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Chapter 3: Problem 17

Solve. $$|3 x+2|=1$$

Short Answer

Expert verified
The solutions are \( x = -\frac{1}{3} \) and \( x = -1 \).

Step by step solution

01

Understand the Absolute Value Equation

The equation given is \( |3x + 2| = 1 \). Remember that an absolute value equation \( |A| = B \) can be split into two separate equations: \( A = B \) and \( A = -B \).
02

Split into Two Cases

Split the given absolute value equation \( |3x + 2| = 1 \) into two separate equations: \( 3x + 2 = 1 \) and \( 3x + 2 = -1 \).
03

Solve the First Equation

Solve \( 3x + 2 = 1 \).Subtract 2 from both sides: \[ 3x = 1 - 2 \] \[ 3x = -1 \]Divide by 3: \[ x = -\frac{1}{3} \]
04

Solve the Second Equation

Solve \( 3x + 2 = -1 \).Subtract 2 from both sides: \[ 3x = -1 - 2 \] \[ 3x = -3 \]Divide by 3: \[ x = -1 \]
05

Provide the Solutions

The solutions to the equation \( |3x + 2| = 1 \) are \( x = -\frac{1}{3} \) and \( x = -1 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solving Equations
An equation is a mathematical sentence that shows the equality of two expressions. To solve an equation, we aim to find the values of the variables that make the equation true. When dealing with equations, ensure each step logically follows the next.

For solving equations effectively:
  • Identify the type of equation
  • Isolate the variable of interest
  • Perform operations systematically

Applied to the original exercise, the equation is an absolute value equation, which we'll explore next.
Absolute Value Functions
The absolute value of a number is its distance from zero on the number line, represented by \( |x| \). It is always non-negative.

To solve absolute value equations like \( |A| = B \):
  • Recognize that \( A \) can be either \( B \) or \( -B \)
  • Set up two separate equations
  • Solve each equation independently

For instance, in our exercise, \( |3x + 2| = 1 \), can be split into:

\( 3x + 2 = 1 \)
\( 3x + 2 = -1 \)
Precalculus
Precalculus is an advanced form of secondary school mathematics that prepares students for calculus. It includes topics like algebra, trigonometry, and functions. Understanding absolute value functions and solving equations are foundational skills in precalculus.

To tackle precalculus problems:
  • Review fundamental algebraic principles
  • Understand function behavior
  • Practice, practice, practice

In the context of our absolute value equation, it's crucial to revisit key algebraic techniques to solve for \( x \), as it sets the stage for more complex calculus concepts.

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Most popular questions from this chapter

In business, profit is the difference between revenue and cost; that is, $$ \begin{array}{l} \text { Total profit }=\text { Total revenue }-\text { Total cost, } \\ P(x)=R(x)-C(x) \end{array} $$ where \(x\) is the number of units sold. Find the maximum profit and the number of units that must be sold in order to yield the maximum profit for each of the following. $$R(x)=20 x-0.1 x^{2}, C(x)=4 x+2$$

Determine whether the statement is true or false. The conjugate of a product is the product of the conjugates of the individual complex numbers.

Fill in the blank with the correct term. Some of the given choices will not be used. distance formula, midpoint formula, function, relation, \(x\) -intercept, y-intercept, perpendicular, parallel ,horizontal lines, vertical lines,symmetric with respect to the \(x\) -axis, symmetric with respect to the \(y\) -axis, symmetric with respect to the origin, increasing, decreasing, constant \(\mathrm{A}(\mathrm{n})\)___ is a correspondence such that each member of the domain corresponds to exactly one member of the range.

Solve. $$2 t^{2}+(t-4)^{2}=5 t(t-4)+24$$

Solve. $$\left(x-\frac{1}{5}\right)\left(x^{2}-\frac{1}{4}\right)+\left(x-\frac{1}{5}\right)\left(x^{2}+\frac{1}{8}\right)=0$$
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