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Chapter 4: Problem 66

Solve each inequality. Graph the solution set and write it using interval notation. $$ \frac{1}{3} x+1<4+5 x $$

Short Answer

Expert verified
Solution: \( x > -\frac{9}{14} \); Interval notation: \((-\frac{9}{14}, \infty)\).

Step by step solution

01

Move Variable Term

Subtract \( \frac{1}{3}x \) from both sides of the inequality to gather like terms on one side. This gives you: \[ 1 < 4 + 5x - \frac{1}{3}x \] Combine the terms on the right: \( 5x - \frac{1}{3}x = \frac{15}{3}x - \frac{1}{3}x = \frac{14}{3}x \). The new inequality is \( 1 < 4 + \frac{14}{3}x \).
02

Isolate the Variable

Subtract 4 from both sides to move the constant to the left side, giving you: \[ 1 - 4 < \frac{14}{3}x \] Simplify the left side to get: \[ -3 < \frac{14}{3}x \]
03

Solve for x

To solve for \( x \), multiply both sides by \( \frac{3}{14} \) (the reciprocal of \( \frac{14}{3} \)) to isolate \( x \): \[ -3 \cdot \frac{3}{14} < x \] Calculating the left side gives: \[ -\frac{9}{14} < x \] This implies that \( x > -\frac{9}{14} \).
04

Write Solution in Interval Notation

The solution in interval notation is written with round brackets since the inequality is strict (not including \( -\frac{9}{14} \)): \( (-\frac{9}{14}, \infty) \).
05

Graph the Solution Set

On a number line, represent the solution set by an open circle at \( -\frac{9}{14} \) and shade all numbers to the right, extending towards positive infinity, indicating all numbers greater than \( -\frac{9}{14} \).

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Key Concepts

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

Graphing Inequalities
Graphing inequalities is a way to visually represent the set of possible solutions to an inequality on a number line. When graphing a strict inequality like \( x > -\frac{9}{14} \), follow these steps to create an accurate representation:
  • First, locate the number \(-\frac{9}{14}\) on the number line. Since this is not a whole number or a simple fraction, you might approximate its position between \(-1\) and \(0\).
  • Use an open circle at \(-\frac{9}{14}\) to show that this point is not included in the solution. An open circle indicates that the endpoint is not part of the solution set.
  • Shade or draw a line to the right of this point to indicate that all numbers greater than \(-\frac{9}{14}\) are included in the solution set.
This method helps you intuitively understand the range of values that satisfy the inequality, making it easier to comprehend how different values relate to the inequality in question.
Interval Notation
Interval notation is a concise way of expressing a range of numbers that satisfy an inequality. It uses brackets and parentheses to show whether endpoints are included. Here鈥檚 how you can write the solution to the inequality \( x > -\frac{9}{14} \) using interval notation:
  • Since the inequality is strict (\( x \) is greater than but not equal to \( -\frac{9}{14} \)), use a parenthesis "(" at \(-\frac{9}{14}\) to indicate that it is not included in the solution set.
  • The interval goes to infinity (\( \infty \)), because there is no upper limit to the values \( x \) can take. Remember, infinity is not a number, so it is always represented with a parenthesis.
  • Combine these to form the interval: \( (-\frac{9}{14}, \infty) \).
This notation succinctly captures the solution set in a format that is widely used in mathematics to describe continuous ranges of values.
Algebraic Manipulation
Algebraic manipulation is a crucial skill for solving inequalities like \( \frac{1}{3} x + 1 < 4 + 5 x \). It involves rearranging the inequality to isolate the variable and find its potential values. Let's break down the process:
  • First, move all terms involving the variable to one side. This involves subtracting \( \frac{1}{3} x \) from both sides so that all terms with \( x \) are together, leaving you with \( 1 < 4 + \frac{14}{3} x \).
  • Next, isolate the variable by moving constants to the opposite side. Subtract \(4\) from both sides to simplify, resulting in \(-3 < \frac{14}{3} x \).
  • Finally, solve for \( x \). This involves multiplying both sides by the reciprocal of \( \frac{14}{3} \), which is \( \frac{3}{14} \), achieving \( x > -\frac{9}{14} \).
Through these steps, we balance both sides of the inequality and systematically reduce it until the variable is isolated. Such manipulation is foundational for solving more complex mathematical problems.

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Most popular questions from this chapter

Railroad Crossing. The warning sign in the illustration is to be painted on the street in front of a railroad crossing. If \(y\) is \(30^{\circ}\) more than twice \(x,\) find \(x\) and \(y\)

Solve: $$ \left\\{\begin{array}{l} 2 a+3 b+2 c=7 \\ a+2 b-c=4 \\ 3 a-b+c=5 \end{array}\right. $$

Solve the inequality in part a. Graph the solution set and write it in interval notation. Then use your answer to part a to determine the solution set for the inequality in part b. (No new work is necessary!) Graph the solution set and write it in interval notation. a. \(\frac{x-3}{2} \leq \frac{1}{2}-\frac{x-5}{4}\) b. \(\frac{x-3}{2}>\frac{1}{2}-\frac{x-5}{4}\)

Solve each equation and inequality. For the inequalities, graph the solution set and write it using interval notation. $$ |0.5 x+1|<-23 $$

Solve each equation. See Example 5. $$ |2 x+1|=|3(x+1)| $$
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