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Chapter 2: Problem 44

Solve the inequality. Then graph the solution set. $$\frac{x+12}{x+2}-3 \geq 0$$

Short Answer

Expert verified
The solution is \(x \in (-鈭�,-2) \cup (3,鈭�)\)

Step by step solution

01

Simplify the Inequality

First, simplify the inequality to a standard form. Bring all terms to one side of the inequality and set it to zero: \[ \frac{x+12}{x+2} - 3 \geq 0 \]\[ \frac{x+12}{x+2} \geq 3 \]\[ \frac{x+12}{x+2} - 3 \geq 0\]This simplifies to\[ \frac{x+12 - 3(x+2)}{x+2} \geq 0 \]\[ \frac{x+12 - 3x - 6}{x+2} \geq 0 \]\[ \frac{-2x+6}{x+2} \geq 0 \]
02

Identify Critical Points

Identify the values of \(x\) for which the inequality equals to zero or undefined, these are the critical points. The critical points are at\[ -2x + 6 = 0 \] which simplifies to \(x = 3\), and \(x = -2\), where the denominator is zero and the whole expression is undefined.
03

Test Intervals

Test the sign of the inequality in each interval determined by the critical points (-2,3). The intervals are (-鈭�, -2), (-2, 3) and (3, 鈭�). If the sign of the inequality in a particular interval is positive, that means the inequality holds true for all values in that interval and vice versa. The inequality holds true for intervals (-鈭�, -2) and (3, 鈭�).
04

Graph the Solution

Draw a number line and plot the intervals of each solution on it. We mark 'x=-2' and 'x=3' on the number line. For 'x< -2', we make an open dot at -2 and draw a line to left denoting all values less than -2 satisfies the inequality. For 'x > 3', we make an open dot at x=3 and draw a line to right denoting all values greater than 3 satisfies the inequality.

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Most popular questions from this chapter

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