Ch. 1 - Equations and Inequalities

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

Blitzer 8th Edition

Ch. 1 - Equations and Inequalities

Problem 92

Chapter 2, Problem 92

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

Verified step by step guidance

Start with the given equations: $y = |2 - 3x|$ and $y = 13$. Since both expressions equal $y$, set them equal to each other: $|2 - 3x| = 13$.

Recall that the absolute value equation $|A| = B$ (where $B > 0$) splits into two cases: $A = B$ or $A = -B$. Apply this to $|2 - 3x| = 13$ to get two equations: $2 - 3x = 13$ and $2 - 3x = -13$.

Solve the first equation $2 - 3x = 13$ by isolating $x$: subtract 2 from both sides to get $-3x = 11$, then divide both sides by $-3$ to find $x$.

Solve the second equation $2 - 3x = -13$ similarly: subtract 2 from both sides to get $-3x = -15$, then divide both sides by $-3$ to find $x$.

The solutions from both cases give all values of $x$ that satisfy the original equation $|2 - 3x| = 13$.

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Key Concepts

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

Absolute Value Function

The absolute value of a number represents its distance from zero on the number line, always yielding a non-negative result. For an expression like |2 - 3x|, it means the value inside the bars can be positive or negative, but the output is always positive or zero.

Solving Absolute Value Equations

To solve equations involving absolute values, set the expression inside the absolute value equal to both the positive and negative of the given value. For |2 - 3x| = 13, solve 2 - 3x = 13 and 2 - 3x = -13 separately to find all possible x values.

Linear Equations

Linear equations are algebraic expressions of the first degree, meaning variables are not raised to any power other than one. After removing the absolute value, solving 2 - 3x = ±13 involves isolating x using basic algebraic operations like addition, subtraction, multiplication, and division.

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Related Practice

Textbook Question

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

Textbook Question

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

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

Textbook Question

Solve each equation in Exercises 83–108 by the method of your choice. (2x - 5)(x + 1) = 2

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Solve each equation in Exercises 83–108 by the method of your choice. (3x - 4)² = 16

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Textbook Question

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