General Numbers:

In algebra as in arithmetic the quantities with which we deal are numbers. But whereas in arithmetic the numbers used are represented by figures (1, 2, 3, 4, etc.), each of which has a definite value, in algebra they are represented by other symbols (usually letters of the alphabet) which may have any numerical values assigned to them or whose numerical values are to be found. Such quantities are called general numbers.

Processes of Algebra:

We assume that the four fundamental processes, indicated by the symbols +, −, ×, ÷, will have the same meaning when applied to general numbers that they had in arithmetic.

Algebraic Expressions: 

Any combination of numbers (numerical or general) and the four processes is called an algebraic expression. Parts of such expressions which are separated from each other by the signs + or − are called terms.

Illustration: 2a + 3b - 5c is an algebraic expression consisting of three terms.

Note When no sign precedes a term the sign + is understood.

Simple and Multiple Expressions:

If an algebraic expression consists of one term, it is called a simple expression. If it contains two or more terms, it is called a multiple expression.

 Such algebraic expressions are also distinguished otherwise. Simple expressions are called monomials. An expression of two terms, such as a + b, is called a binomial, and an expression of three terms, such as a -b +c, is called a trinomial. In general, a multiple expression is called a multinomial.

Products:

When two or more quantities are multiplied together the result is called the product of the quantities.

 Since, in using letters to represent numbers, there is confusion between the multiplication sign and the letters x and X , we will often indicate multiplication in algebra either (1) by using a dot halfway up between the two letters, or (2) by writing the letters one after the other with no symbol between them. Thus, the product of a and b may be written either as a × b, a∙b, or as ab.

Factors:

Each of the quantities multiplied together to form a product is called a factor of the product. Thus 3, a, and 6 are the factors of the product 3 ab. 3 is called the numerical factor, and a and b are called literal factors.

Coefficients:

The numerical factor of an expression is called the coefficient of the remaining factors. Thus, in the expression 3ab, 3 is the coefficient. In a more general sense, 3a may be considered as the literal coefficient of b.

Note: The quantity a has the same meaning as 1a. Hence, if the coefficient of a quantity is not explicitly written, it is understood to be 1.

Powers and Exponents:

A power of a quantity is the product obtained in multiplying the quantity by itself any number of times. It is expressed by writing to the right and above the quantity an index which indicates the number of factors to be taken. Thus

    a × a is called the second power of a, and is written a^2;

    a × a × a is called the third power of a, and is written a^3; and so on.

The small number which expresses the power of any quantity is called an exponent. Thus, 2, 4, 6 are the exponents of a^2, a^4, a^6 respectively.

          a^2 is usually read “a square”;
          a^3 is usually read “a cube”;
          a^4 is usually read ‘‘a to the fourth”; and so on.

Note: The quantity a has the same meaning as a^1. Hence, if the power of a quantity is not explicitly written, it is understood to be 1.

Example 1 What is the difference in meaning between 2a and a^2?

Solution: By 2a, we mean the product of 2 and a. By a^2, we mean the product of a and a.
Thus, if a = 5,
          2a=2*a=2*5=10,
whereas,         a^2=a*a=5*5 = 25.

Example 2 If p = 1, find the value of 4p^5.

Solution:4p^5=4*p*p*p*p*p=4*1*1*1*1*1=4

Note: Observe that every power of 1 is 1.

Example 3 If x = 2, y = 5, z = 7, find the value of xy + yz-xz.

Solution:

          xy =2*5= 10;
          yz=5*7 = 35;
          xz=2*7= 14.

Hence the value of the expression equals 10 + 35 -14 = 31.

Note: Each term is evaluated before adding and subtracting.

Example 4 If a = 5, b = 3, c = 1, evaluate 4(a-b)(b + c).

Solution:

          (a - b) = (5 - 3) = 2;
          (b+c)=(3+1 =4,

Hence, 4(a -b)(b + c) = 4*2*4 = 32.

Note: The parenthesis ( ) is used to indicate a single quantity.

Formulas

A rule or principle expressed in algebraic terms is called a formula. In our study of arithmetic, we learned of many rules for the computation of unknown quantities. A few of these rules are as follows:

1. The area of a rectangle equals the product of its length and its width:

                A = lw.

2. The area of a triangle equals one-half the product of a base and its corresponding altitude:

                A = 1/2bh.

3. The area of a circle equals the product of {pi} ({pi}=22/7) and the square of the radius:

                A ={pi}r^2.

4. The perimeter (or boundary) of a rectangle equals the sum of twice the width and twice the length:

                P=2l+2w.

5. The circumference of a circle equals the product of {pi} and the diameter:

                C = {pi}d.

6. The area of a trapezoid equals the product of one-half the altitude and the sum of the bases:

                A= 1/2h(b1+b2).

7. Distance traveled equals the product of the rate of travel and the time:

                D = RT.

8. Simple interest on money invested equals the product of the principal invested, the rate of interest per year, and the time in years:

                I = PRT.

9. The total value of a set of identical objects equals the product of the number of objects and the price per object:

                V = np.

Evaluation of Formulas

By the process of substituting definite numbers for the letters of a formula and then applying the algebraic operation indicated, we may find the value of the unknown quantity sought.

Example 1 Find the area of a triangle having a base 6 feet long, if the corresponding altitude is 4 feet.

Solution: Substituting the numbers into the formula for the area of a triangle, we have

            A=1/2*6*4= 12 square feet.

Example 2 Find the area of a circle whose radius is 7 feet.

Solution: Substituting the numbers into the formula for the area of a circle, we have

            A={pi}7^2 = 22/7*7*7 = 154 square feet.

Example 3 Find the area of a trapezoid whose bases are 8 feet and 12 feet, and whose altitude is 3 feet.

Solution: Substituting the numbers into the formula for the area of a trapezoid, we have

            A =1/2*3(8+ 12) = 1/2*3*20 = 30 square feet.

Operations of Algebra Applied to General Numbers

Addition: Just as
            2 ft.+ 3 ft. = 5 ft.,
            $2 + $3 = $5;
So             2a+3a= 5a.

We think of the literal part of the general number as indicating the nature of the quantities being added, and the numerical coefficients are actually added.

Note It is essential that the literal parts be the same in order to add in this manner. Terms having the same literal parts are called similar or like terms.

Subtraction: Just as
            5 ft. − 3 ft. = 2 ft.,
            $5 − $3 = $2;
So            5a-3a=2a.

As for addition, the literal part of the general number indicates the type of quantities being subtracted and the numerical coefficients are actually subtracted.

Multiplication: Just as
            3 × 2 ft. = 6 ft.,
            3 × $2 = $6;
So             3 × 2a= 6a.

One needs only to recall the meaning of multiplication to verify these statements. For example, 3 × $2 means $2 + $2 + $2, which, by addition, is $6. So, too, 3 × 2a means 2a +2a+ 2a, which adds to 6a.

It is to be noted in the above multiplication that only one of the two factors contains letters. The case in which both factors contain letters will be discussed later.

Division: Just as
            6 ft. ÷ 3 = 2 ft.,
            $6 ÷ 3 = $2
So           6a ÷ 3 = 2a.

The quotient obtained in division is by definition a quantity which when multiplied by the divisor yields the dividend. That this condition is satisfied in the above examples is evident by comparing with the examples under Multiplication.

Note Division in algebra is denoted more often by writing the dividend and divisor as a fraction, rather than by using the symbol ÷. For example, the above relationships, indicating division, could be written as follows:

            6 ft./3 = 2 ft., $6/3 = $2, 6a/3 = 2a

Equations

In the study of arithmetic we solved many problems of “missing numbers.” Having four fundamental operations at our disposal these problems always fell into one of four basic groups, illustrated as follows:

I. What number increased by 3 equals 6?

II. What number multiplied by 3 equals 6?

III. What number decreased by 3 equals 6?

IV. What number divided by 3 equals 6?

To answer these queries, we would probably picture the questions in our minds in the following form:

I. ?+3 =6, II. 3 × ? = 6, III. ? − 3 = 6, IV. ?/3 = 6.

In algebra, we shall answer such questions by actually carrying through our mental processes. That is, we let x (or any letter) represent the unknown quantity which we seek. Then, we actually write down what we pictured in our minds, namely:

I. x+3=6, II. 3x=6, III. X-3=6, IV. x/3=6.

A statement of equality between two expressions such as each of these represents is called an equation. The terms to the left of the sign of equality are said to form the left member or left side of the equation and the terms to the right are said to form the right member or right side of the equation. A number which when substituted for the unknown in an equation, reduces both members to the same number is said to satisfy the equation. A value of the unknown, which satisfies an equation, is called a root or solution of the equation.

Thus, the roots of the four equations given above are respectively:

I. x=3, II. x=2, III. x=9, IV. x=18.

Fundamental Principle Concerning Equations

Consider the numerical equality 6 = 6. No matter what number we either add to or subtract from both members of this equality, the resulting numbers will be equal.

Illustrations: (1) 6 = 6,           given equality;

         6+ 3 = 6 + 3,   adding 3 to both members;

         9=9        the results are equal.

        (2) 6 = 6, given equality;

         6- 3 = 6-3,       subtracting 3 from both members;

         3 = 3,          the results are equal.

Also, no matter by what number we multiply or divide both members of this equality, the resulting numbers will be equal. (Division by zero excluded.)

Illustrations: (1) 6 = 6,        given equality:

         6 × 3 = 6 × 3,     multiplying both members by 3;

         18 = 18,       the results are equal.

        (2) 6 = 6,        given equality;

         6 ÷ 3 = 6 ÷ 3,     dividing both members by 3;

         2 = 2,        the results are equal.

This apparently obvious property of equalities is of fundamental importance.

        

Solution of Equations

Consider the four equations of previous examples. Each of these may be solved for the unknown x by application of the above principle, thus

I.         x+3 = 6,          given equation;

        x+3-3 = 6 - 3,      subtracting 3 from both sides;

              x=3,       simplifying.

II.            3x = 6,         given equation;

         (3x)/3 = 6/3,           dividing both sides by 3;

             x = 2,          simplifying.

III.        x-3 = 6,          given equation;

        x-3+3 = 6 + 3,      adding 3 to both sides;

              x=9,       simplifying.

IV.           x/3 = 6,          given equation;

        3*x/3 = 3*6,            multiplying both sides by 3;

            x = 18,         simplifying.

In general, if a and b represent any numbers whatsoever, then these four basic types of equations and their solutions are as follows:

        

Having obtained what we believe to be a solution in the above described manner, it is important to see if it actually does satisfy the given equation. This is called the process of checking the solution.

Example 1 Solve the equation x + 8 = 15 and check your result.

Solution:

      x+8 = 15,     given equation;

      x= 15-8,       subtracting 8 from both sides;

      x= 7,         simplifying.

Check:
      7+ 8 = 15,      substituting 7 for x in the given equation:

      15 = 15,         simplifying.

Example 2 Solve the equation x/4 = 31/2 and check your result.

Solution:

      x/4 = 31/2,     given equation;

      x = 4∙31/2,      multiplying both sides by 4;

      x= 14,      simplifying.

Check:
      14/4 = 31/2,      substituting 14 for x in the given equation:

      31/2 = 31/2,     simplifying.

Equations Requiring More Than One Operation

To solve some equations requires the use of more than one of the four basic methods just described.

Example 1 Solve the equation 2x+ 3 = 11 and check your result.

Solution:

        2x+ 3 = 11,      given equation;

              2x = 8,      subtracting 3 from both sides;

                x = 4,       dividing both sides by 2.

Check:

        2*4+3 = 11,     substituting 4 for x in the given equation;

          8+3=11,     simplifying terms;

            11 = 11,       combining terms.

Example 2 Solve the equation 5x-7 = 5 + x and check your result.

Solution:

        5x-7 = 5 + x,      given equation;

            5x = 12 + x,     adding 7 to both sides:

            4x = 12,       subtracting x from both sides;

            x = 3,         dividing both sides by 4.

Check:

        5*3-7=5+ 3,     substituting 3 for x in the given equation;

           15-7=5+ 3,       simplifying terms;

               8 = 8,       combining terms.

Algebraic Representations

To solve algebraic problems which are stated in words the chief difficulty that one has, in the beginning, is to express the conditions of the problem by means of symbols. A statement proposed in general numbers will frequently be found puzzling, while a similar arithmetical statement would present no difficulty whatsoever.

 Thus, the answer to the statement find a number which is  x more than a, may not be self-evident, whereas the answer to a similar arithmetical statement find a number which is 3 more than 5 is determined with no difficulty at all. Just as the number which is 3 more than 5 is 5 + 3 = 8, so the number which is x more than a is a + x.

 In general, when in doubt as to the exact meaning of any statement involving letters, choose suitable numbers for the letters and investigate the meaning of the statement with respect to these numbers.

Example 1 By how much does x exceed 12?

Solution: Take a numerical instance: “By how much does 15 exceed 12?” Clearly the answer is 3, which is obtained by subtracting 12 from 15.

In the same manner then, x exceeds 12 by x-12.

Example 2 If x is an integer, what is the next larger integer?

Note The whole numbers 1, 2, 3, 4, etc., are called integers.

Solution: Numerical instance: “If 17 is an integer, what is the next larger integer?” The answer is evidently 18, which is obtained by adding 1 to 17.

Likewise, the next larger integer following x is x + 1.

In the following examples, the choice of numerical instances is left to the student.

Example 3 What number is 10% more than h?

Solution: Here we must increase h by 10% of h. Since the numerical value of 10% is .1, the increase is to be .1 of h or .1h. Hence, the number is h + .1h or 1.1h. Always simplify representations in this manner when it is possible to do so.

Note In algebra, percentages should always be reduced to their equivalent numerical values. Thus 7% = .07 or 7/100, r% = .01r or r/100.

Example 4 Jack is twice as old as Bill. If Bill is y years old, how old was Jack three years ago?

Solution: Since Jack is now twice as old as Bill, his present age is 2y years. Three years ago, he was 3 years younger or 2y-3 years old.

Example 5 How many cents are there in q quarters and d dimes?

Solution: Since each quarter contains 25 cents, q quarters will contain q times as many cents or 25q cents. Also since each dime contains 10 cents, d dimes will contain d times as many cents or 10d cents. Taken together, there will be a total of 25q + 10d cents.

Formulas Made from Statements

Fundamentally a formula expresses a certain type of relationship which exists between two or more varying quantities. To make a formula, we need to know something of this relationship. This we may ascertain either from the nature of the quantities themselves or from an explicit statement regarding their relationship. The following examples illustrate these methods of making formulas:

Example 1 Obtain a formula for the total area (A) of a cube.

Solution: Each face of a cube is a square. If we let the letter e represent the length of each edge of the cube, then the area of each face of the cube is represented by e^2. Since, however, there are six faces to a cube, the total area is six times the area of one face; or

                A = 6e^2.

Example 2 What is the cost (C) in dollars of 50 books at x dimes each?

Solution: Since one dime is valued at 1/10 of a dollar, x dimes are valued at x times as much or x/10 dollars. Thus, the cost of one book is x/10 dollars. The cost of 50 books is fifty times the cost of one book, hence,

                C= 50*x/10 =5x,

where C is understood to be expressed in dollars.

Example 3 What is the area (A) in square feet of a rectangle whose length is x yards and whose width is y feet?

Solution: Since 1 yard contains 3 feet, x yards will contain x times as many feet, or 3x feet. The area in square feet is the product of the length and width, both expressed in feet. Hence,

                A = 3xy.

Example 4 Make a formula for the total cost (C) in cents of a telegram, when the first ten words of the telegram cost 35 cents and each additional word costs 5 cents. (Assume the telegram to contain more than 10 words.)

Solution: Let the letter n represent the number of words in the telegram. The first ten of these words as a group cost 35 cents and the remaining n-10 of the words each costs 5 cents. Hence, the cost of the remaining words is 5(n-10) and the total cost is given:

                C=35+5(n-10)

Stated Problems

A written or verbal statement expressing some condition or conditions of equality which exist among one or more unknown quantities is called a staled problem.

The method of procedure in solving such problems is as follows:

(1) Represent some one of the unknown quantities by a symbol — such as x.

(2) Give the algebraic representations of the other unknown quantities in terms of the unknown x.

(3) From a condition of the problem, obtain a simple equation involving the unknown quantities.

(4) Solve the equation obtained in (3) for the unknown x. Having x, the other unknowns may be determined by substitution in (2).

Note Unknown quantities are usually represented by the last letters of the alphabet.

Example 1 The larger of two numbers is 7 more than the smaller, and their sum is 41. Find the numbers.

Solution: Let x = the smaller number,

then x+ 7 = the larger number.

Since the sum of these two unknown numbers must equal 41, we must have x+x+7 = 41.

Hence, 2x + 7 = 41, combining terms;

2x = 34, subtracting 7 from both sides;

x = 17, dividing both sides by 2.

Therefore, x+ 7 = 24, substituting 17 for x in x + 7.

Answer: The numbers are 17 and 24.

Check: 24 is 7 more than 17; and the sum of 17 and 24 does equal 41.

Example 2 The length of a rectangle is 3 feet less than twice its width, and the perimeter of the rectangle is 36 feet. Find the dimensions of the rectangle.

Solution: Let x = the width in feet,

then 2x-3 = the length in feet.

              

Since the perimeter is the total distance around a closed figure, we must have

          x+2x-3+x+2x-3=36

Hence, 6x-6 = 36, combining terms;

6x = 42, adding 6 to both sides;

x= 7, dividing both sides by 6.

Therefore, 2x -3 = 11, substituting 7 for x in 2x-3.

Answer: The dimensions of the rectangle are 7 feet by 11 feet.

Check: 3 less than twice 7 does equal 11; and the perimeter, 7 +11+ 7 + 11, does equal 36.

Note The student will find that geometrical problems are more easily understood when a figure is drawn.