In Exercises 51–58, solve each compound inequality. - 11 \u003c 2x - 1 ≤ - 5

Start by writing the compound inequality as two separate inequalities combined: $$11 \u003c 2x - 1$$ and $$2x - 1 \leq -5. $$Solve the first inequality $$11 \u003c 2x - 1 by $$adding 1 to both sides to isolate the term with $$x$$: $$11 + 1 \u003c 2x$$ which simplifies to $$12 \u003c 2x. $$Next, divide both sides of $$12 \u003c 2x by 2 to $$solve for $$x$$: $$\frac{12}{2} \u003c x$$ which simplifies to $$6 \u003c x. $$Now solve the second inequality $$2x - 1 \leq -5 by $$adding 1 to both sides: $$2x \leq -5 + 1$$ which simplifies to $$2x \leq -4. $$Divide both sides of $$2x \leq -4 by 2 to $$solve for $$x$$: $$x \leq \frac{-4}{2}$$ which simplifies to $$x \leq -2.$$

Ch. 1 - Equations and Inequalities

Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook

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Blitzer 8th Edition

Ch. 1 - Equations and Inequalities

Problem 55

Chapter 2, Problem 55

In Exercises 51–58, solve each compound inequality. - 11 < 2x - 1 ≤ - 5

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Start by writing the compound inequality as two separate inequalities combined: \$11 < 2x - 1$ and $2x - 1 \leq -5$.

Solve the first inequality \$11 < 2x - 1$ by adding 1 to both sides to isolate the term with $x$: $11 + 1 < 2x$ which simplifies to $12 < 2x$.

Next, divide both sides of \$12 < 2x$ by 2 to solve for $x$: \(\frac{12}{2} < x$ which simplifies to \)6 < x$.

Now solve the second inequality $2x - 1 \leq -5$ by adding 1 to both sides: $2x \leq -5 + 1$ which simplifies to $2x \leq -4$.

Divide both sides of $2x \leq -4$ by 2 to solve for $x$: $x \leq \frac{-4}{2}$ which simplifies to $x \leq -2$.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Compound Inequalities

Compound inequalities involve two inequalities combined into one statement, often connected by 'and' or 'or'. Solving them requires finding values that satisfy both inequalities simultaneously, typically represented as a range or union of intervals.

Solving Linear Inequalities

Solving linear inequalities involves isolating the variable on one side by performing inverse operations, similar to solving equations, but remembering to reverse the inequality sign when multiplying or dividing by a negative number.

Interval Notation and Graphing Solutions

After solving inequalities, solutions are often expressed in interval notation, which concisely describes all values that satisfy the inequality. Graphing these intervals on a number line helps visualize the solution set clearly.

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Interval Notation

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Textbook Question

Exercises 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)

Solve each equation in Exercises 47–64 by completing the square. $x^{2} - 5 x + 6 = 0$

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Textbook Question

Solve each formula for the specified variable. Do you recognize the formula? If so, what does it describe? 1/R = 1/R₁ + 1/R₂ for R₁

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