ALGEBRAIC OPERATIONS AND EQUATIONS
The previous articles have been concerned with the arithmetic operations involving addition, subtraction, multiplication, and division. These operations have been applied to whole numbers and fractions. Algebra involves the extension of these principles to symbols that are used to represent numbers. The symbols are used to write arithmetic statements for formulas into which many sets of specific numerical values can be substituted. Therefore, algebra may be called generalized arithmetic. For example, the area of a rectangle equals the length times the width. This is represented by the algebraic expression A = lw.
Area = length x width
The above statement for the area of a rectangle is always true. To compute a numerical value for the area of a specific rectangle, the numerical values of the length, l, and the width, w, are substituted and the indicated multiplication carried out. Letters are used to represent numbers. Since these letters represent numbers, they are subject to the same rules developed for whole numbers and fractions.
Before taking up the study of algebra, it is necessary to become familiar with the concept of signed numbers. The numbers that are used to describe the number of objects in a group, and the counting numbers, are always positive numbers. That is, they are always greater than zero. However, there are many occasions when negative numbers, numbers less than zero, must be used. These numbers arise when describing measurement in a direction opposite to the positive numbers. For example, if assigning a value of +3 to a point which is 3 feet above the ground, what numbers should be assigned to a point which is 3 feet below the ground? Perhaps the most familiar example of the use of negative numbers is the measurement of temperature, where temperatures below an arbitrary reference level are assigned negative values.
Every number has a sign associated with it. The plus (+) sign indicates a positive number; the minus (-) sign indicates a negative number. When no sign is given, a plus sign is implied. The fact that the plus and minus signs are also used for the arithmetic operations of addition and subtraction should not be the cause for confusion since their meanings are equivalent.
Every number has an absolute value, regardless of its sign. The absolute value for a number is indicated by a pair of vertical lines enclosing the number. The absolute value indicates the distance from zero, without regard to direction. The number +5 is 5 units from zero, in the positive direction. The number -5 is also 5 units from zero, in the negative direction. The absolute value of each of these numbers is 5.
The absolute value of -5, written as |-5|, = 5.
|-5| = 5
The absolute value of +5, written as |+5|, is also 5.
|+5| = 5
|-5| = |+5| = 5
Performing the operations of addition, subtraction, multiplication, and division are more easily visualized if the numbers are placed on a number line. The positive numbers are greater than zero, and lie to the right of zero on the number line. The negative numbers are less than zero, and lie to the left of the zero on the number line.
-4 -1 -2 -10 +1 +2 +3 +4
The number line extends an infinite distance in each direction and therefore includes all numbers. The process of addition can be considered as counting to the right on the number line. For example when adding 1+2 the location of +1 must be found on the number line. In this example 2 units must be counted to the right, for adding. The result is +3. To illustrate further, when adding -2 and +4 the location of -2 on the number line must first be found. Counting 4 units to the right ends up at +2.
The number line is useful for illustrating the principles of addition, but it clearly would be inconvenient to use in the case of large numbers. Consequently, the following rules were developed to govern the addition process:
1. Add -37 and -16.
Since these numbers have like signs, add their absolute values and prefix the common sign.
|-37| = 37 37
|-16| = 16 +16
2. Add -37 and +16.
Since these numbers have unlike signs, take the difference between their absolute values and prefix the sign of the number having the greater absolute value.
|-37| = 37 37
|+16| = 16 -16
3. Add +32, -16, -19, -12 and +14.
First, combine the positive and negative numbers separately.
Then combine the results: (+46) + (-47) = -1
The subtraction of signed numbers is governed by only one rule. To subtract two signed numbers, change the sign of the subtrahend and then add the two numbers as shown below.
Subtract 7 from 11. (11) - (7) = (11) + (-7) = 4
Subtract -7 from 11. (11) - (-7) = (11) + (+7) = 18
Subtract -7 from -11. (-11) - (-7) = (-11) + (+7) = -4
Subtract 7 from -11. (-11) - (7) = (-11) + (-7) = -18
In each of these examples, the subtrahend was changed, then the rules for addition were followed.
(+) x (+) = (+) (+) x (-) = (-)
(-) x (-) = (+) (-) x (+) = (-)
(+) ÷ (+) = (+) (+) ÷ (-) = (-)
(-) ÷ (-) = (+) (-) ÷ (+) = (-)
Remember that multiplication is really a short form of addition. When multiplying +4 by -3, the number -3 is added four times. That is, (-3) + (-3) + (-3) + (-3) = -12. Also, since division is a short form of subtraction, the number -6 is subtracted from -24 four times in order of reach 0, i.e., 24 - (-6) - (-6) - (-6) - (-6) = 0. Although the process could be repeated for the multiplication and division of any signed numbers, usage of the two rules will produce equivalent results.
An algebraic term is "a combination of numbers and letters (literal numbers) linked by multiplication or division." There is no limit to the number of quantities in the term. The following are all algebraic terms:
1. 3 6. (2xz)/ y
2. 3x 7. 3abcxy2
3. 4ab 8. (5bcx2yz)/ 3a
4. (7x2)\\ y 9. 19/ abcxy
5. 11xyz 10. X2
An algebraic expression is "a sum or difference of algebraic terms." There is no limit as to the number of terms in the expression.
The following are all algebraic expressions:
1. 3 + 4ab
2. 11xyz + (2xz)/y
3. 7x2/y + 19/abcxy + y2 + (2xy)/ z
The factors of a term are each of the numbers or letters that, when combined (multiplied or dividend), produce the term. For example:
Notice that numbers are often referred to as factors, as was done in example three above. Numbers may have many factors.
Prime numbers are those which have only themselves and 1 as factors.
Examples: 3, 7, 11, 13, and 17 are prime numbers. Their only factors are themselves and 1.
When considering the factors of numbers, only the factors which are whole numbers are considered. In the example above, the numbers 3.5 and 2 are factors of 7 in the sense that 3.5 x 2 = 7, but of course 3.5 is not a whole number.
The word coefficient is used for the numerical factors in an algebraic term. In the term 5xy, the number 5 is the coefficient of xy. The coefficient need not be a whole number.
As we said before, algebraic expressions may have any number of terms. These expressions have special names, which indicate the number of terms contained in them. A monomial is an algebraic expression consisting of only one term. A polynomial is an algebraic expression, which contains more than one term. Special names for polynomials are binomials, consisting of two terms, and trinomials, consisting of three terms.
Suppose there are two terms, 2 and 3, and their sum is desired: 2 + 3. The sum, of course, is represented by the single symbol 5. This really means that 2 units plus 3 units is equal to 5 units. If literal numbers are used instead of numerical ones, they can be added or subtracted provided that they have the same units. The literal number x added to the literal number 2x will result in the number 3x. Algebraic terms having exactly the same letter parts are called like terms, and may be added or subtracted by adding or subtracting the numerical coefficients. For example:
1. Add 3x2, -2x2, and 7x2.
(3x2), (-2x2), and (7x2) = 8x2
2. Add (3x2 - 2xy + 5y2) and (x2 + y2).
3x2 - 2xy + 5y2
+ x2 + y2
4x2 - 2xy = 6y2
Notice that the like terms are added or subtracted directly. Unlike terms cannot be combined even when they contain several of the same letters. The terms must be exactly the same. For example:
In the second example above, the terms are very similar but one has a factor of x3 and the other a factor of x2. This prevents them from being combined.
Whenever algebraic expressions are written out horizontally, there will always be confusion regarding just what operations are to be performed. Consider, for example, the expression 3 + x ÷ 2 + 5. This can have many different interpretations. It could mean add 3 and x, and divide the sum by 7, or divide x by 7 and add it to 3, or add 3 and x, divide by 2, and add 5, etc. Parentheses or brackets are used to group the quantities in the exact order in which the arithmetic operations are to be performed. In the example above, the expression to mean add 3 and x, and divide the sum by 7, is written:
(3 + x) ÷ (2 + 5)
This indicates that the entire quantity (3 + x) is to be divided by the entire quantity (2 + 5). Other examples are:
The following rules govern the use of parentheses:
When the parentheses are used in algebraic expressions, they are used in closed pairs. Such expressions are evaluated by working from the innermost to the outermost closed pair, as shown in the following examples:
1. Evaluate (3xy) + (3x2 - 5xy) - (4xy + 3z).
(3xy) + (3x2 - 5xy) - (4xy + 3z)
Rule 1 Rule 2
= 3xy + 3x2 - 5xy - 4xy -3z
= 3x2 - 3z - 6xy = 3(x2 - z - 2xy)
2. Evaluate 3x - 2(5y -(4x +2))
3x - 2(5y - (4y +2))
= 3x -2(5y - 4x -2)
= 3x -10y + 8x =4
= 11x - 10y + 4
Note the points in the examples where the rules have been applied.
The last section covered algebraic terms and how they can be combined by addition or subtraction only when they are exactly alike, except for their numerical coefficients. However, unlike terms can be multiplied or divided directly as shown below.
To multiply or divide terms that include the same literal number (letter), the rules which apply to this type of operation must be introduced. This subject is treated in much greater detail elsewhere, but for now it is sufficient to state The following rule:
For factors which are the same base raised to a power (exponent), the factors are multiplied by adding the exponents and are divided by subtracting the exponents. The following example demonstrates this rule:
1. Multiply a3 by a4. a3 = a . a . a a4 = a . a . a . a a3 x a4 = (a . a . a)(a . a . a . a)
= (a . a . a . a . a . a . a)
In applying this rule to a number that does not have an exponent associated with it, that number is treated as having an exponent of +1. For example:
1. Multiply x2 and x3.
(x2)(x3) = x2+3 = x5
2. Divide x5 by x2.
(x5) ÷ (x2) = x5-2 = x3
3. Multiply (x2yz) and (x5y2).
(x2yz)(x5y2) = x2+5y1+2z = x7y3z
Multiplication of monomials follows the following three steps:
Step 1. Determine the signs of the product by the rule for the multiplication of signed numbers.
Step 2. Multiply the numerical coefficients of the factors to get the numerical coefficient of the product.
Step 3. Multiply the literal parts by writing down all literal factors and adding the exponents of like literal factors.
1. (3abx2) (-4bcx3)
-(3) (4) ab1+1cx2+3
2. (7x2yz3) (4xy2z3)
= (7) (4) x2+1y1+2z3+3
The division of monomials is done in exactly the same way, except the rules for division rather than multiplication is used.
Step 1. Determine the sign of the quotient by the rule for the division of signs.
Step 2. Divide the numerical coefficients to get the numerical coefficient of the quotient.
Step 3. Divide the literal parts by writing down all the literal factors of the dividend and subtracting the exponents of like factors in the divisor.
1. Divide 4x2yz by 2xab.
((4x2yz)/ (2xab)) = 4/2 (((x2 - 1)(y1)(z1))/ (ab)) = 2 ((xyz)/ (ab))
2. Divide (-36a2bx3y4z) by 4abxy4.
The multiplication and division of monomials involves a straight-forward multiplication and division of terms. The multiplication of polynomials is only slightly more complicated. To multiply polynomials, multiply each term of one polynomial by each term of the other, and then combine any like terms. As an example, multiply the two numbers 6 and 8. The answer is 48. If instead, the number 6 is represented as (4 + 2) and the number 8 as (5 + 3), the product can also be written as (4 + 2)(5 + 3), a product of two polynomials. Each term of one polynomial is multiplied by each term of the other and combined.
(4 + 2)(5 + 3) = (4 x 5) + (4 x 3) + (2 x 5) + (2 x 3)
= 20 + 12 + 10 + 6
The multiplication process is done in exactly the same way for algebraic polynomials.
Example: Multiply (2x2 + x - 3) and (6x2 - 2x - 5).
(2x2 + x - 3)(6x2 - 2x - 5)
= (2x2)(6x2) - (2x2)(2x) - (2x2)(5) + x(6x2) - (x)(2x) - (x)(5)
- 3(6x2) + 3(2x) + (3)(5)
= 12x4 - 4x3 - 10x2 + 6x3 - 2x2 - 5x - 18x2 + 6x + 15
Combining like terms:
= 12x4 + 2x3 - 30x2 + x + 15
Factoring is the reverse of multiplication. For example, the numbers 2 and 3 are factors of the number 6, since 2 x 3 = 6. In the same way, the factors of the polynomial are two (or more) other polynomials which when multiplied yield the original polynomial. In the example at the end of the last section, the two factors of the polynomial (12x4 + 2x3 - 30x2 + x + 15) are (2x2 + x - 3) and (6x2 - 2x - 5). Therefore, the process of factoring involves finding those factors as shown below.
Certain algebraic expressions can be factored by extracting a factor which is common to each term. The following rules apply:
Factor: (9x3 + 6x2 + 3x)
The greatest common factor of each term is 3x. Dividing 3x into each term yields 3x2 + 2x + 1. Therefore, the two factors are:
(3x) (3x2 + 2x + 1)
The factoring process can be checked simply by multiplying the two factors together and obtaining the original polynomial.
Certain types of binomials can be factored immediately. If the binomial is composed of the difference of two perfect squares, the two factors will be the sum and difference of the square roots of each term in the binomial. This is shown below by factoring
(81x4 - 4)
(81x4 - 4) is the difference between two squares. The square root of 81x4 is 9x2; the square root of 4 is 2. Therefore, the two factors are:
(9x2 + 2) and (9x2 - 2)
Therefore, (81x4 - 4) = (9x2 + 2)(9x2 - 2)
Note that this rule does not apply to a binomial, which is the sum of two squares, only the difference of two squares.
The most complicated polynomial usually dealt with is the trinomial, in particular, trinomials which are in the form ax2 + bx + c. The a, b, and c are positive or negative numbers. There is a straight forward procedure which will yield the factors. First, multiply a and c, to obtain the product ac. Then list all the factors of ac.
Example: Factor (4x2 - 10x -6)
Here, a = 4, c = -6.
The product of a and c is (4)(-6) = -24
The factors of -24 are:
The pair of factors whose algebraic sum is equal to b is then found. For example, in the [removed]4x2 - 10x -6), b = -10. The factors of -24 whose algebraic sum is -10 are -12, 2 since -12 + 2 = -10.
Example: The factors of (4x2 - 10x - 6) are:
The two factors obtained can be checked by multiplication.
For another example: Find the two factors of (3x2 + 4x - 15).
1. Find factors of (3) (-15) = -45
2. Find the two factors whose algebraic sum is +4. They are 9, -5, since 9 + (-5) = 4.
3. The factors are:
Every trinomial can be factored in this matter. There may be cases, however, where the numbers may not turn out to be whole numbers, but the factoring process will still be valid.
All operations that were developed with regard to numerical fractions apply directly to algebraic fractions. If the numerator and denominator of any fraction is multiplied or divided by the same quantity, the value of the fraction is unchanged. If the fractions are to be added or subtracted, they must have the same denominators. Finally, the fractions should be reduced to their simplest form. These rules apply to algebraic fractions because the letters simply represent numbers, and so they must obey the same rules.
Algebraic fractions can frequently be reduced by factoring both the numerator and denominator, and then dividing out any common factors. For example:
Reduce the fraction:
The numerator of the fraction is the difference between two squares, and so can be factored as (x + 3) (x -3).
The denominator can be factored according to the method developed for factoring trinomials. Its two factors are: (3x + 1)(x -3).
The fraction is now:
Since the factor (x -3) is common to both numerator and denominator, it can be divided out, leaving:
(x + 3)
(3x + 1)
Fractions are added or subtracted by finding the lowest common denominator and then combining the numerators. To demonstrate this add the fractions:
The lowest common denominator is (x + 3)(x -3). Rewriting each fraction with the LCD as its denominator:
The procedure is just the same as for numerical fractions.
When dealing with algebraic expressions and numerical expressions, use has been made of parentheses to group terms so that the operations of addition, subtraction, multiplication, and division are done in the proper order. When expressions are combined, the parentheses are removed, and care must be taken to properly account for the algebraic signs.
In an algebraic expression, parentheses which are preceded by a minus (-) sign may be removed by changing the sign of each term in the parentheses. For example, -(2x2 + 3x - 7) becomes -2x2 - 3x + 7. The expression is equivalent to + (-1)(2x2 + 3x -7).
Parentheses which are preceded by a plus (+) sign may be removed without changing the sign of any term in the parentheses. + (2x2 + 3x - 7) would be equal to 2x2 + 3x - 7.
If the parentheses are preceded by some quantity other than -1 or +1, this indicates that the multiplication is to be performed if the parentheses are to be removed. This is really just an application of the rule for multiplying polynomials, where each term of one polynomial multiplies every term of the other. For example:
-3(2x2 + 3x -7)
= -(6x2 + 9x - 21)
= -6x2 - 9x + 21
In connection with any fraction, there are always three signs associated with it:
If any two of these three signs are changed, the value of the fraction will not be changed. The following example is used to demonstrate this.
Use parentheses to show explicitly the three signs:
All of these fractions have the same value. What is really being done is multiplying the numerator and denominator by the same quantity, in this case -1. This operation does not change the value of a fraction.
An equation is "a mathematical statement which says that two quantities are equal." The equality is expressed by the equal (=) sign between the two quantities. Thus, the statement 3 = 3 is an equation. It says that the quantity to the left of the (=) sign is equal to the quantity to the right of the (=) sign. Similarly, the statement x = 3 is an equation. It says that the variable represented by the letter x has the value of 3.
There are two types of equations, identities and conditional equations. An identity is an equation which is always true no matter what numerical value is given to the letter which represents a variable, or unknown.
3x + 5x = 8x is an identity since it is true for all values of x.
The equations which are dealt with most of the time are conditional equations. These are equations which are true only for some particular value or values of the unknown.
3x + 5 = 8 is a conditional equation since it is true only for the value of x = 1.
One of the fundamental purposes of algebra is to determine the value of the unknown which makes an equation true; that is, which "satisfies" the equation. In the example above, it can be determined that the value x = 1 satisfies the equation by replacing the letter x by the number 1 and performing the indicated arithmetical operations. Algebra provides the mechanism whereby the equation can be solved; that is, its root may be determined.
The equality sign (=) that separates two equal quantities allows certain operations to be performed in our efforts to solve an equation. For any equation, the following statements are always true:
The four axioms for solving algebraic equations may be summarized by one general principle. Whatever operation is performed on one side of an equation, the same operation must be performed on the other side of the equation if the equation is to remain true. For example:
1. Solve the equation 4x + 3 = 19.
Step 1. Using Axiom 2, subtract 3 from both sides of the equation.
4x + 3 - 3 = 19 - 3
4x = 16
Step 2. Using Axiom 4, divide both sides of the equation by 4.
(4/4)X - 8 = 2
2. Solve the equation.
(1/4)X- 8 = 2
Step 1. Using Axiom 1, add 8 to both sides of the equation.
¼X - 8 + 8 = 2 + 8
¼X = 10
Step 2. Using Axiom 3, multiply both sides of the equation by 4.
(4)¼X = 10(4)
X = 40
In each of these examples, the axioms are used to isolate the unknown on one side of the equation.
There is a shorter method available for applying the addition and subtraction Axioms 1 and 2 to algebraic equations. Any term may be transposed or transferred from one side of an equation to the other provided its sign is changed. For example:
In the equation 5x + 4 = 7, the 4 can be transposed to the other side of the equation by changing the sign.
5x + 4 = 7
5x = 7 - 4
5x = 3
This corresponds to applying Axiom 2, subtracting 4 from both sides of the equation.
All equations studied so far contained only whole numbers. In addition to whole numbers, an equation may also contain fractions, either common fractions or decimal fractions. A fractional equation is an equation that contains fractions. The unknown may be located anywhere in the equation.
is a fractional equation.
Fractional equations are solved exactly as any other equation, using the four axioms. Generally, the equation is first cleared of the fractions. As shown below, operations are performed that remove all of the fractions.
Solve the following fractional equation:
Step 1: Multiply each term by 8x, which is the lowest common denominator.
5x – 72 = 4x - 6
Step 2: Transpose (- 72) and (4x).
5x – 4x = -6 + 72
x = 66
As the final step in the solution of any equation, the root should be substituted back into the equation to ensure that it makes the equation true. If the root does not check, then an error has been made during the solution.
Check to see if x = 66 is a root of the equation.
The root checks.
The concept of ratio and proportion are one which are used almost every day. If the quantities listed in a recipe will produce servings for eight, it is instinctively known that to make servings for four, each quantity should be cut in half. What has been done is to form a ratio of the number of servings. A ratio is a comparison of two like quantities by division. It is often indicated by a colon (:), although it is really a fraction.
The ratio of $25 to $5 is $25 ÷ $5 which equals 5. This is often indicated as:
$25 : $5 and is read $25 is to $5. It can also be written as a fraction:
A proportion is a statement of equality between two equal ratios.
or: $5 : $25 = 2 lb : 10 lb
Notice that the ratios are comparisons of different amounts of the like quantities, dollars and pounds, respectively. The proportion simply states that the ratio of $5 to $25 is equal to the ratio between 2 pounds and 10 pounds. When the proportion is written out in the colon form, there is a relationship between the various terms. The first and fourth terms in a proportion are called the extremes. The second and third terms are called the means. For example:
In the proportional $5 : $25 = 2 lb : 10 lb, $5 and 10 lb are the extremes, $25 and 2 lb are the means.
In any proportion, the product of the means equals the product of the extremes.
$5 : $25 = 2 lb : 10 lb
The product of the means is [$25] [2 lb] = 50 dollar-pounds.
The product of the extremes is [$5] [10 lb] = 50 dollar-pounds.
This can be seen clearly if the proportion is written in fractional form.
1. Multiply the equation by $25.
2. Multiply the equation by 10 lb.
$5 x 10 lb = 2 lb x $25
50 $-lb = 50 $-lb
This principle permits finding any missing term in a proportion. For example:
3. Find the missing term in the proportion 5 : x = 4 : 15.
The product of the extreme is:   = 75
The product of the means is:  [x] = 4x
4x = 75
This could also be solved in fractional form.
4. Find the missing term in the proportion 5 : x = 4 : 15.
The concepts of ratio and proportion are useful in solving problems such as the example below.
If 5 pounds of apples cost 80 cents, how much will 7 pounds cost?Using x for the cost of 7 pounds of apples, the following proportion can be written.
The product of the extremes is (5)(x) = 5x.
The product of the means is (7)(80) = 560.
Equate these two products and solve the resulting equation.
5x = 560
x = 112
The unit of x is cents. Thus, 7 pounds of apples cost 112 cents or $1.12.
Previous sections have been concerned with the development of the mathematical tools required for equation solving. The problems encountered in everyday life, however, are rarely stated in equation form. These problems are stated in words, and they must be translated into the appropriate mathematical equations in order to find the solution. This is a two-step process:
Methods for accomplishing Step 2 have already been studied. In this section, the methods for accomplishing Step 1 will be presented.
Before attempting to solve any word problem, the problem must be understood completely. It is frequently beneficial to draw a picture of the physical situation described by the problem. The drawing should be labeled with the known and unknown quantities. At the very least, these quantities should be listed in some logical order.
After ensuring that the problem is understood, it can be solved by following these five fundamental steps:
Step 1. Let some letter, such as x, represent one of the unknowns. There will always be a choice of which unknown can be called x. There is no one right choice. For example, suppose that it is given that an orange costs 3 cents more than an apple. If x equals the cost of the orange, the cost of the apple will be x - 3. On the other hand, if x equals the cost of the apple, then the cost of the orange is x + 3. Either choice is correct.
Step 2. Express the other unknowns, using the information given, in terms of x. In doing this, it is helpful to look for certain words that indicate algebraic operations. The words sum and total signify addition; the words difference and less than signify subtraction; the words times and multiples of signify multiplication; the words divided by and per signify division. The words same as and equal to signify equality.
Step 3. Write an equation. Make the equation say in symbols exactly what the problem says in words. This involves reading the problem carefully to understand exactly what is being asked.
Step 4. Solve the equation using the methods discussed in previous sections.
Step 5. Check the solution by substituting it into the equations. For example:
A family took a trip and traveled a total of 820 miles in three days. They drove twice as many miles the second day as on the first. The third day they drove 60 miles less than they did on the second day. Find the distance traveled each day.
Before solving this word problem, exactly what must be known is required to be found, that is, what are the unknowns? For this example, draw a diagram of the problem and write in the knowns and unknowns.
Now proceed through the following five steps to solve the problem:
Step 1. Let x = the number of miles driven on the first day. x could = the number of miles driven on the second or third day, but for the moment x = the number of miles driven the first day.
Step 2. It is given that they drove twice as many miles on the second day as the first. Therefore, if x is the number of miles driven on the first day, then 2x is the number driven on the second day. It is also given that on the third day, they drove 60 miles less than on the second day. If they drove 2x miles on the second day, they drove 2x - 60 miles on the third day. All of the unknowns have been expressed in terms of x.
Step 3. Write an equation, which relates the unknowns. Total miles driven = miles driven on the first day + miles driven on second day - miles driven on third day. The total miles driven is 820. Therefore, x + (2x) + (2x - 60) = 820
Step 4. Solve the equation. x + 2x + 2x – 60 = 820
5x = 820 + 60
5x = 880
x = 176 miles
This is the number of miles driven on the first day.
First day: x = 176 miles
Second day: 2x = 352 miles
Third day: 2x – 60 = 292 miles
Step 5. Check the answers. x + 2x + 2x – 60 = 820
176 + 352 + 292 = 820
820 = 820
In solving this problem, let x equal the number of miles driven on the first day. There is nothing magical about this choice. Just to illustrate the point, solve the problem in a slightly different manner. Let y be the number of miles driven on the second day. Then will be the miles driven on the first day, and y - 60 will be the miles driven on the third day. The equation will now be:
Solve this equation and see that the answer is the same as before. Try another example:
A man is 2 years older than three times his son’s age. Ten years from now he will only be twice as old as his son. What are their ages now?
Step 1. Let x = the son’s age now.
Step 2. Then, 3x + 2 = the father’s age now.
Step 3. The father’s age ten years from now will be two times the son’s age at that time.
(3x + 2) + 10 = 2(x + 10)
Step 4. (3x + 2) + 10 = 2(x + 10)
3x + 12 = 2x + 20 x = 8
Son’s age now: x = 8
Father’s age now: 3x + 2 - 3(8) + 2 = 26
Algebraic problems involving money can become confusing because there is a tendency to use quantity and value interchangeably. For example, a quantity of 5 dimes has a value of 50 cents, whereas a quantity of 5 nickels has a value of 25 cents. The quantities are the same, but the values are different. The general relationship used to solve algebraic word problems involving money is:
Total Value = Sum of [(Quantity) times (Value per Quantity)]
Example: The total value of five pennies, two nickels, three dimes, and four quarters is:
[(5) x $0.01)] + [(2) x ($0.05)] + [(3) x ($0.10)] + [(4) x ($0.25)] = Total Value
$0.05 + $0.10 + $0.30 + $1.00 = $1.45
“How many dimes?” is a different question than “how much in dimes?”
A total of 6,000 tickets were sold for a basketball game. Tickets were priced at $2.00 and $2.80 each. A total of $14,264 was collected. How many of each price ticket were sold?
Step 1. Let x = the number of $2.00 tickets sold.
Step 2. Then, 6,000 - x = the number of $2.80 tickets sold.
Step 3. Total Collected = (Number of $2.00 Tickets Sold) x ($2.00) + (Number of $2.80 Tickets Sold) x ($2.80)
$14,264 = (x)($2.00) + (6,000 - x)($2.80)
Step 4. $14,264 = 2x + 16,800 - 2.8x
2.8x - 2x = $16,800 - $14,264
0.8x = 2,536
x = 3,170
Number of $2.00 Tickets Sold: x = 3,170
Number of $2.80 Tickets Sold: 6,000 - x = 2,830
Step 5. Check:
(3,170)($2.00) + (2,830)($2.80) = $14,264
$6,340 + $ 7,924 = $14,264
$14,264 = $14,264
Of course, the problem could also have been solved by letting x equal the number (quantity) of $2.80 tickets sold. Note that the quantity of tickets sold is different than the value of the tickets.
There is a large variety of problems that involve travel and travel times. In order to solve these problems, a relationship between distance, speed and time is needed. This relationship is:
Distance = (Speed) x (Time)
D = vt
If a car is moving at a uniform speed of 60 miles per hour for two hours, the distance traveled is 120 miles. d = vt d = (60 mph)(2 hr) d = 120 miles
Algebraic word problems involving uniform motion are solved using this general relationship and following the steps for solving any algebraic word problem. This is shown in the example below.
A man takes a trip of 675 miles, part of the trip by train at 60 mph and the rest of the trip by car at 50 mph. If the entire trip takes 12 hours, how far has he traveled by each mode of transportation?
As always, before attempting to solve the problem, make sure what is being asked is understood.
Step 1. Let x = the number of miles traveled by train.
Step 2. 675 miles total are traveled with x miles traveled by train, and the remainder traveled by car. Thus, the number of miles traveled by car is (675 - x).
The number of hours traveled by train equals the distance traveled divided by the speed. Thus, the number of hours traveled by train equals
The total number of hours traveled is 12 with hours traveled by train and the remainder traveled by car. Thus, the total number of hours traveled by car is:
and the distance traveled by car is:
Step 3. Distance traveled by train + Distance traveled by car = 675.
Step 4. Distance traveled by train: x = 450 miles
Distance traveled by car: 675 - x = 675 - 450 = 225 miles.
Step 5. Check
450 + 225 = 675 This corresponds to the total distance traveled.
675 = 675
This example may seem complicated, but if the work is arranged in a logical, step-by-step manner, the solution process becomes straightforward.