**Find all solutions to the equation.**

\[\dfrac{x+4}{x-1}=\dfrac{-x^2+3x+12}{(x+3)(x-1)}\]

Choose all answers that apply:

Choose all answers that apply:

We are given the equation:

\[
\dfrac{x+4}{x-1} = \dfrac{-x^2 + 3x + 12}{(x+3)(x-1)}
\]

### Step 1: Eliminate the denominator
The denominator \( x - 1 \) appears on both sides of the equation, so as long as \( x \neq 1 \) (to avoid division by zero), we can multiply both sides by \( x - 1 \) to eliminate the denominator:

\[
x + 4 = \dfrac{-x^2 + 3x + 12}{x + 3}
\]

### Step 2: Multiply both sides by \( x + 3 \)
To eliminate the denominator on the right-hand side, multiply both sides of the equation by \( x + 3 \):

\[
(x + 4)(x + 3) = -x^2 + 3x + 12
\]

### Step 3: Expand both sides
Expand both sides of the equation:

- Left-hand side:
\[
(x + 4)(x + 3) = x^2 + 3x + 4x + 12 = x^2 + 7x + 12
\]

- Right-hand side:
\[
-x^2 + 3x + 12
\]

Thus, the equation becomes:
\[
x^2 + 7x + 12 = -x^2 + 3x + 12
\]

### Step 4: Move all terms to one side
Move all terms to the left-hand side:

\[
x^2 + 7x + 12 + x^2 - 3x - 12 = 0
\]

Combine like terms:

\[
2x^2 + 4x = 0
\]

### Step 5: Factor the equation
Factor out the common factor \( 2x \):

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

### Step 6: Solve for \( x \)
Now, solve for \( x \) by setting each factor equal to zero:

1. \( 2x = 0 \) gives \( x = 0 \)
2. \( x + 2 = 0 \) gives \( x = -2 \)

### Step 7: Check for extraneous solutions
We must check that the solutions do not make any denominator in the original equation equal to zero.

The original equation has denominators \( x - 1 \) and \( (x + 3)(x - 1) \). Therefore, \( x \neq 1 \) and \( x \neq -3 \).

The solutions \( x = 0 \) and \( x = -2 \) do not make the denominators zero, so they are valid solutions.

### Final Answer
The solutions to the equation are:

\[
x = 0 \quad \text{and} \quad x = -2
\]

Would you like a more detailed verification or help with similar problems?

Here are some related questions:
1. What would happen if the equation had complex solutions?
2. How do extraneous solutions arise when solving rational equations?
3. How can you graphically interpret the solutions to rational equations?
4. What are some common mistakes when solving rational equations?
5. How do we approach equations with multiple fractions on both sides?

**Tip:** Always check that potential solutions do not cause any denominator to be zero before finalizing them!

Learn how to solve the rational equation (x+4)/(x-1) = (-x^2 + 3x + 12)/((x+3)(x-1)) with a detailed step-by-step solution. This problem covers key algebraic concepts, including rational equations and quadratic equations. Suitable for students in Grades 10-12.

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