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13 Solve: 5 = 3. X - 3 X = +

13 Solve: 5 = 3. X - 3 X = +

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Solve: + 2. 5 X + 2 This equation contains a rational expression that has a variable in the denominator. We begin by asking, "What value(s) of x make that denominator 0?" If a number makes the denominator of a rational expression 0, that number cannot be a solution of the equation because division by 0 is undefined. 7 We note that x cannot be -2, because this would produce a 0 in the denominator of х+ 2 Since the denominators of the rational expressions in the equation are 5 and x + 2, we multiply both sides by the LCD, 5(x + 2), to clear the equation of fractions. 3 +x+ 2 This is the equation to solve. = 2 5 7 5(x + 2) ( E+z) = 5(x + 2)(2) Write each side of the equation within parentheses and then multiply both sides by the LCD. X + 2 sK* + 2(}) + s* + 2(,2) 7 5(x + 2)() + 5(x + 2)(- 5(x + 2)(2) On the left side, distribute the multiplication by 5(x + 2). 7 1 X + 2 (x + 2)(5 + 5 (x+2} 5(x + 2)(2) On the left side, simplify: 1 and X + 2 (X+2} 1. 1 3(x + 2) + 5(7) (x + 2) Simplify each side. The resulting equation does not contain any fractions. We now solve this linear equation for x. 3x + 6 + 35 = 10x + Use the distributive property and simplify. Зх + 10x + 20 Combine like terms. %3D -7x = Subtract 10x and 41 from both sides. X = Divide both sides by –7. The solution is 3 and the solution set is {3}. To check, we substitute 3 for x in the original equation and simplify. 7 Check: + This is the original equation. 5 x + 2 7 + Substitute. 5 3 + 2 3 7 5 2 = True. +
Transcribed Image Text:Solve: + 2. 5 X + 2 This equation contains a rational expression that has a variable in the denominator. We begin by asking, "What value(s) of x make that denominator 0?" If a number makes the denominator of a rational expression 0, that number cannot be a solution of the equation because division by 0 is undefined. 7 We note that x cannot be -2, because this would produce a 0 in the denominator of х+ 2 Since the denominators of the rational expressions in the equation are 5 and x + 2, we multiply both sides by the LCD, 5(x + 2), to clear the equation of fractions. 3 +x+ 2 This is the equation to solve. = 2 5 7 5(x + 2) ( E+z) = 5(x + 2)(2) Write each side of the equation within parentheses and then multiply both sides by the LCD. X + 2 sK* + 2(}) + s* + 2(,2) 7 5(x + 2)() + 5(x + 2)(- 5(x + 2)(2) On the left side, distribute the multiplication by 5(x + 2). 7 1 X + 2 (x + 2)(5 + 5 (x+2} 5(x + 2)(2) On the left side, simplify: 1 and X + 2 (X+2} 1. 1 3(x + 2) + 5(7) (x + 2) Simplify each side. The resulting equation does not contain any fractions. We now solve this linear equation for x. 3x + 6 + 35 = 10x + Use the distributive property and simplify. Зх + 10x + 20 Combine like terms. %3D -7x = Subtract 10x and 41 from both sides. X = Divide both sides by –7. The solution is 3 and the solution set is {3}. To check, we substitute 3 for x in the original equation and simplify. 7 Check: + This is the original equation. 5 x + 2 7 + Substitute. 5 3 + 2 3 7 5 2 = True. +
음 + Solve: X - 3 X = 3. II
Transcribed Image Text:음 + Solve: X - 3 X = 3. II
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