Multiple ChoiceSolve the Equation. 3(2−5x)=4x+253\left(2-5x\right)=4x+253(2−5x)=4x+25342views27rank
Multiple ChoiceSolve the equation. Then state whether it is an identity, conditional, or inconsistent equation. x4+16=x3\frac{x}{4}+\frac16=\frac{x}{3}4x+61=3x240views5rank3comments
Multiple ChoiceSolve the equation. Then state whether it is an identity, conditional, or inconsistent equation. −2(5−3x)+x=7x−10-2\left(5-3x\right)+x=7x-10−2(5−3x)+x=7x−10247views6rank2comments
Multiple ChoiceSolve the equation. Then state whether it is an identity, conditional, or inconsistent equation. 5x+17=8x+12−3(x+4)5x+17=8x+12-3\left(x+4\right)5x+17=8x+12−3(x+4)159views
Multiple ChoiceSolve the equation.92+14(x+2)=34x\frac92+\frac14\left(x+2\right)=\frac34x29+41(x+2)=43x276views7rank
Multiple ChoiceSolve the equation. Then state whether it is an identity, conditional, or inconsistent equation. 5x+17=8x+12−3(x+4)5x+17=8x+12-3\left(x+4\right)252views5rank
Textbook QuestionIn Exercises 1–34, solve each rational equation. If an equation has no solution, so state. 1/x + 2 = 3/x227views
Textbook QuestionSolve each equation. A= 24f / B(p+1), for f (approximate annual interest rate)299views
Textbook QuestionIn Exercises 1–34, solve each rational equation. If an equation has no solution, so state. (x−2)/2x + 1 = (x+1)/x218views
Textbook QuestionDecide whether each statement is true or false. The solution set of 2x+5=x -3 is {-8}.272views
Textbook QuestionSolve each problem. If x represents the number of pennies in a jar in an applied problem, which of the following equations cannot be a correct equation for finding x? (Hint:Solve the equations and consider the solutions.) A. 5x+3 =11 B.12x+6 =-4 C.100x =50(x+3) D. 6(x+4) =x+24250views
Textbook QuestionIn Exercises 1–34, solve each rational equation. If an equation has no solution, so state. 3/(x+1) = 5/(x−1)210views
Textbook QuestionIn Exercises 1–26, solve and check each linear equation. 11x - (6x - 5) = 40364views
Textbook QuestionDecide whether each statement is true or false. The equation 5x=4x is an example of a contradiction.341views
Textbook QuestionIn Exercises 1–34, solve each rational equation. If an equation has no solution, so state. (x−6)/(x+5) = (x−3)/(x+1)200views
Textbook QuestionIn Exercises 1–14, simplify the expression or solve the equation, whichever is appropriate. 3x/4 - x/3 + 1 = 4x/5 - 3/20193views
Textbook QuestionIn Exercises 1–14, simplify the expression or solve the equation, whichever is appropriate. 4x-2(1-x)=3(2x+1)-5187views
Textbook QuestionIn Exercises 1–34, solve each rational equation. If an equation has no solution, so state. 1 − 4/(x+7) = 5/(x+7)185views
Textbook QuestionIn Exercises 15–35, solve each equation. Then state whether the equation is an identity, a conditional equation, or an inconsistent equation. 2x-5 = 7270views
Textbook QuestionIn Exercises 15–35, solve each equation. Then state whether the equation is an identity, a conditional equation, or an inconsistent equation. 7(x-4) = x + 2257views
Textbook QuestionIn Exercises 15–35, solve each equation. Then state whether the equation is an identity, a conditional equation, or an inconsistent equation. 2(x-4)+3(x+5)=2x-2268views
Textbook QuestionIn Exercises 1–26, solve and check each linear equation. 2(x - 1) + 3 = x - 3(x + 1)291views
Textbook QuestionIn Exercises 1–34, solve each rational equation. If an equation has no solution, so state. 6/x − x/3 = 1202views
Textbook QuestionIn Exercises 1–26, solve and check each linear equation. 2 - (7x + 5) = 13 - 3x211views
Textbook QuestionIn Exercises 15–35, solve each equation. Then state whether the equation is an identity, a conditional equation, or an inconsistent equation. 7x + 13 = 2(2x-5) + 3x + 23255views
Textbook QuestionIn Exercises 1–26, solve and check each linear equation. 16 = 3(x - 1) - (x - 7)249views
Textbook QuestionIn Exercises 1–34, solve each rational equation. If an equation has no solution, so state. 1/x−1 + 1/x+1 = 2/x²−1185views
Textbook QuestionIn Exercises 15–35, solve each equation. Then state whether the equation is an identity, a conditional equation, or an inconsistent equation. (3x+1)/3 - 13/2 = (1-x)/4289views
Textbook QuestionExercises 27–40 contain linear equations with constants in denominators. Solve each equation. x/3 = x/2 - 2264views
Textbook QuestionExercises 27–40 contain linear equations with constants in denominators. Solve each equation. 20 - x/3 = x/2242views
Textbook QuestionExercises 27–40 contain linear equations with constants in denominators. Solve each equation. x/5 - 1/2 = x/6297views1rank
Textbook QuestionIn Exercises 1–34, solve each rational equation. If an equation has no solution, so state. 2/(x+3) − 5/(x+1) = (3x+5)/(x²+4x+3)192views
Textbook QuestionExercises 27–40 contain linear equations with constants in denominators. Solve each equation. 3x/5 = 2x/3 + 1220views
Textbook QuestionDetermine whether each equation is an identity, a conditional equation, or a contradic-tion. Give the solution set. 1/2(6x+20) = x+4 +2(x+3)973views
Textbook QuestionIn Exercises 1–34, solve each rational equation. If an equation has no solution, so state. 3y/(y²+5y+6) + 2/(y²+y−2) = 5y/(y²+2y−3)202views
Textbook QuestionDetermine whether each equation is an identity, a conditional equation, or a contradic-tion. Give the solution set. 2(x-8) = 3x-16658views1comments
Textbook QuestionIn Exercises 15–35, solve each equation. Then state whether the equation is an identity, a conditional equation, or an inconsistent equation. 3-5(2x + 1) - 2(x-4) = 0503views
Textbook QuestionExercises 27–40 contain linear equations with constants in denominators. Solve each equation. (x + 3)/6 = 3/8 + (x - 5)/4313views
Textbook QuestionExercises 27–40 contain linear equations with constants in denominators. Solve each equation. 5 + (x - 2)/3 = (x + 3)/8234views
Textbook QuestionDetermine whether each equation is an identity, a conditional equation, or a contradic-tion. Give the solution set. -0.6(x-5)+0.8(x-6) = 0.2x - 1.8356views
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)264views
Textbook QuestionExercises 27–40 contain linear equations with constants in denominators. Solve each equation. 3x/5 - (x - 3)/2 = (x + 2)/3240views
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. 4/x = 5/2x + 3268views
Textbook QuestionSolve each formula for the specified variable. Assume that the denominator is not 0 if variables appear in the denominator. P=2l+2w,for w (perimeter of a rectangle)196views
Textbook QuestionSolve each formula for the specified variable. Assume that the denominator is not 0 if variables appear in the denominator. F = GMm/r², for m (force of gravity)256views
Textbook QuestionSolve each formula for the specified variable. Assume that the denominator is not 0 if variables appear in the denominator. s = 1/2gt², for g (distance traveled by a falling object)231views
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. 5/2x - 8/9 = 1/18 - 1/3x399views
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. (x - 2)/2x + 1 = (x + 1)/x881views
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. 1/(x - 1) + 5 = 11/(x - 1)525views
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. 3/(x + 4) - 7 = - 4/(x + 4)256views
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. 3/(2x - 2) + 1/2 = 2/(x - 1)358views
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. 3/(x + 2) + 2/(x - 2) = 8/(x + 2)(x - 2)409views
Textbook QuestionIn Exercises 61–66, find all values of x satisfying the given conditions. y1 = 5(2x - 8) - 2, y2 = 5(x - 3) + 3, and y1 = y2.433views
Textbook QuestionIn Exercises 61–66, find all values of x satisfying the given conditions. y1 = (x - 3)/5, y2 = (x - 5)/4, and y1 - y2 = 1.235views
Textbook QuestionIn Exercises 61–66, find all values of x satisfying the given conditions. y1 = (2x - 1)/(x^2 + 2x - 8), y2 = 2/(x + 4), y3 = 1/(x - 2), and y1 + y2 = y3.300views
Textbook QuestionIn Exercises 67–70, find all values of x such that y = 0. y = 2[3x - (4x - 6)] - 5(x - 6)690views
Textbook QuestionIn Exercises 67–70, find all values of x such that y = 0. y = (x + 6)/(3x - 12) - 5/(x - 4) - 2/3351views
Textbook QuestionIn Exercises 67–70, find all values of x such that y = 0. y = 1/(5x + 5) - 3/(x + 1) + 7/5307views
Textbook QuestionIn Exercises 71–78, solve each equation. Then determine whether the equation is an identity, a conditional equation, or an inconsistent equation. 4x + 7 = 7(x + 1) - 3x503views
Textbook QuestionIn Exercises 71–78, solve each equation. Then determine whether the equation is an identity, a conditional equation, or an inconsistent equation. 4(x + 5) = 21 + 4x299views
Textbook QuestionExercises 73–75 will help you prepare for the material covered in the next section. Simplify: √18 - √8259views
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 QuestionIn Exercises 71–78, solve each equation. Then determine whether the equation is an identity, a conditional equation, or an inconsistent equation. 10x + 3 = 8x + 3230views
Textbook QuestionIn Exercises 71–78, solve each equation. Then determine whether the equation is an identity, a conditional equation, or an inconsistent equation. 5x + 7 = 2x + 7215views
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. 4/(x - 2) + 3/(x + 5) = 7/(x + 5)(x - 2)319views
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. 4x/(x + 3) - 12/(x - 3) = (4x^2 + 36)/(x^2 - 9)239views
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. 4/(x^2 + 3x - 10) - 1/(x^2 + x - 6) = 3/(x^2 - x - 12)219views
Textbook QuestionRetaining the Concepts. Solve and determine whether 8(x - 3) + 4 = 8x - 21 is an identity, a conditional equation, or an inconsistent equation.216views
Textbook QuestionEvaluate x^2 - x for the value of x satisfying 4(x - 2) + 2 = 4x - 2(2 - x).454views
Textbook QuestionIn Exercises 99–106, solve each equation. 5 - 12x = 8 - 7x - [6 ÷ 3(2 + 5^3) + 5x]262views
Textbook QuestionIn Exercises 99–106, solve each equation. 4x + 13 - {2x - [4(x - 3) - 5]} = 2(x - 6)204views