1. Explain using an algebraic argument why the inequalities 􏰃􏰂1 < 􏰃􏰂3 and

   􏰃􏰂5 < 0 are equivalent (􏰃􏰀1)(􏰃􏰂2)

􏰃􏰀1 􏰃􏰂2
 

To show why the inequalities

\[
\frac{1}{x-1} < \frac{3}{x+2}
\]

and

\[
\frac{5}{(x-1)(x+2)} < 0
\]

are equivalent, we'll analyze them step by step.

### Step 1: Start with the first inequality
The first inequality is:

\[
\frac{1}{x-1} < \frac{3}{x+2}
\]

To compare these two fractions, we'll subtract the right-hand side from the left-hand side to get a single inequality involving zero:

\[
\frac{1}{x-1} - \frac{3}{x+2} < 0
\]

### Step 2: Find a common denominator
The common denominator between \(x-1\) and \(x+2\) is \((x-1)(x+2)\). Let's rewrite the left-hand side with this common denominator:

\[
\frac{(x+2) - 3(x-1)}{(x-1)(x+2)} < 0
\]

Now, expand the numerator:

\[
\frac{x + 2 - 3x + 3}{(x-1)(x+2)} < 0
\]

Simplify the numerator:

\[
\frac{-2x + 5}{(x-1)(x+2)} < 0
\]

### Step 3: Analyze the resulting expression
The inequality we now have is:

\[
\frac{-2x + 5}{(x-1)(x+2)} < 0
\]

This is algebraically equivalent to the inequality:

\[
\frac{5 - 2x}{(x-1)(x+2)} < 0
\]

Since \(-2x + 5\) can be factored or rewritten as \(5 - 2x\), the inequality simplifies to:

\[
\frac{5}{(x-1)(x+2)} < 0
\]

### Step 4: Multiply both sides by -1
To get an equivalent expression where the numerator is positive, multiply the entire inequality by \(-1\):

\[
\frac{-5}{(x-1)(x+2)} > 0
\]

But this does not change the nature of the inequality; the critical points (where the expression changes sign) remain the same. Therefore, the sign of the entire expression depends on the sign of the denominator. 

Thus, we get:

\[
\frac{5}{(x-1)(x+2)} < 0
\]

### Step 5: Conclude equivalence
Both inequalities depend on the same expression \(\frac{5}{(x-1)(x+2)} < 0\), meaning that they will have the same solution set. The sign of the expression will change when the product \((x-1)(x+2)\) changes sign.

### Conclusion
Since both inequalities reduce to examining the sign of \(\frac{5}{(x-1)(x+2)}\), they are algebraically equivalent. They both represent the same condition on \(x\) for the inequality to hold, proving their equivalence.

Would you like to explore further details or ask any questions?

### 8 Related Questions:
1. How can finding a common denominator help in comparing two rational expressions?
2. What steps are essential when subtracting fractions with different denominators in an inequality?
3. How does multiplying an inequality by a negative number affect the inequality sign?
4. What are the critical points in a rational inequality, and how are they determined?
5. Can rational inequalities be solved by graphing the related function?
6. How can you verify the solution set of a rational inequality?
7. What are the possible pitfalls when simplifying inequalities involving fractions?
8. How does the inequality \(\frac{5}{(x-1)(x+2)} < 0\) compare to a simpler inequality like \(x^2 - 4 < 0\)?

### Tip:
Always keep track of domain restrictions in rational inequalities, as they can cause division by zero, which is undefined and may lead to incorrect solutions.

Explore why the inequalities 1/(x-1) < 3/(x+2) and 5/((x-1)(x+2)) < 0 are algebraically equivalent. This detailed analysis covers step-by-step reasoning and algebraic manipulations involved. Suitable for Grades 10-12 students studying inequalities in algebra.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation