Solve each rational inequality. Give the solution set in interval notation. (2x-3)(3x+8)/(x-6)3≥0
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- 0. Review of Algebra4h 18m
- 1. Equations & Inequalities3h 18m
- 2. Graphs of Equations1h 43m
- 3. Functions2h 17m
- 4. Polynomial Functions1h 44m
- 5. Rational Functions1h 23m
- 6. Exponential & Logarithmic Functions2h 28m
- 7. Systems of Equations & Matrices4h 5m
- 8. Conic Sections2h 23m
- 9. Sequences, Series, & Induction1h 22m
- 10. Combinatorics & Probability1h 45m
1. Equations & Inequalities
Linear Inequalities
Problem 105
Textbook Question
Answer the following. Why can 3 not be in the solution set of 14x+9 / x-3 < 0? (Do not solve the inequality.)
Verified step by step guidance1
Identify the expression given in the inequality: \(\frac{14x + 9}{x - 3} < 0\).
Recognize that the denominator \(x - 3\) cannot be zero because division by zero is undefined in mathematics.
Set the denominator equal to zero to find values that are not allowed: \(x - 3 = 0\).
Solve for \(x\): \(x = 3\).
Conclude that \(x = 3\) cannot be in the solution set because it makes the denominator zero, which is undefined, so it must be excluded from the domain of the inequality.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Domain Restrictions in Rational Expressions
A rational expression is undefined where its denominator equals zero. Since division by zero is undefined, any value that makes the denominator zero cannot be part of the solution set. Here, x = 3 makes the denominator zero, so it must be excluded.
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Inequalities Involving Rational Expressions
When solving inequalities with rational expressions, the solution set includes values that satisfy the inequality and are within the domain. Values that cause division by zero are excluded, even if they satisfy the inequality algebraically.
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Rationalizing Denominators
Exclusion of Points from Solution Sets
Points that make the expression undefined are excluded from the solution set regardless of the inequality. This ensures the solution set only contains valid inputs where the expression is defined and the inequality holds true.
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