Multiple ChoiceSolve the equation. 5x−23x=4+3x\frac{5}{x}-\frac{2}{3x}=4+\frac{3}{x}x5−3x2=4+x3193views3rank
Multiple ChoiceSolve the equation. −5x+4−3=x−1x+4\frac{-5}{x+4}-3=\frac{x-1}{x+4}x+4−5−3=x+4x−1182views4rank
Textbook QuestionIn Exercises 1–34, solve each rational equation. If an equation has no solution, so state. (7x−4)/5x = 9/5 − 4/x167views
Textbook QuestionIn Exercises 1–14, simplify the expression or solve the equation, whichever is appropriate. 3(2x-5)-2(4x+1)=-5(x+3)-2137views
Textbook QuestionIn Exercises 1–34, solve each rational equation. If an equation has no solution, so state. 8/x²−9 + 4/x+3 = 2/x−3204views
Textbook QuestionIn Exercises 1–26, solve and check each linear equation. 2(x - 1) + 3 = x - 3(x + 1)290views
Textbook QuestionIn Exercises 1–26, solve and check each linear equation. 25 - [2 + 5y - 3(y + 2)] = - 3(2y - 5) - [5(y - 1) - 3y + 3]260views
Textbook QuestionIn Exercises 15–35, solve each equation. Then state whether the equation is an identity, a conditional equation, or an inconsistent equation. 2x/3 = 6 - x/4232views
Textbook QuestionExercises 27–40 contain linear equations with constants in denominators. Solve each equation. 20 - x/3 = x/2241views
Textbook QuestionIn Exercises 1–34, solve each rational equation. If an equation has no solution, so state. 4/(x²+3x−10) + 1/(x²+9x+20) = 2/(x²+2x−8)171views
Textbook QuestionExercises 27–40 contain linear equations with constants in denominators. Solve each equation. 3x/5 - x = x/10 - 5/2160views
Textbook QuestionDetermine whether each equation is an identity, a conditional equation, or a contradic-tion. Give the solution set. -6(2x+1) - 3(x-4) = -15x+1233views
Textbook QuestionSolve each formula for the specified variable. Assume that the denominator is not 0 if variables appear in the denominator. I=Prt,for P (simple interest)261views
Textbook QuestionExercises 41–60 contain rational equations with variables in denominators. For each equation, a. write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation. 7/2x - 5/3x = 22/3241views
Textbook QuestionSolve each formula for the specified variable. Assume that the denominator is not 0 if variables appear in the denominator. z = x-μ/σ, for x (standardized value)203views
Textbook QuestionExercises 41–60 contain rational equations with variables in denominators. For each equation, a. write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation. 2/(x - 2) = x/(x - 2) - 2237views
Textbook QuestionIn Exercises 61–66, find all values of x satisfying the given conditions. y1 = 5/(x + 4), y2 = 3/(x + 3), y3 = (12x + 19)/(x^2 + 7x + 12). and y1 + y2 = y3.223views
Textbook QuestionIn Exercises 71–78, solve each equation. Then determine whether the equation is an identity, a conditional equation, or an inconsistent equation. 5x + 9 = 9(x + 1) - 4x298views
Textbook QuestionExercises 73–75 will help you prepare for the material covered in the next section. Rationalize the denominator: (7 + 4√2)/(2 - 5√2).255views
Textbook QuestionThe equations in Exercises 79–90 combine the types of equations we have discussed in this section. Solve each equation. Then state whether the equation is an identity, a conditional equation, or an inconsistent equation. 2/x + 1/2 = 3/4202views
Textbook QuestionEvaluate x^2 - (xy - y) for x satisfying 3(x + 3)/5 = 2x + 6 and y satisfying - 2y - 10 = 5y + 18.521views
Textbook QuestionIn Exercises 99–106, solve each equation. - 2{7 - [4 -2(1 - x) + 3]} = 10 - [4x - 2(x - 3)]208views