Activity #3

A. Find the solution for each given rational equation.

1. 3/(x + 1) = 2/(x - 3)

2. (2x)/(x + 1) + 5/(2x) = 2

3. (x ^ 2 - 10)/(x - 1) = (- 14 - 5x)/(x - 1)

Let's solve each of the given rational equations step by step.

### 1. Solve the equation: \(\frac{3}{x + 1} = \frac{2}{x - 3}\)

To solve this equation, we can cross multiply:

\[
3(x - 3) = 2(x + 1)
\]

Now, distribute and simplify both sides:

\[
3x - 9 = 2x + 2
\]

Next, subtract \(2x\) from both sides:

\[
x - 9 = 2
\]

Add 9 to both sides:

\[
x = 11
\]

So, the solution is \(x = 11\).

### 2. Solve the equation: \(\frac{2x}{x + 1} + \frac{5}{2x} = 2\)

To solve this, first find a common denominator for the terms on the left side:

The common denominator is \((x + 1)(2x)\):

\[
\frac{2x \cdot 2x}{(x + 1)(2x)} + \frac{5(x + 1)}{(x + 1)(2x)} = 2
\]

Simplify the numerators:

\[
\frac{4x^2 + 5(x + 1)}{(x + 1)(2x)} = 2
\]

Distribute the 5:

\[
\frac{4x^2 + 5x + 5}{(x + 1)(2x)} = 2
\]

Cross multiply to eliminate the fraction:

\[
4x^2 + 5x + 5 = 2 \cdot (x + 1)(2x)
\]

Expand the right-hand side:

\[
4x^2 + 5x + 5 = 4x^2 + 2x
\]

Subtract \(4x^2\) from both sides:

\[
5x + 5 = 2x
\]

Subtract \(2x\) from both sides:

\[
3x + 5 = 0
\]

Subtract 5 from both sides:

\[
3x = -5
\]

Divide by 3:

\[
x = -\frac{5}{3}
\]

So, the solution is \(x = -\frac{5}{3}\).

### 3. Solve the equation: \(\frac{x^2 - 10}{x - 1} = \frac{-14 - 5x}{x - 1}\)

Since the denominators are the same, we can set the numerators equal to each other:

\[
x^2 - 10 = -14 - 5x
\]

Move all terms to one side:

\[
x^2 + 5x + 4 = 0
\]

Factor the quadratic equation:

\[
(x + 4)(x + 1) = 0
\]

Set each factor equal to zero:

\[
x + 4 = 0 \quad \text{or} \quad x + 1 = 0
\]

Solve for \(x\):

\[
x = -4 \quad \text{or} \quad x = -1
\]

So, the solutions are \(x = -4\) and \(x = -1\).

### Summary of Solutions:
1. \(x = 11\)
2. \(x = -\frac{5}{3}\)
3. \(x = -4\) or \(x = -1\)

Would you like any further details or explanations? Here are some related questions you might find useful:

1. What happens if the equation has no solution?
2. How do we handle extraneous solutions in rational equations?
3. Can you provide examples of rational equations with undefined solutions?
4. How do we solve rational equations with different denominators?
5. What are the steps to check if a solution is valid?

**Tip:** Always check if the solution makes the denominator zero in the original equation. If it does, the solution is not valid.

Explore how to solve complex rational equations with detailed step-by-step solutions. This page covers examples like 3/(x + 1) = 2/(x - 3), (2x)/(x + 1) + 5/(2x) = 2, and (x^2 - 10)/(x - 1) = (-14 - 5x)/(x - 1). Suitable for Grades 9-12.

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