The following rational equation has denominators that contain variables.

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.

[tex]\[
\frac{2}{x} = \frac{13}{8x} + 3
\][/tex]

a. What is/are the value or values of the variable that make(s) the denominators zero?

[tex]\[
x = \square
\][/tex]

(Simplify your answer. Use a comma to separate answers as needed.)



Answer :

To tackle this problem, let's break it down into steps:

### Part a: Finding the Restrictions on the Variable

The given equation is:
[tex]\[ \frac{2}{x} = \frac{13}{8x} + 3 \][/tex]

We need to determine the values of [tex]\( x \)[/tex] that make any of the denominators equal to zero.

1. Denominator from the first term: [tex]\(\frac{2}{x}\)[/tex]
[tex]\[ x \neq 0 \][/tex]

2. Denominator from the second term: [tex]\(\frac{13}{8x}\)[/tex]
[tex]\[ 8x \neq 0 \][/tex]
[tex]\[ x \neq 0 \][/tex] (since 8 is a non-zero constant)

Thus, the value of [tex]\( x \)[/tex] that makes the denominator zero is:
[tex]\[ x = 0 \][/tex]

### Part b: Solving the Equation

Knowing that [tex]\( x = 0 \)[/tex] is a restriction and cannot be a solution, let's solve the equation:

[tex]\[ \frac{2}{x} = \frac{13}{8x} + 3 \][/tex]

To eliminate the denominators, we can multiply every term in the equation by the least common multiple (LCM) of [tex]\( x \)[/tex] and [tex]\( 8x \)[/tex], which is [tex]\( 8x \)[/tex]:

[tex]\[ 8x \cdot \frac{2}{x} = 8x \cdot \frac{13}{8x} + 8x \cdot 3 \][/tex]

Simplify each term:
[tex]\[ 8 \cdot 2 = 8 \cdot \frac{13}{8} + 8x \cdot 3 \][/tex]
[tex]\[ 16 = 13 + 24x \][/tex]

Now, solve for [tex]\( x \)[/tex]:

[tex]\[ 16 = 13 + 24x \][/tex]

Subtract 13 from both sides:

[tex]\[ 16 - 13 = 24x \][/tex]
[tex]\[ 3 = 24x \][/tex]

Divide by 24:

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

### Conclusion

a. What is/are the value or values of the variable that make a denominator zero?
[tex]\[ x = 0 \][/tex]

b. Solution to the equation keeping the restrictions in mind:
[tex]\[ x = \frac{1}{8} \][/tex]

So, the correct solution to the problem is [tex]\( x = \frac{1}{8} \)[/tex], which is within the allowable range since it does not make the denominators zero.