Questions: College Algebra Hannah Ociesa 01/11/25 8:26 PM This quiz: 40 point(s) Question 1 of 20 This question: 2 point(s) possible The following rational equation has denominators that contain variables. For this equation, a. Write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation. 4/x = 29/(8x) + 3 a. What is/are the value or values of the variable that make(s) the denominators zero? x= (Simplify your answer. Use a comma to separate answers as needed.) b. Solve the equation. Select the correct choice below and, it necessary, fill in the answer box to complete your choice.

Solution Steps

The denominators in the equation are \( x \) and \( 8x \). To find the values of \( x \) that make the denominators zero, set each denominator equal to zero and solve for \( x \):

1. \( x = 0 \)
2. \( 8x = 0 \implies x = 0 \)

Thus, the only value of \( x \) that makes the denominators zero is \( x = 0 \). This is the restriction on the variable.

\[ x = \boxed{0} \]

The given equation is:

\[ \frac{4}{x} = \frac{29}{8x} + 3 \]

To solve for \( x \), first eliminate the denominators by multiplying both sides of the equation by the least common denominator (LCD), which is \( 8x \):

\[ 8x \cdot \frac{4}{x} = 8x \cdot \left( \frac{29}{8x} + 3 \right) \]

Simplify each term:

\[ 8 \cdot 4 = 29 + 24x \]

\[ 32 = 29 + 24x \]

Subtract \( 29 \) from both sides:

\[ 32 - 29 = 24x \]

\[ 3 = 24x \]

Divide both sides by \( 24 \):

\[ x = \frac{3}{24} = \frac{1}{8} \]

Check that \( x = \frac{1}{8} \) does not violate the restriction \( x \neq 0 \). Since \( \frac{1}{8} \neq 0 \), the solution is valid.

Final Answer