Problem 57
Question
Solve each inequality in Exercises 57-84 by first rewriting each one as an equivalent inequality without absolute value bars. Graph the solution set on a number line. Express the solution set using interval notation. $$|x|<3$$
Step-by-Step Solution
Verified Answer
The solution to the inequality \(|x|<3\) is \(-3
1Step 1: Rewrite the absolute value inequality without absolute value bars
An absolute value inequality \(|x|<3\) can be rewritten as \( -3-3\). This is because the definition of an absolute value is that it is the distance a number is from zero on the number line, which can be either positive or negative.
2Step 2: Solve the inequalities
Solving the inequality \(x<3\) gives all the real numbers less than 3, and solving \(x>-3\) gives all the real numbers more than -3. Since we need to satisfy both inequalities simultaneously, the solution would be the intersection of both of these sets. Thus, the solution of the inequality \(|x|<3\) is the set of all x such that \(-3
3Step 3: Graph the solution set on the number line
The solution set can be graphed on a number line by marking both -3 and 3 on the number line and shading the region between them. This is because all the values between -3 and 3 are in solution set.
4Step 4: Express the solution set using interval notation
The solution set \(-3
Key Concepts
Understanding the Solution SetInterval NotationGraphing the Solution Set on a Number Line
Understanding the Solution Set
The solution set of an inequality refers to the set of all possible values that satisfy the inequality. In the case of the absolute value inequality \(|x| < 3\), we find its solution set by rewriting it into two separate inequalities:
This forms the solution set \(-3 < x < 3\), which includes all numbers between -3 and 3 but not -3 and 3 themselves. This solution set gives us the range of permissible values for \(x\) that make the original inequality true.
- \(x < 3\)
- \(x > -3\)
This forms the solution set \(-3 < x < 3\), which includes all numbers between -3 and 3 but not -3 and 3 themselves. This solution set gives us the range of permissible values for \(x\) that make the original inequality true.
Interval Notation
Interval notation is a convenient way of expressing sets of numbers, especially useful for solution sets derived from inequalities. For the inequality \(-3 < x < 3\), the solution set is expressed in interval notation as \((-3, 3)\).
- The parenthesis \((\) and \()\) indicate that -3 and 3 are not included in the set (this is called an "open interval").
- If the interval were \([-3, 3]\), it would mean -3 and 3 are included, shown by using brackets \([\) and \()\) (this is a "closed interval").
Graphing the Solution Set on a Number Line
The number line is a fundamental tool for visualizing the solution set. To graph the solution set \(-3 < x < 3\), we begin by marking the points -3 and 3 on the number line. However, because these points are not included in the solution set (as denoted by the open interval \((-3, 3)\)), we use open circles for -3 and 3.
Next, we shade the region between these two points, representing all the values that \(x\) can take.
Next, we shade the region between these two points, representing all the values that \(x\) can take.
- The open circles show that the endpoint numbers (in this case, -3 and 3) are not part of the solution set.
- The shaded part in between illustrates all the real numbers that satisfy the inequality \(-3 < x < 3\).
Other exercises in this chapter
Problem 57
Solve each equation in Exercises \(55-64\) using the quadratic formula. $$ x^{2}+5 x+3=0 $$
View solution Problem 57
In Exercises \(57-76,\) solve each formula for the specified variable. Do you recognize the formula? If so, what does it describe? \(A=l w\) for \(w\)
View solution Problem 57
In Exercises \(51-58,\) determine whether each equation is an identity, a conditional equation, or an inconsistent equation. $$ \frac{2 x}{x-3}=\frac{6}{x-3}+4
View solution Problem 58
Solve each equation in by making an appropriate substitution. $$ \left(y-\frac{10}{y}\right)^{2}+6\left(y-\frac{10}{y}\right)-27=0 $$
View solution