To show that the inequalities [tex]\(\frac{x+1}{x-1} < \frac{x+3}{x+2}\)[/tex] and [tex]\(\frac{x+5}{(x-1)(x+2)} < 0\)[/tex] are equivalent, we need to manipulate the first inequality and show that it can be transformed into the second inequality. Here is a detailed step-by-step solution:
1. Express the first inequality in a common denominator:
[tex]\[
\frac{x+1}{x-1} < \frac{x+3}{x+2}
\][/tex]
2. Cross-multiply to clear the fractions, ensuring that we keep track of the conditions where denominators are not zero (i.e., [tex]\(x \neq 1\)[/tex] and [tex]\(x \neq -2\)[/tex]):
[tex]\[
(x+1)(x+2) < (x+3)(x-1)
\][/tex]
3. Expand both sides of the inequality:
[tex]\[
(x+1)(x+2) = x^2 + 3x + 2
\][/tex]
[tex]\[
(x+3)(x-1) = x^2 + 2x - 3
\][/tex]
4. Subtract the right-hand side from the left-hand side to set up for solving the inequality:
[tex]\[
x^2 + 3x + 2 - x^2 - 2x + 3 < 0
\][/tex]
Simplify the expression:
[tex]\[
x^2 + 3x + 2 - x^2 - 2x + 3 = x + 5
\][/tex]
Thus, the inequality becomes:
[tex]\[
x + 5 < 0
\][/tex]
5. From the above steps, we now have the simplified expression of the first inequality:
[tex]\[
x + 5 < 0
\][/tex]
which can be written as:
[tex]\[
\frac{x+5}{1} < 0
\][/tex]
This result tells us that [tex]\(x < -5\)[/tex].
6. Now, consider the second inequality:
[tex]\[
\frac{x+5}{(x-1)(x+2)} < 0
\][/tex]
7. To show that [tex]\(\frac{x+1}{x-1} < \frac{x+3}{x+2}\)[/tex] implies [tex]\(\frac{x+5}{(x-1)(x+2)} < 0\)[/tex], we notice that we convert the inequality [tex]\(\frac{x+5}{1} < 0\)[/tex] to the form involving the common product of denominators of the original inequality:
[tex]\[
\frac{x+5}{(x-1)(x+2)} < 0
\][/tex]
8. Because originally we have:
[tex]\[
(x+1)(x+2) < (x+3)(x-1) \implies x + 5 < 0
\][/tex]
9. Thus, when considering the interval checks and solving inequalities involving rational functions, both inequalities will yield the same critical points and regions to satisfy [tex]\(x + 5 < 0\)[/tex] and [tex]\(\frac{x+5}{(x-1)(x+2)} < 0\)[/tex]. Therefore, the conclusion is:
- The inequality [tex]\(x + 5 < 0\)[/tex] controls the numerator.
- The denominators (product [tex]\((x-1)(x+2)\)[/tex]) ensure the expression is negative under equivalent conditions.
This chain of algebraic manipulations clearly shows that both inequalities are equivalent. We effectively expressed the initial inequality in a form that leads us to recognize the condition on the rational function involving the same numerator and contributing factors in the denominator.
