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The image contains a section of a math worksheet titled "SOLVING RATIONAL FUNCTIONS," labeled as "Seatwork No. 3". The task is to solve for the variable in the following rational equations and to check the solutions:

1. \( \frac{10}{3} = \frac{4}{x} + 3 \)
2. \( x = \frac{6}{x - 1} \)
3. \( \frac{16}{x^2 - 16} = 2 \)

### Let's solve these step by step:

#### 1. \( \frac{10}{3} = \frac{4}{x} + 3 \)
Subtract 3 from both sides:
\[
\frac{10}{3} - 3 = \frac{4}{x}
\]
Convert \( 3 \) into a fraction with denominator 3:
\[
\frac{10}{3} - \frac{9}{3} = \frac{4}{x}
\]
This simplifies to:
\[
\frac{1}{3} = \frac{4}{x}
\]
Cross-multiply:
\[
x = 12
\]

#### 2. \( x = \frac{6}{x - 1} \)
Multiply both sides by \( x - 1 \) to eliminate the denominator:
\[
x(x - 1) = 6
\]
Expand the left-hand side:
\[
x^2 - x = 6
\]
Rearrange the equation:
\[
x^2 - x - 6 = 0
\]
Factor the quadratic equation:
\[
(x - 3)(x + 2) = 0
\]
Thus, \( x = 3 \) or \( x = -2 \).

However, we must check for extraneous solutions. Substituting \( x = 3 \) and \( x = -2 \) into the original equation:
- For \( x = 3 \): \( 3 = \frac{6}{3 - 1} = 3 \), which is correct.
- For \( x = -2 \): \( -2 = \frac{6}{-2 - 1} = -2 \), which is correct as well.

Both solutions are valid.

#### 3. \( \frac{16}{x^2 - 16} = 2 \)
First, recognize that \( x^2 - 16 = (x - 4)(x + 4) \), so rewrite the equation:
\[
\frac{16}{(x - 4)(x + 4)} = 2
\]
Multiply both sides by \( (x - 4)(x + 4) \):
\[
16 = 2(x - 4)(x + 4)
\]
Divide both sides by 2:
\[
8 = (x - 4)(x + 4)
\]
Expand the right-hand side:
\[
8 = x^2 - 16
\]
Rearrange the equation:
\[
x^2 = 24
\]
Take the square root of both sides:
\[
x = \pm \sqrt{24} = \pm 2\sqrt{6}
\]

Would you like further clarification on any of these solutions? Here are 5 related questions to consider:

1. What steps are involved in solving rational equations in general?
2. How can you identify extraneous solutions in rational equations?
3. What is the importance of factoring in solving rational equations?
4. How can you simplify complex fractions involving polynomials?
5. How do you verify the solutions of a rational equation?

**Tip:** Always check for extraneous solutions when solving rational equations, as the variable may create restrictions that are not valid in the original equation.

Solve for the value of the variable in the following rational equations and check the solutions:
1. 10/3 = 4/x + 3
2. x = 6/(x - 1)
3. 16/(x^2 - 16) = 2

This guide demonstrates solving rational equations using techniques like cross-multiplication, factoring, and solving quadratic equations. Examples include solving equations like 10/3 = 4/x + 3 and x = 6/(x - 1), appropriate for students in Grades 9-10.

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