Problem 61 - Calculus Early Transcendentals: Pearson New International Edition

Question

Solve the inequalities. $$ 2|2 x-3|<|x+10| $$

Step-by-Step Solution

Answer

Solution: \(\frac{3}{2} \leq x < \frac{16}{3}\).

Step 1: Determine Cases for Absolute Values

To solve \(2|2x-3|<|x+10|\), recognize that there are two absolute values. The expression \(|a|\) equals \(a\) if \(a\geq0\) and \(-a\) if \(a<0\). We have two cases for each expression involving absolute values.

Step 2: Case 1 - Both Expressions are Non-Negative

Assume \(2x-3 \geq 0\) and \(x+10 \geq 0\) which become \(x \geq \frac{3}{2}\) and \(x \geq -10\), respectively. This makes \(x \geq \frac{3}{2}\). Under these conditions, the original inequality becomes \(2(2x-3) < x+10\).

Step 3: Solve the Inequality of Case 1

For \(2(2x-3) < x+10\), distribute and simplify the inequality: \(4x - 6 < x + 10\). Solve for \(x\) by rearranging terms: \(4x - x < 10 + 6\) which simplifies to \(3x < 16\). This gives \(x < \frac{16}{3}\). Combine with \(x \geq \frac{3}{2}\) to get \(\frac{3}{2} \leq x < \frac{16}{3}\).

Step 4: Case 2 - Both Expressions are Negative

Assume \(2x-3 < 0\) and \(x+10 < 0\) which become \(x < \frac{3}{2}\) and \(x < -10\), respectively. Since \(x < \frac{3}{2}\) and also \(x < -10\) imply \(x < -10\), the overlap between these conditions does not provide any further restriction as it fits entirely within \(x < -10\). The inequality \(2(-(2x-3)) < -(x+10)\) transforms into \(-4x+6 < -x-10\).

Step 5: Solve the Inequality of Case 2

For \(-4x+6 < -x-10\), rearrange the terms: \(-4x + x < -10 - 6\). Simplify to \(-3x < -16\), which results in \(x > \frac{16}{3}\). However, since there is no overlap with \(x < -10\), this case does not provide a solution within a valid range for \(x\). If calculated properly, it directly excludes any valid solution, hence remaining cases should be checked.

Step 6: Validate All Solutions Together

The valid solution region is where all conditions from individual cases overlap. From valid regions in both cases, conclude that the set \(\frac{3}{2} \leq x < \frac{16}{3}\) works.

Key Concepts

Absolute Value Expressions

Absolute value expressions often appear in mathematical problems, especially in inequalities. An absolute value measures the distance of a number from zero on the number line. It's always non-negative. The notation for absolute value is vertical bars, like this: \(|x|\).
For any real number \(a\), the absolute value \(|a|\) yields two possible cases. The expression equals \(a\) if \(a\) is non-negative, meaning \(a \geq 0\). Conversely, if \(a\) is negative, \(|a|\) becomes \(-a\). This duality is crucial for solving inequalities involving absolute values.
When dealing with expressions containing absolute values, always prepare to split the problem into different scenarios based on the sign of the variables inside the absolute values. For example, in the equation \(2|2x-3|\), the term inside the absolute value is \(2x-3\). You need to check:

• If \(2x-3 \geq 0\), then replace \(|2x-3|\) with \(2x-3\).
• If \(2x-3 < 0\), then replace \(|2x-3|\) with \(-2x+3\).

This comprehensive approach helps in tackling both the positive and negative scenarios of the expression, leading to a complete solution.

Inequality Solutions

Solving inequalities involves finding the range of values that satisfy given expressions. Much like solving equations, inequalities require rearranging and simplifying terms to isolate the variable. However, some distinct rules apply when dealing with inequalities:

• Adding or subtracting the same number from both sides maintains the direction of the inequality.
• Multiplying or dividing both sides by a positive number keeps the inequality direction unchanged.
• Multiplying or dividing both sides by a negative number reverses the inequality direction.

It's essential to apply these rules carefully. Consider the inequality \(4x - 6 < x + 10\) from the example. First, simplify the inequality by bringing like terms together: \(4x - x < 10 + 6\), which becomes \(3x < 16\).
This can then be simplified to \(x < \frac{16}{3}\). When looking at solutions of inequalities, ensure to verify the overlap between solutions from each case, determining the valid range for the variable.

Case Analysis in Inequalities

Case analysis is crucial when working with inequalities that involve absolute value expressions. Given the nature of absolute values, it is imperative to dissect the inequalities into numerous cases based on the sign of expressions within absolute value brackets.
Consider the problem \(2|2x-3| < |x+10|\), which contains two absolute value expressions. When facing such problems, analyze:

• What happens when each expression is positive.
• What happens when each expression is negative.

The method involves:

• Defining conditions for each scenario: For instance, \(2x-3 \geq 0\) implies \(x \geq \frac{3}{2}\), while \(x+10 \geq 0\) results in \(x \geq -10\).
• Solving the inequality under these assumptions.
• Checking the resulting values of \(x\) to ensure they fit the defined conditions.

Each case can lead to a different range of solutions. By the end, cross-verifying all scenarios ensures the solution fulfills the original problem constraints, arriving at the correct interval or set of possible solutions.