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Solve the inequality and specify the answer using interval notation. $$\left|\frac{x-2}{3}\right|<4$$

Short Answer

Expert verified
The solution is \((-10, 14)\).

Step by step solution

01

Understanding Absolute Inequalities

The inequality \( \left|\frac{x-2}{3}\right| < 4 \) suggests that the expression \( \frac{x-2}{3} \) lies between -4 and 4. This means we split the absolute value inequality into two separate inequalities: \( \frac{x-2}{3} < 4 \) and \( \frac{x-2}{3} > -4 \).
02

Solve \( \frac{x-2}{3} < 4 \)

To solve \( \frac{x-2}{3} < 4 \), first eliminate the fraction by multiplying both sides by 3, giving \( x-2 < 12 \). Add 2 to both sides to isolate \( x \): \( x < 14 \).
03

Solve \( \frac{x-2}{3} > -4 \)

To solve \( \frac{x-2}{3} > -4 \), multiply both sides by 3 to get \( x-2 > -12 \). Add 2 to both sides to isolate \( x \): \( x > -10 \).
04

Combine the Solutions

Combine the results from the two separate inequalities (\( x < 14 \) and \( x > -10 \)) to find the solution set. The solution is the intersection of the two conditions: \( -10 < x < 14 \).
05

Express in Interval Notation

Finally, express the solution in interval notation. The solution \( -10 < x < 14 \) is represented as \( (-10, 14) \).

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Key Concepts

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

Absolute Value Inequalities
Absolute value inequalities, like \( \left| \frac{x-2}{3} \right| < 4 \), tell us how far an expression can be from zero. The absolute value \( |a| \) of any number \( a \) is its distance from zero, regardless of direction on the number line.
In this context, the inequality shows us that the value of \( \frac{x-2}{3} \) must remain within 4 units of zero. This is why we split it into two inequalities: the expression is greater than \(-4\) and less than \(4\). This provides two separate "limits" for \( \frac{x-2}{3} \) to solve for \( x \).

The process:
  • If we have an inequality of the form \( |A| < B \), it translates into \( -B < A < B \).
  • It creates a range of values that \( A \) can take.
  • The absolute value just measures distance from zero, hence we consider both the negative and positive limits.
Interval Notation
Interval notation is a concise way of expressing a range of values solutions, especially useful after solving inequalities. In interval notation, we use brackets or parentheses to show which numbers are included or excluded in a set.
  • Parentheses \(( )\) are used to indicate that an endpoint is not included in the interval, which is called an open interval.
  • Brackets \([ ]\) imply that an endpoint is included, known as a closed interval.
Using our solution \(-10 < x < 14\) from the inequality, it means \( x \) can be any number between \(-10\) and \(14\), but not exactly those numbers.
So, in interval notation, this is written as \((-10, 14)\).

Think of interval notation like a shorthand substitute for describing parts of the number line:
  • It allows mathematicians and students to communicate inequalities quickly.
  • It is important in expressing solutions clearly and precisely, especially for computer calculations.
Solving Inequalities
Solving inequalities involves finding all possible values of a variable that satisfy the inequality condition. Unlike equations, inequalities often produce ranges or intervals as solutions rather than single values.
Here's a brief guide to solve inequalities like this one:
  • Start by manipulating the inequality just like you would an equation. Perform operations to isolate the variable.
  • Remember when multiplying or dividing by a negative number, flip the inequality sign.
  • For absolute inequalities, split them into two separate inequalities, one for the greater than and one for the less than condition.
In our example, we broke the absolute inequality into two parts and solved them like regular linear inequalities. Then, we found a common solution that satisfies both conditions.
This highlighted the intersection of solutions, giving us the possible range for \( x \).

Understanding how to solve inequalities is key because:
  • They describe scenarios where multiple outcomes are possible.
  • They form the basis for constraints in optimization and real-world problems.
Always check your solutions by substituting back, ensuring they meet the original inequality conditions.

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Most popular questions from this chapter

Suppose that after studying a corporation's records, a business analyst predicts that the corporation's monthly revenues \(R\) for the near future can be closely approximated by the equation $$\begin{aligned}R=&-0.0217 x^{5}+0.626 x^{4}-6.071 x^{3}+25.216 x^{2} \\\&-57.703 x+159.955 \quad(1 \leq x \leq 12)\end{aligned}$$ where \(R\) is the revenue (in thousands of dollars) for the month \(x,\) with \(x=1\) denoting January, \(x=2\) denoting February, and so on. (a) According to this model, for which months will the monthly revenue be no more than \(\$ 80,000 ?\) Hint: You need to solve the inequality \(R \leq 80 .\) Round each key number to the nearest integer. (Why?) (b) For which months, if any, will the monthly revenue be at least \(\$ 120,000 ?\)

Solve the inequalities. Suggestion: A calculator may be useful for approximating key numbers. $$\frac{2 x}{x-2}<3$$

Use the discriminant to determine how many real roots each equation has. $$2 x^{2}-6 x+5=0$$

For which values of \(b\) will the equation \(x^{2}+b x+1=0\) have real solutions?

Given two positive numbers \(a\) and \(b,\) we define the geometric mean (G.M.) and the arithmetic mean (A.M.) as follows: $$\text { G.M. }=\sqrt{a b} \quad \text { A.M. }=\frac{a+b}{2}$$ (a) Complete the table, using a calculator as necessary so that the entries in the third and fourth columns are in decimal form. $$\begin{array}{rrrr} & & \sqrt{a b} & (a+b) / 2 & \text { Which is larger, } \\ a & b & (G . M .) & (A . M .) & G . M . \text { or } A . M . ? \\ \hline 1 & 2 & & & \\ 1 & 3 & & & \\ 1 & 4 & & & \\ 2 & 3 & & & \\ 3 & 4 & & & \\ 5 & 10 & & & \\ 9 & 10 & & & \\ 99 & 100 & & & \\ 999 & 1000 & & & & \\ \hline \end{array}$$ (b) Prove that for all nonnegative numbers \(a\) and \(b\) we have $$\sqrt{a b} \leq \frac{a+b}{2}$$ Hint: Use the following property of inequalities: If \(x\) and \(y\) are nonnegative, then the inequality \(x \leq y\) is equivalent to \(x^{2} \leq y^{2}\) (c) Assuming that \(a=b\) (and that \(a\) and \(b\) are nonnegative), show that inequality (1) becomes an equality. (d) Assuming that \(a\) and \(b\) are nonnegative and that \(\sqrt{a b}=\frac{a+b}{2},\) show that \(a=b\) Remark: Parts (b) through (d) can be summarized as follows. For all nonnegative numbers \(a\) and \(b,\) we have \(\sqrt{a b} \leq \frac{a+b}{2},\) with equality holding if and only if \(a=b\) This result is known as the arithmetic- geometric mean inequality for two numbers. The mini project at the end of this section shows an application of this result.

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