Solve Inequality: 5/(x+1) >= 3/(x-2)

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the wild world of mathematics, specifically tackling a tricky inequality: $\frac{5}{x+1} \geq \frac{3}{x-2}$. This kind of problem can look intimidating at first glance, but don't sweat it! With a systematic approach, we can break it down and find the solution set. We'll walk through each step, making sure you understand the 'why' behind it all. So, grab your thinking caps, and let's get this solved!

Understanding Inequalities and Potential Pitfalls

Before we jump into solving, let's talk about what inequalities are and why they sometimes throw us for a loop. Unlike equations where we're looking for specific values of 'x' that make both sides equal, inequalities ask for a range of values that make one side greater than or equal to (or less than or equal to) the other. This means our answer won't be a single number, but an interval or a set of intervals on the number line. The real trick with rational inequalities (those with variables in the denominator, like ours) is handling those denominators. You cannot just multiply both sides by a denominator like $(x+1)$ or $(x-2)$ without considering its sign. If it's positive, the inequality direction stays the same; if it's negative, you have to flip the inequality sign. This is a common mistake, and it's why we often use a different strategy: bringing everything to one side and finding a common denominator.

Step-by-Step Solution to $\frac{5}{x+1} \geq \frac{3}{x-2}$

Alright, let's get down to business. Our goal is to find the values of 'x' that satisfy $\frac{5}{x+1} \geq \frac{3}{x-2}$.

Step 1: Bring all terms to one side.

To make things easier and avoid multiplying by potentially negative values, we'll move the $\frac{3}{x-2}$ term to the left side. This gives us:

$$\frac{5}{x+1} - \frac{3}{x-2} \geq 0$$

Step 2: Find a common denominator.

The common denominator for $(x+1)$ and $(x-2)$ is simply their product: $(x+1)(x-2)$. Now, we rewrite each fraction with this new denominator:

$$\frac{5(x-2)}{(x+1)(x-2)} - \frac{3(x+1)}{(x+1)(x-2)} \geq 0$$

Step 3: Combine the fractions.

Now that they have the same denominator, we can combine the numerators:

$$\frac{5(x-2) - 3(x+1)}{(x+1)(x-2)} \geq 0$$

Step 4: Simplify the numerator.

Let's expand and simplify the top part:

$$\frac{(5x - 10) - (3x + 3)}{(x+1)(x-2)} \geq 0$$

$$\frac{5x - 10 - 3x - 3}{(x+1)(x-2)} \geq 0$$

$$\frac{2x - 13}{(x+1)(x-2)} \geq 0$$

Step 5: Identify critical points.

Critical points are the values of 'x' that make the numerator or the denominator equal to zero. These are the points where the expression might change its sign.

• Numerator: $2x - 13 = 0 \implies 2x = 13 \implies x = \frac{13}{2} = 6.5$
• Denominator: $(x+1) = 0 \implies x = -1$
• Denominator: $(x-2) = 0 \implies x = 2$

So, our critical points are $x = -1$, $x = 2$, and $x = 6.5$. These points divide the number line into four intervals: $(-\infty, -1)$, $(-1, 2)$, $(2, 6.5)$, and $(6.5, \infty)$.

Step 6: Test the intervals.

We need to pick a test value within each interval and plug it into our simplified inequality $\frac{2x - 13}{(x+1)(x-2)} \geq 0$ to see if it holds true. Remember, the critical points themselves need special consideration.

• Interval 1: $(-\infty, -1)$ Let's pick $x = -2$.

  $$\frac{2(-2) - 13}{(-2+1)(-2-2)} = \frac{-4 - 13}{(-1)(-4)} = \frac{-17}{4}$$

  This is negative ($< 0$), so this interval is not part of our solution.

• Interval 2: $(-1, 2)$ Let's pick $x = 0$.

  $$\frac{2(0) - 13}{(0+1)(0-2)} = \frac{-13}{(1)(-2)} = \frac{-13}{-2} = 6.5$$

  This is positive ($

0$), so this interval is part of our solution.

• Interval 3: $(2, 6.5)$ Let's pick $x = 3$.

  $$\frac{2(3) - 13}{(3+1)(3-2)} = \frac{6 - 13}{(4)(1)} = \frac{-7}{4}$$

  This is negative ($< 0$), so this interval is not part of our solution.

• Interval 4: $(6.5, \infty)$ Let's pick $x = 7$.

  $$\frac{2(7) - 13}{(7+1)(7-2)} = \frac{14 - 13}{(8)(5)} = \frac{1}{40}$$

  This is positive ($

0$), so this interval is part of our solution.

Step 7: Consider the endpoints (critical points).

Our inequality is $\geq 0$, meaning we include values that make the expression equal to zero.

• The numerator being zero gives us $x = 6.5$. Since $6.5$ makes the expression equal to 0, and $0 \geq 0$ is true, $x = 6.5$ is included in our solution.
• The denominator being zero gives us $x = -1$ and $x = 2$. Division by zero is undefined, so these values are excluded from our solution.

The Final Solution Set

Combining our findings from testing the intervals and considering the endpoints, our solution set consists of the interval $(-1, 2)$ and the interval $[6.5, \infty)$.

In interval notation, this is: $(-1, 2) \cup [6.5, \infty)$

Looking at the options provided:

A. $(-\infty,-1] \cup[2, \infty)$ B. $[-1,2]$ C. $[-1,2] \cup(6.5, \infty)$ D. $(-1,2) \cup[6.5, \infty)$

Our solution matches option D.

There you have it, guys! Solving rational inequalities might seem like a puzzle, but by systematically bringing terms together, finding common denominators, identifying critical points, and testing intervals, we can confidently find the correct solution. Keep practicing these, and they'll become second nature. If you found this breakdown helpful, make sure to check out more math articles here on Plastik Magazine!