1 Copyright © Cengage Learning. All rights reserved. 2. Equations and Inequalities 2.1 Equations

2 Equations An equation (or equality) is a statement that two quantities or expressions are equal. Equations are employed in every field that uses real numbers. As an illustration, the equation d = rt, or distance = (rate)(time), is used in solving problems involving an object moving at a constant rate of speed. If the rate r is 45 mi  hr (miles per hour), then the distance d (in miles) traveled after time t (in hours) is given by d = 45t.

3 Equations The following chart applies to a variable x, but any other variable may be considered. The abbreviations LS and RS in the second illustration stand for the equation’s left side and right side, respectively.

4 Equations An algebraic equation in x contains only algebraic expressions such as polynomials, rational expressions, radicals, and so on. An equation of this type is called a conditional equation if there are numbers in the domains of the expressions that are not solutions. For example, the equation x 2 = 9 is conditional, since the number x = 4 (and others) is not a solution. If every number in the domains of the expressions in an algebraic equation is a solution, the equation is called an identity.

5 Equations The most basic equation in algebra is the linear equation, defined in the next chart, where a and b denote real numbers. The illustration in the preceding chart indicates a typical method of solving a linear equation.

6 Equations Following the same procedure, we see that if ax + b = 0, then x =, provided a ≠ 0. Thus, a linear equation has exactly one solution.

7 Example 1 – Solving a linear equation Solve the equation 6x – 7 = 2x + 5. Solution: The equations in the following list are equivalent: 6x – 7 = 2x + 5 (6x – 7) + 7 = (2x + 5) + 7 6x = 2x + 12 add 7 given simplify

8 Example 1 – Solution 6x – 2x = (2x + 12) – 2x 4x = 12 x = 3 Check: x = 3 LS: 6(3) – 7 = 18 – 7 = 11 RS: 2(3) + 5 = = 11 Since 11 = 11 is a true statement, x = 3 checks as a solution. cont’d subtract 2x simplify divide by 4 simplify

9 Equations The next example illustrates that a seemingly complicated equation may simplify to a linear equation.

10 Example 2 – Solving an equation Solve the equation (8x – 2)(3x + 4) = (4x + 3)(6x – 1). Solution: The equations in the following list are equivalent: (8x – 2)(3x + 4) = (4x + 3)(6x – 1) 24x x – 8 = 24x x – 3 26x – 8 = 14x – 3 multiply factors given subtract 24x 2

11 Example 2 – Solution 12x – 8 = –3 12x = 5 x = Hence, the solution of the given equation is. cont’d subtract 14x add 8 divide by 12

12 Equations If an equation contains rational expressions, we often eliminate denominators by multiplying both sides by the lcd of these expressions. If we multiply both sides by an expression that equals zero for some value of x, then the resulting equation may not be equivalent to the original equation, as illustrated in the next example.

13 Example 3 – An equation with no solutions Solve the equation. Solution: (x – 2) = (1)(x – 2) + (x – 2) 3x = (x – 2) + 6 3x = x + 4 multiply by x – 2 given simplify

14 Example 3 – Solution 2x = 4 x = 2 Check: x = 2 LS: Since division by 0 is not permissible, x = 2 is not a solution. Hence, the given equation has no solutions. cont’d subtract x divide by 2

15 Equations In the process of solving an equation, we may obtain, as a possible solution, a number that is not a solution of the given equation. Such a number is called an extraneous solution or extraneous root of the given equation. In Example 3, x = 2 is an extraneous solution (root) of the given equation.

16 Equations The following guidelines may also be used to solve the equation in Example 3. In this case, observing guideline 2 would make it unnecessary to check the extraneous solution x = 2. Guidelines for Solving an Equation Containing Rational Expressions 1. Determine the lcd of the rational expressions. 2. Find the values of the variable that make the lcd zero. These are not solutions, because they yield at least one zero denominator when substituted into the given equation.

17 Equations 3. Multiply each term of the equation by the lcd and simplify, thereby eliminating all of the denominators. 4. Solve the equation obtained in guideline The solutions of the given equation are the solutions found in guideline 4, with the exclusion of the values found in guideline 2. We shall follow these guidelines in the next example.

18 Example 4 – An equation containing rational expressions Solve the equation Solution: Guideline 1: Rewriting the denominator 2x – 4 as 2(x – 2), we see that the lcd of the three rational expressions is 2(x – 2)(x + 3). Guideline 2: The values of x that make the lcd 2(x – 2)(x + 3) zero are 2 and –3, so these numbers cannot be solutions of the equation.

19 Example 4 – Solution Guideline 3: Multiplying each term of the equation by the lcd and simplifying gives us the following: 2(x – 2)(x + 3) – 2(x – 2)(x + 3) = 2(x – 2)(x + 3) 3(x + 3) – 10(x – 2) = 4(x + 3) 3x + 9 – 10x + 20 = 4x + 12 cont’d cancel like factors multiply factors

20 Example 4 – Solution Guideline 4: We solve the last equation obtained in guideline 3. 3x – 10x – 4x = 12 – 9 – 20 –11x = –17 x = cont’d combine like terms subtract 4x, 9, and 20 divide by –11

21 Example 4 – Solution Guideline 5: Since is not included among the values (2 and –3) that make the lcd zero (guideline 2), we see that x = is a solution of the given equation. We shall not check the solution x = by substitution, because the arithmetic involved is complicated. It is simpler to carefully check the algebraic manipulations used in each step. cont’d

22 Equations Formulas involving several variables occur in many applications of mathematics. Sometimes it is necessary to solve for a specific variable in terms of the remaining variables that appear in the formula, as the next example illustrates.

23 Example 5 – Relationship between temperature scales The Celsius and Fahrenheit temperature scales are shown on the thermometer in Figure 2. The relationship between the temperature readings C and F is given by C = (F – 32). Solve for F. Figure 2

24 Example 5 – Solution To solve for F we must obtain a formula that has F by itself on one side of the equals sign and does not have F on the other side. We may do this as follows: C = (F – 32) C = F – 32 C + 32 = F F = C + 32 given multiply by add 32 equivalent equation

25 Equations We can make a simple check of our result in Example 5 as follows. Start with C = (F – 32) and substitute 212 (an arbitrary choice) for F to obtain 100 for C. Now let C = 100 in F = C + 32 to get F = 212. Again, this check does not prove we are correct, but certainly lends credibility to our result.