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College Algebra Sixth Edition James Stewart Lothar Redlin Saleem Watson P Prerequisites P.8 Solving Basic Equations Overview Equations are the basic mathematical tool for solving real-world problems. In this section, we learn • How to solve equations. Equations An equation is a statement that two mathematical expressions are equal. • For example: 3+5=8 Variables Most equations that we study in algebra contain variables, which are symbols (usually letters) that stand for numbers. • In the equation 4x + 7 = 19, the letter x is the variable. • We think of x as the “unknown” in the equation. • Our goal is to find the value of x that makes the equation true. Solving the Equation The values of the unknown that make the equation true are called the solutions or roots of the equation. • The process of finding the solutions is called solving the equation. Equivalent Equations Two equations with exactly the same solutions are called equivalent equations. • To solve an equation, we try to find a simpler, equivalent equation in which the variable stands alone on one side of the equal sign. Properties of Equality We use the following properties to solve an equation. • A, B, and C stand for any algebraic expressions. • The symbol means “is equivalent to.” Properties of Equality These properties require that you perform the same operation on both sides of an equation when solving it. • Thus if we say “add 4” when solving an equation, that is just a short way of saying “add 4 to each side of the equation.” Solving Linear Equations Linear Equation The simplest type of equation is a linear equation, or first-degree equation. • It is an equation in which each term is either a constant or a nonzero multiple of the variable. Linear Equation—Definition A linear equation in one variable is an equation equivalent to one of the form ax + b = 0 where: • a and b are real numbers. • x is the variable. Linear Equations These examples illustrate the difference between linear and nonlinear equations. Linear Equations Nonlinear Equations 4x – 5 = 3 x2 + 2x = 8 Contains the square of the variable 2x = ½x – 7 Reason for Being Nonlinear x 6x 0 Contains the square root of the variable x – 6 = x/3 3/x – 2x = 1 Contains the reciprocal of the variable E.g. 1—Solving a Linear Equation Solve the equation 7x – 4 = 3x + 8 • We solve this by changing it to an equivalent equation with all terms that have the variable x on one side and all constant terms on the other. E.g. 1—Solving a Linear Equation 7x – 4 = 3x + 8 (Given equation) (7x – 4) + 4 = (3x + 8) + 4 (Add 4) 7x = 3x + 12 (Simplify) 7x – 3x = (3x + 12) – 3x (Subtract 3x) 4x = 12 (Simplify) ¼ . 4x = ¼ . 12 (Multiply by ¼) x=3 (Simplify) Checking Your Answer It is important to check your answer. We do so in many examples. • In these checks, LHS stands for “left-hand side” and RHS stands for “right-hand side” of the original equation. Checking Your Answer We check the answer of Example 1. x = 3: • LHS = 7(3) – 4 = 17 • RHS = 3(3) + 8 = 17 • LHS = RHS E.g. 2—Solving an Equation That Involves Fractions Solve the equation x 2 3 x 6 3 4 • The LCD of the denominators 6, 3, and 4 is 12. • So, we first multiply each side of the equation by 12 to clear denominators. E.g. 2—Solving an Equation Involving Fractions x 2 3 12 12 x 4 6 3 2x 8 9 x 8 7x 8 x 7 An Equation that Simplifies to a Linear Equation In the next example, we solve an equation that doesn’t look like a linear equation, but it simplifies to one when we multiply by the LCD. E.g. 3—An Equation Involving Fractional Expressions Solve the equation 1 1 x 3 2 x 1 x 2 x x 2 • The LCD of the fractional expressions is (x + 1)(x – 2) = x2 – x – 2. • So, as long as x ≠ –1 and x ≠ 2, we can multiply both sides by the LCD. E.g. 3—Equation Involving Fractional Expressions 1 1 x 3 ( x 1)( x 2) ( x 1)( x 2) 2 x 1 x 2 x x 2 ( x 2) ( x 1) x 3 2x 1 x 3 x4 Extraneous Solutions It is always important to check your answer. • Even if you never make a mistake in your calculations. • This is because you sometimes end up with extraneous solutions. Extraneous Solutions Extraneous solutions are potential solutions that do not satisfy the original equation. • The next example shows how this can happen. E.g. 4—An Equation with No Solution Solve the equation 5 x 1 2 x4 x4 • First, we multiply each side by the common denominator, which is x – 4. E.g. 4—An Equation with No Solution 5 x 1 ( x 4) 2 ( x 4) x 4 x 4 2( x 4) 5 x 1 2x 8 5 x 1 2x 3 x 1 2x x 4 x4 E.g. 4—An Equation with No Solution Now, we try to substitute x = 4 back into the original equation. • We would be dividing by 0, which is impossible. • So this equation has no solution. Extraneous Solutions The first step in the preceding solution, multiplying by x – 4, had the effect of multiplying by 0. • Do you see why? • Multiplying each side of an equation by an expression that contains the variable may introduce extraneous solutions. • That is why it is important to check every answer. Solving Power Equations Power Equations Linear equations have variables only to the first power. Now let’s consider some equations that involve squares, cubes, and other powers of the variable. • Such equations will be studied more extensively in Sections 1.6 and 1.7. Power Equation Here we just consider basic equations that can be simplified into the form Xn = a • Equations of this form are called power equations • They are solved by taking radicals of both sides of the equation. Solving a Power Equation The power equation Xn = a has the solution X= na if n is odd X = n a if n is even and a ≥ 0 If n is even and a < 0, the equation has no real solution. Examples of Solving a Power Equation Here are some examples of solving power equations. • The equation x5 = 32 has only one real solution: x = 5 32 = 2. • The equation x4 = 16 has two real solutions: x = 4 16 = ±2. Examples of Solving a Power Equation Here are some examples of solving power equations. • The equation x5 = –32 has only one real solution: x = 5 32 = –2. • The equation x4 = –16 has no real solutions 4 because 16 does not exist. E.g. 5—Solving Power Equations Solve each equation. (a) x2 – 5 = 0 (b) (x – 4)2 = 5 E.g. 5—Solving Power Equations Example (a) x2 – 5 = 0 x2 = 5 x=± 5 The solutions are x = 5 and x = – 5 . E.g. 5—Solving Power Equations Example (b) We can take the square root of each side of this equation as well. (x – 4)2 = 5 x–4=± 5 x=4± 5 (Take the square root) (Add 4) The solutions are x = 4 + 5 and x = 4 – 5 . E.g. 6—Solving Power Equations Find all real solutions for each equation. (a) x3 = – 8 (b) 16x4 = 81 E.g. 6—Solving Power Equations Example (a) Since every real number has exactly one real cube root, we can solve this equation by taking the cube root of each side. (x3)1/3 = (–8)1/3 x = –2 E.g. 6—Solving Power Equations Example (b) Here we must remember that if n is even, then every positive real number has two real nth roots. • A positive one and a negative one. • If n is even, the equation xn = c (c > 0) has two solutions, x = c1/n and x = –c1/n. E.g. 6—Solving Power Equations 81 x 16 4 1/ 4 81 (x ) 16 3 x 2 4 1/ 4 Example (b) Equations with Fractional Power The next example shows how to solve an equation that involves a fractional power of the variable. E.g. 7— Solving an Equation with a Fractional Power Solve the equation 5x2/3 – 2 = 43 • The idea is to first isolate the term with the fractional exponent, then raise both sides of the equation to the reciprocal of that exponent. • If n is even, the equation xn/m = c has two solutions, x = cm/n and x = –cm/n. E.g. 7— Solving an Equation with a Fractional Power 5x2/3 – 2 = 43 5x2/3 = 45 x2/3 = 9 x = ±93/2 x = ±27 The solutions are x = 27 and x = –27. Solving for One Variable in Terms of Others Several Variables Many formulas in the sciences involve several variables. It is often necessary to express one of the variables in terms of the others. • In the next example, we solve for a variable in Newton’s Law of Gravity. E.g. 8—Solving for One Variable in Terms of Others Solve for the variable M in mM F G 2 r • The equation involves more than one variable. • Still, we solve it as usual by isolating M on one side and treating the other variables as we would numbers. E.g. 8—Solving for One Variable in Terms of Others Gm F 2 M (Factor M from RHS) r 2 2 r r Gm Gm F Gm r 2 M (Multiply by reciprocal of 2 r F M Gm r 2F The solution is M . Gm ( Simplify) Gm r 2 ) E.g. 9—Solving for One Variable in Terms of Others The surface area A of the closed rectangular box shown can be calculated from the length l, the width w, and the height h according to the formula: A = 2lw + 2wh + 2lh • Solve for w in terms of the other variables in the equation. E.g. 9—Solving for One Variable in Terms of Others The equation involves more than one variable. Still, we solve it as usual by isolating w on one side, treating the other variables as we would numbers. E.g. 9—Solving for One Variable in Terms of Others A = (2lw + 2wh) + 2lh (Collect terms involving w) A – 2lh = 2lw + 2wh (Subtract 2lh) A – 2lh = (2l + 2h)w (Factor w from RHS) A 2l h w 2l 2h (Divide by 2l + 2h) A 2l h The solution is w 2l 2h

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