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Ch. 1 - Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
All textbooksBlitzer 8th EditionCh. 1 - Equations and InequalitiesProblem 91
Chapter 2, Problem 91

In Exercises 59–94, solve each absolute value inequality. 12<∣−2x+67∣+3712 < \(\left\)| -2x + \(\frac{6}{7}\) \(\right\)| + \(\frac{3}{7}\)

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Start by isolating the absolute value expression on one side of the inequality. Subtract \( \frac{3}{7} \) from both sides to get: \[ 12 - \frac{3}{7} < \left| -2x + \frac{6}{7} \right| \]
Simplify the left side by finding a common denominator and performing the subtraction: \[ \text{Calculate } 12 - \frac{3}{7} = \frac{84}{7} - \frac{3}{7} = \frac{81}{7} \]. So the inequality becomes: \[ \frac{81}{7} < \left| -2x + \frac{6}{7} \right| \]
Rewrite the inequality to the standard form for absolute value inequalities: \[ \left| -2x + \frac{6}{7} \right| > \frac{81}{7} \]
Recall that for \( |A| > B \) where \( B > 0 \), the solution splits into two inequalities: \[ A < -B \quad \text{or} \quad A > B \]. Apply this to get: \[ -2x + \frac{6}{7} < -\frac{81}{7} \quad \text{or} \quad -2x + \frac{6}{7} > \frac{81}{7} \]
Solve each inequality separately for \( x \). For each, subtract \( \frac{6}{7} \) from both sides, then divide by \( -2 \), remembering to reverse the inequality sign when dividing by a negative number.

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Key Concepts

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

Absolute Value Inequalities

An absolute value inequality involves expressions within absolute value bars and requires finding all values of the variable that satisfy the inequality. The absolute value represents the distance from zero, so inequalities often split into two cases to solve.
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Isolating the Absolute Value Expression

Before solving an absolute value inequality, isolate the absolute value term on one side of the inequality. This often involves subtracting or adding constants and combining like terms to simplify the inequality for further analysis.
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Solving Compound Inequalities

Absolute value inequalities can translate into compound inequalities, such as 'less than' or 'greater than' forms. Understanding how to split and solve these compound inequalities is essential to find the solution set for the variable.
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Related Practice
Textbook Question

In Exercises 91–100, find all values of x satisfying the given conditions. y = |2 - 3x| and y = 13

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Textbook Question

Solve each equation in Exercises 92–93 by making an appropriate substitution. x^4 - 5x^2 + 4 = 0

Textbook Question

The 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/(x2 + 3x - 10) - 1/(x2 + x - 6) = 3/(x2 - x - 12)

Textbook Question

In Exercises 85–90, find the x-intercepts of the graph of each equation. Then use the x-intercepts to match the equation with its graph. [The graphs are labeled (a) through (f).] y = 2(x + 2)^2 + 5(x + 2) - 3

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Textbook Question

The 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) = (4x2 + 36)/(x2 - 9)

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Textbook Question

Solve each equation in Exercises 83–108 by the method of your choice. (2x + 3)(x + 4) = 1