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Chapter 17: Problem 22

Solve the given inequalities. Graph each solution. $$4<6 R+2 \leq 16$$

Short Answer

Expert verified
\( \frac{1}{3} < R \leq \frac{7}{3} \). Graph with an open circle at \( \frac{1}{3} \), closed at \( \frac{7}{3} \).

Step by step solution

01

Set Up the Compound Inequality

The given compound inequality is \( 4 < 6R + 2 \leq 16 \). This means we have two inequalities to solve: \( 4 < 6R + 2 \) and \( 6R + 2 \leq 16 \). We need to solve each inequality separately.
02

Solve the First Inequality

For \( 4 < 6R + 2 \), start by subtracting 2 from both sides: \[ 4 - 2 < 6R \] \[ 2 < 6R \] Next, divide both sides by 6 to solve for \( R \): \[ \frac{2}{6} < R \] Simplify to: \[ \frac{1}{3} < R \].
03

Solve the Second Inequality

For \( 6R + 2 \leq 16 \), start by subtracting 2 from both sides: \[ 6R + 2 - 2 \leq 16 - 2 \] \[ 6R \leq 14 \] Next, divide both sides by 6 to solve for \( R \): \[ R \leq \frac{14}{6} \] Simplify to: \[ R \leq \frac{7}{3} \].
04

Combine the Inequalities

Now combine the two inequalities from Step 2 and Step 3. The solution for \( R \) will satisfy both: \[ \frac{1}{3} < R \leq \frac{7}{3} \].
05

Graph the Solution

To graph \( \frac{1}{3} < R \leq \frac{7}{3} \), draw a number line. Mark an open circle at \( \frac{1}{3} \) to indicate that this boundary is not included, and a closed circle at \( \frac{7}{3} \) to indicate this boundary is included. Shade the region in between the circles.

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Key Concepts

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

Graphing Inequalities
Graphing inequalities involves visually representing a range of solutions on a number line, which helps us understand where a variable, like \( R \), can exist. To successfully graph an inequality:
  • Identify the critical values. These are the values beyond which the inequality changes.
  • Use an open circle to indicate a range that does not include the boundary (e.g., \( < \) or \( > \)).
  • Use a closed circle for boundaries that are included (e.g., \( \leq \) or \( \geq \)).
  • Shade the area between the two critical values if dealing with a compound inequality. This illustrates all numbers that satisfy both conditions.
For the inequality \( \frac{1}{3} < R \leq \frac{7}{3} \), the open circle at \( \frac{1}{3} \) demonstrates that \( R \) cannot equal \( \frac{1}{3} \). Meanwhile, the closed circle at \( \frac{7}{3} \) indicates \( R \) can be exactly \( \frac{7}{3} \). The shaded line in between these points highlights all possible \( R \) values that satisfy both parts of the inequality. This visual representation enables easy interpretation of the solution set.
Solving Linear Inequalities
Solving linear inequalities involves a process similar to solving linear equations, but with careful attention to the inequality signs. Here's how we tackle them:
  • Isolate the variable on one side of the inequality.
  • Be mindful of the direction of the inequality. If you multiply or divide both sides by a negative number, reverse the inequality sign.
  • Simplify fractions and combine like terms as needed to make the solution clearer.
In our problem, we solve two inequalities: \( 4 < 6R + 2 \) and \( 6R + 2 \leq 16 \). By subtracting 2 from both sides in each, we simplify to \( 2 < 6R \) and \( 6R \leq 14 \). Dividing each by 6, we find \( R \) must be between \( \frac{1}{3} \) and \( \frac{7}{3} \). Since no multiplication or division by a negative was needed, the direction of the inequalities in their solutions remains unchanged, maintaining clear logical boundaries for \( R \).
Compound Inequalities
Compound inequalities combine two separate inequalities into a single statement, which narrows the range of possible solutions. They are written using either "and" or "or."
  • "And" indicates that both conditions must be met simultaneously.
  • "Or" suggests that at least one of the conditions must be satisfied for the solution to hold.
In our exercise, the compound inequality \( 4 < 6R + 2 \leq 16 \) uses "and," meaning \( R \) must satisfy both \( \frac{1}{3} < R \) and \( R \leq \frac{7}{3} \). This forms a span of values for \( R \) that lie strictly between \( \frac{1}{3} \) and \( \frac{7}{3} \), including \( \frac{7}{3} \) itself but excluding \( \frac{1}{3} \). Understanding the distinction helps to accurately graph and interpret the solution set for compound inequalities.

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Most popular questions from this chapter

answer the given questions about the inequality \(0 < a < b\) What is wrong with the following sequence of steps? \(a < b, a b < b^{2}, a b-b^{2} < 0, b(a-b) < 0, b < 0\)

Use inequalities involving absolute values to solve the given problems. The moon is at a maximum distance of \(4.07 \times 10^{5} \mathrm{km}\) from earth and at a minimum distance of \(3.57 \times 10^{5} \mathrm{km} .\) Express this as an inequality with absolute values.

Answer the given questions by solving the appropriate inequalities. The weight \(w\) (in \(\mathrm{N}\) ) of an object \(h\) meters above the surface of earth is \(w=r^{2} w_{0} /(r+h)^{2},\) where \(r\) is the radius of earth and \(w_{0}\) is the weight of the object at sea level. Given that \(r=6380 \mathrm{km},\) if an object weighs \(200 \mathrm{N}\) at sea level, for what altitudes is its weight less than \(100 \mathrm{N} ?\)

graph the given inequalities on the number line. \(t \leq-5 \quad\) and \(\quad t \geq-5\)

some applications of inequalities are shown. The velocity \(v\) of an ultrasound wave in soft human tissue may be represented as \(1550 \pm 60 \mathrm{m} / \mathrm{s}\), where the \(\pm 60 \mathrm{m} / \mathrm{s}\) gives the possible variation in the velocity. Express the possible velocities by an inequality.
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