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**SOLVING RATIONAL FUNCTIONS**

**SEATWORK NO. 3**

**B. Solve for the value of the variable in the following rational equations. Remember to check your solutions to verify if it satisfies the given equation.**

1. \(\frac{10}{3} = \frac{4}{x} + 3\)

2. \(x = \frac{6}{x - 1}\)

3. \(\frac{16}{x^2 - 16} = \frac{2}{x - 4}\)

Let's solve each of these equations step-by-step.

### 1. Solving the equation: \(\frac{10}{3} = \frac{4}{x} + 3\)

First, isolate \(\frac{4}{x}\) by subtracting 3 from both sides:

\[
\frac{10}{3} - 3 = \frac{4}{x}
\]

Convert 3 to a fraction with a denominator of 3:

\[
\frac{10}{3} - \frac{9}{3} = \frac{4}{x}
\]

This simplifies to:

\[
\frac{1}{3} = \frac{4}{x}
\]

Cross-multiply to solve for \(x\):

\[
1 \cdot x = 3 \cdot 4 \implies x = 12
\]

So, \(x = 12\).

### 2. Solving the equation: \(x = \frac{6}{x - 1}\)

Cross-multiply to get rid of the fraction:

\[
x(x - 1) = 6
\]

Expand and rearrange into a quadratic equation:

\[
x^2 - x - 6 = 0
\]

Factor the quadratic:

\[
(x - 3)(x + 2) = 0
\]

Set each factor to zero and solve for \(x\):

\[
x - 3 = 0 \implies x = 3, \quad x + 2 = 0 \implies x = -2
\]

Both \(x = 3\) and \(x = -2\) are solutions. However, we must check if they satisfy the original equation:

- For \(x = 3\): \[ 3 = \frac{6}{3 - 1} \implies 3 = 3 \] (valid)
- For \(x = -2\): \[ -2 = \frac{6}{-2 - 1} \implies -2 = \frac{6}{-3} \implies -2 = -2 \] (valid)

Thus, both \(x = 3\) and \(x = -2\) are correct solutions.

### 3. Solving the equation: \(\frac{16}{x^2 - 16} = \frac{2}{x - 4}\)

Factor the denominator \(x^2 - 16\) as a difference of squares:

\[
x^2 - 16 = (x - 4)(x + 4)
\]

Rewrite the equation:

\[
\frac{16}{(x - 4)(x + 4)} = \frac{2}{x - 4}
\]

Cross-multiply:

\[
16(x - 4) = 2(x - 4)(x + 4)
\]

Divide both sides by \(x - 4\), assuming \(x \neq 4\):

\[
16 = 2(x + 4)
\]

Solve for \(x\):

\[
16 = 2x + 8 \implies 8 = 2x \implies x = 4
\]

However, \(x = 4\) makes the denominator zero in the original equation, which is undefined. Therefore, there are **no valid solutions** for this equation.

---

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### Related Questions:
1. How do you solve rational equations with multiple variables?
2. What are the common mistakes to avoid when solving rational equations?
3. Can rational functions have more than one solution?
4. What are the real-world applications of solving rational equations?
5. How do you handle extraneous solutions in rational equations?

### Tip:
When solving rational equations, always check for values that make the denominator zero, as they can invalidate potential solutions.

Solve the following rational equations:
1. (10/3) = (4/x) + 3
2. x = 6 / (x - 1)
3. (16/(x^2 - 16)) = (2/(x - 4))

Learn how to solve rational equations such as (10/3) = (4/x) + 3 and x = 6 / (x - 1). This problem covers key algebraic concepts, including rational equations, factoring, and checking for extraneous solutions. Suitable for students in Grades 9-12.

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