Problem 70 - Algebra and Trigonometry

Question

Solve each inequality in Exercises \(65-70\) and graph the solution set on a real number line. $$ \frac{x^{2}-3 x+2}{x^{2}-2 x-3}>0 $$

Step-by-Step Solution

Answer

The solution of the inequality \(\frac{(x-1)(x-2)}{(x-3)(x+1)}>0\) is \(x<-1\), \(13\). Graphed on a real number line, the intervals \((-∞,-1)\), \((1,2)\), and \((3, ∞)\) are part of the solution set.

Step 1: Factor the Quadratic Expressions

Factor the numerator and denominator. The numerator \(x^{2}-3 x+2\) can be factored as \( (x-1)(x-2)\) and the denominator \(x^{2}-2x-3\) can be factored as \( (x-3)(x+1)\). Thus, the inequality can be rewritten as \[\frac{(x-1)(x-2)}{(x-3)(x+1)}>0\].

Step 2: Find the Critical Values

The inequality is equal to 0 when the numerator is 0, that is, \(x=1\) or \(x=2\). The inequality is undefined where the denominator is 0, that is, \(x=-1\) or \(x=3\). Hence, the critical values are \(x= -1, 1, 2\), and \(3\).

Step 3: Determine the Sign Values

Divide the real number line into regions by the critical values, and determine the sign of the inequality in each region. Choose a representative from each region and substitute it into the inequality to check if the region satisfies the inequality. If \(x<-1\), the inequality is positive. If \(-13\), the inequality is positive.

Step 4: Graph the Solution Set

The solution set is all \(x\) values that satisfy the inequality or make it positive. Hence, the solution set is \(x<-1\), \(13\). On the real number line, mark the intervals \((-∞,-1)\), \((1,2)\), and \((3, ∞)\), indicating that they are part of the solution set.