Concept explainers
In Exercises 1–12, solve each equation.
1. −5 + 3(x + 5) = 2(3x − 4)
To solve: The equation
Answer to Problem 1MC
The solution set of the equation
Explanation of Solution
The given equation is
Simplify the algebraic expression on each side of the equation.
Obtain an equivalent equation by isolating x on one side and the constant terms on the other side as follows.
On further simplifications,
Thus, the solution of the equation is 6.
Replace x with 6 in the original solution and check the correctness of the solution.
The left hand side of the equation is,
The right hand side of the equation is,
That is,
Thus, the solution set of the equation
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Chapter 1 Solutions
College Algebra Essentials (5th Edition)
- In Exercises 105–107, solve each equation using a graphing utility. Graph each side separately in the same viewing rectangle. The solutions are the x-coordinates of the intersection points. 105. |x + 1|| 106. 13(x + 4)| = 12 107. 12x – 3| = 19 – 4x|arrow_forwardFor Exercises 8–10, a. Simplify the expression. Do not rationalize the denominator. b. Find the values of x for which the expression equals zero. c. Find the values of x for which the denominator is zero. 4x(4x – 5) – 2x² (4) 8. -6x(6x + 1) – (–3x²)(6) (6x + 1)2 9. (4x – 5)? - 10. V4 – x² - -() 2)arrow_forwardIn Exercises 65–69, solve each equation. 2x 65. 3 1 + 1 6. 66. 2 10 2 2x 67. 3 68. 4 2 + 3 4 Зх + 1 13 1 - x 69. 3 2 4 6 ||arrow_forward
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- In Exercises 3-5, solve each equation. 3. 5 - 2(3 – x) = 2(2x + 5) + 1 3x 4. 5 *+4- 4 = 5. 3x – 4 = 2(3x + 2) – 3xarrow_forwardSolve 6(x + 8)+4= 5(z +5) for a Submit Questionarrow_forwardRewriting an Expression In Exercises 51–54,rewrite the quadratic portion of the algebraic expressionas the sum or difference of two squares by completingthe square.arrow_forward
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