Let's solve the problem.

We are given the equation:

$\frac{12x^2 + 6x - 23}{bx - 3} = -2x - \frac{23}{bx - 3},$

and need to determine the value of $b$ such that this equation holds for all $x \neq \frac{3}{b}$.

Step-by-Step Solution:

1. Eliminate the denominator $bx - 3$

Multiply through by $bx - 3$, which is valid for $x \neq \frac{3}{b}$:

$12x^2 + 6x - 23 = (-2x)(bx - 3) - 23.$

2. Expand the terms on the right-hand side:

Distribute $-2x$ on the right-hand side:

$12x^2 + 6x - 23 = -2x \cdot bx + 6x - 23.$

Simplify:

$12x^2 + 6x - 23 = -2bx^2 + 6x - 23.$

3. Compare coefficients of $x^2$, $x$, and constants:

Since this equation must hold for all $x$, compare the coefficients of $x^2$, $x$, and the constant term:

• Coefficient of $x^2$: $12 = -2b$, so: $b = -6.$

• Coefficient of $x$: $6 = 6$ (this matches, so it's consistent).

• Constant term: $-23 = -23$ (this also matches).

Final Answer:

The value of $b$ is:

$\boxed{-6}.$

Would you like a detailed explanation of any step? 😊

Related Questions:

1. How do we compare coefficients to solve equations?
2. What happens if $b = 0$ in the given equation?
3. Can you generalize this approach for equations with similar fractions?
4. How does multiplying by the denominator affect the equation?
5. What restrictions does $x \neq \frac{3}{b}$ impose on the solution?

Tip:

When solving equations involving fractions, always consider restrictions on the variable due to denominators before simplifying!