10 to 19 and Subtrahend = 0 to 9, otherwise the procedure will fail. The students will memorize the basic facts of subtraction by

the repetitive use of this algorithm.

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Linear equations: Acronyms: Left Hand Side = LHS, Right Hand Side = RHS.

Sentences that have a quantitative connotation can be expressed as an algebraic sentence, that

is, an equation. To enhance the concept that algebra is a language the students’ must see, read

and compare the quantitative English sentence in close proximity to its algebraic equation

written in translational symmetry form. The examples are designed to show this comparison. As

the students’ progress in their arithmetic skills, the equations can be written in asymmetric form

and the unique solution found by reformulation to achieve symmetry. The examples below are

equations written in translational symmetry form; the unique solution is obtained by the one-to-

one matching of terms. At first, the students need only identify the value of the unknown literal.

x?1?3?1 ? x?3 a?2a?1?0?2a?1 ? a?0

b?2b?3?2(3) ? b?3 c?2c?3c?c?2c?3(1) ? c?1

3(D?4)?3(5?4) ? D?5 3(E?4)?5?3(6?4)?5 ? E?6

1313Z6??? ? Y?4 3(?7)?8?3(?7)?8 ? Z?6Y44455

222x2x22?6X?8?2?6X?8 ? X?2?6x??6(3) ? x?3 X33

Example Problems and Solutions:

??Note: In all problems, the statement of the problem (English sentence) must be in close

proximity to the algebraic equation to demonstrate to the students that Algebra is a Language.

a) A salesperson earns a salary of \$100.00 per week plus a commission of \$2.00 for each item

sold. Find the number of items sold if the earnings were \$500.00 in one week.

English sentence: \$2 times the number of items sold, plus the salary, is equal to the earnings

?Algebraic sentence: \$2 the items sold n + \$100 equals earnings \$500

?Algebraic equation: 2 n + 100 = 500 ? Asymmetric equation

??

??K-? Formulation: 2 n + 100 = 2 200 + 100, n = 200, from one-to-one matching ?? One of many formulations that require the students to ???2 n + 100 = 2 (4 50) + 100 perform arithmetic operations to achieve symmetry ????Solution:

???2 n + 100 = 500 ????? RHS: Reformulate to achieve translational symmetry

?2 n + 100 = (500 100) + 100 ? RHS: add 0 = 100 + 100 and group terms (500 100)

500?1001???2 n + 100 = ? multiply (500 100) by to match coefficient 2?()?1001?2?22

??2?200?100??2 n + 100 = ? RHS: evaluate (500 100) 2 = 200

?? n = 200 ? n = 200 items sold that week, from one-to-one matching of terms

????

??????

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b) The cost to rent an automobile is \$37 dollars a day plus 20 cents per mile. If the final bill is

\$60.00, how many miles were driven?

English sentence: \$0.20 times the number of miles driven, plus the daily cost, is equal to \$60

?Algebraic sentence: \$0.20 m + \$37 = \$60

?Algebraic equation: 0.20 m + 37 = 60 ? Asymmetric equation

????K-? Formulation: 0.20 m + 37 = 0.20 115 + 37, m = 115, from one-to-one matching

??One of many formulations that ??0.20 m + 37 = 0.20 (75 + 40) + 37 ? require students to do arithmetic Solution: to achieve translational symmetry ????

?0.20 m + 37 = 60 ? RHS: Reformulate to achieve translational symmetry

?????0.20 m + 37 = (60 37) + 37 ? RHS: add 0 = 37 + 37 to match term and group (60 37)

60?371???0.20 m + 37 = ? RHS: multiply by to match coefficient 0.20?()?371?0.20?0.200.20

??0.20?115?37??0.20 m + 37 = ? Evaluate (60 37)0.20 = 115

?? m = 115 miles driven, from one-to-one matching

????

??????c) A tank that is five-eights full contains 145 gallons. Find the tank’s capacity.

English sentence: A tank that is five-eights full contains 145 gallons

5Algebraic sentence: A tank that is full contains contains 145 gallons.8

5Algebraic equation: Let C?The tank's Capacity; equation: ?C?1458

55 ?C??232 ? K?? Formulation; C?232 gallons, from one-to-one matching??88

Solution:??

5?C?145 ? Asymmetric equation; reformulate to achieve translational symmetry 8

558588?C???145 ? RHS: Multiply 145 by 1=? to match coefficent. ?145?23288585555?C??232 ? Translational Symmetry: C = 232 gallons, from one-to-one matching88

One of many formulations that require 55 ? ? ??C??(16?16?24)??arithmetic to achieve translational symmetry 88

??

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Section 2, Systems of equations:

In systems of linear equations, where the variables have a tangible meaning, the equations

will be reformulated to obtain symmetry or to contain a collinear equation. Examples: SE 1 to SE

9 introduces problem solving in two variables.

Note: In this and the following sections modern algebraic notation will be used. 23x??2(M 2) =(M 2) 2 = M 2 + M 2, = 3xx, 4(48) = 448, (v + 6)4 = 4v + 24 Notation: ??

SE 1) John is five years older than Mary. Two years ago, John was twice as old as Mary.

What are their ages today? Reformulate each equation to obtain translational symmetry. ?????????? Let: J and M = John and Mary’s ages today; J – 2 and M 2 = John and Mary’s ages 2 years ago

J = 5 + M ? J 2 = 5 + M 2 + 2 2 ? J 2 = 7 + M 4

?J 2 = (M 2) 2 ? J 2 = M 2 + M 2 ? J 2 = M + M 4

Mary’s age is 7 years old, from one-to-one matching; John’s age is 5 + 7 or 12 years old. Two

years ago, John was 10 and Mary was 5, that is, John was twice as old as Mary. ??

SE 2) A train leaves Boston traveling west at a speed of 30 km/h. Two hours later, another train

leaves Boston traveling in the same direction on a parallel track at 45km/h. At what

elapsed time will the faster train overtake the slower train?

?Fundamental equation from physics: Distance = Velocity Time.

Reformulate to achieve translational symmetry.

TrainA:Distance?30(t?2)?D=30t?60 ? D=30t?15(4) ?? TrainB:Distance?45t?D=30t?15t?D=30t?15t

t = 4 hours, from one-to-one matching

??SE 3) There are two supplementary angles in which one angle is 12 degrees less than 3 times

the other. What is the measure of the angles?

Reformulate to obtain translational symmetry

x?y?180 ? x?y?192?12 ? x?y?4(48)?12 y?3x?12 ? x?y?x?3x?12 ? x?y?4x ?12

y=180?48=132? From one-to-one matching, x?48?

??SE 4) Anti-freeze solution A is 2% methanol, and solution B is 6% methanol. Auto-Parts Inc.,

wants to mix the two to get 60 liters of a solution that is 3.2% methanol. How many liters ???? of each solution is required?

Reformulate to contain a collinear equation with coefficients = 2%

A?B?60 ? 2%A?2%B?2%(60) 2%A?6%B?3.2%(60)?2%A?2%B?4%B?2%(60)?1.2%(60)

1.2%(60) B? = 18 liters of 6% methanol 4%?? ? ? A = 42 liters of 2% methanol A?18?6042?18?60

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SE 6) A \$4800 investments in two corporate bonds earn \$412 in interest the first year. The bonds

pay interest of 8% and 9% per annum. Find the amount invested at each rate of interest.

Reformulate to contain a collinear equation with coefficient ratio = 1: 0.08; 0.08(4800) = 384

E?N?4800?E? N ?4800 0.08E?0.09N?412?0.08E?0.08N?0.01N?384?28

28 = \$2800 at 9% interest; E = \$4800 \$2800 = \$2000 at 8% interest N?0.01

??SE 7) The ground floor of the Empire State building in New York is a rectangle. The perimeter

of the building is 860 ft. The width is 100 ft less than the length. Find the length and width.

Reformulate to contain a collinear equation with coefficient ratio = 2:1

2l?2w?860 ?2l?2w?660?200? 2l?2w?4(165)?200 l?w?100?l?w?w?w?100 ? l? w ?2w ?100

w = 165 ft from one-to-one matching. l = 165 + 100 = 265 ft

??SE 8) Anti-freeze solution A is 2% methanol, and solution B is 6% methanol. Auto-Parts Inc.,

wants to mix the two to get 60 liters of a solution that is 3.2% methanol. How many liters of each

solution is required?

Reformulate to contain a collinear equation with coefficient ratio = 1:2%

A?B?60 ? A? B? 60 2%A?6%B?3.2%(60)?2%A?2%B?4%B?2%(60)?1.2%(60)

1.2%(60) = 18 liters of 6% methanol B?4%??A?60?18 ? A = 42 liters of 2% methanol

SE 9) Cool Mitts, Inc., sold 20 pairs of gloves. Plain leather gloves sold for \$24.95 per pair and

Gold-braided gloves sold for \$37.50 per pair. The company took in \$687.25. How many of each ??kind were sold?

Reformulate to contain a collinear equation with coefficient ratio =1: 24.95; 20(24.95) = 499.00

p?g?20 ? p? g?20

24.95p?37.50g?687.25?24.95p?24.95g?12.55g?499.00?188.25

188.25g??g?15Gold-braided gloves 12.55

p=20?15?p=5 Plainleather gloves

??

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