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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

Solve the given linear programming problems. An oil refinery refines types \(A\) and \(B\) of crude oil and can refine as much as 4000 barrels each week. Type A crude has 2 kg of impurities per barrel, type \(\mathrm{B}\) has \(3 \mathrm{kg}\) of impurities per barrel, and the refinery can handle no more than \(9000 \mathrm{kg}\) of these impurities each week. How much of each type should be refined in order to maximize profits, if the profit is \(\$ 4 /\) barrel for type \(\mathrm{A}\) and \$5/barrel for type B?

For the inequality \(4<9,\) state the inequality that results when the given operations are performed on both members. Add 3.

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Solve the given inequalities. Graph each solution. $$\frac{1}{3} x+2 \geq 1$$

Draw a sketch of the graph of the region in which the points satisfy the given system of inequalities. $$\begin{aligned} &y>x^{2}\\\ &y
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