To determine which statement best describes the solutions to the given equation:
[tex]\[
\frac{x}{3} + \frac{x-1}{x+4} = \frac{x+3}{x+4}
\][/tex]
Let's go through solving the equation step-by-step.
1. Combine the Fractions on the Right Side:
We observe that the denominators [tex]\((x + 4)\)[/tex] are the same for the second term on the left and the term on the right. We can simplify:
[tex]\[
\frac{x}{3} + \frac{(x-1) - (x+3)}{x+4} = 0
\][/tex]
Simplifying the numerator inside the second fraction:
[tex]\[
\frac{x}{3} + \frac{x-1 - x-3}{x+4} = \frac{x}{3} - \frac{4}{x+4}
\][/tex]
2. Set the Entire Expression to Zero:
Now, set the combined fractions to zero:
[tex]\[
\frac{x}{3} - \frac{4}{x+4} = 0
\][/tex]
3. Eliminate the Fractions:
To clear the fractions, multiply through by a common denominator, which would be [tex]\(3(x+4)\)[/tex]:
[tex]\[
x(x+4) - 4 \cdot 3 = 0
\][/tex]
This simplifies to:
[tex]\[
x^2 + 4x - 12 = 0
\][/tex]
4. Solve the Quadratic Equation:
Use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] where [tex]\(a = 1\)[/tex], [tex]\(b = 4\)[/tex], and [tex]\(c = -12\)[/tex]:
[tex]\[
x = \frac{-4 \pm \sqrt{16 + 48}}{2} = \frac{-4 \pm \sqrt{64}}{2} = \frac{-4 \pm 8}{2}
\][/tex]
This gives two solutions:
[tex]\[
x = \frac{4}{2} = 2 \quad \text{and}\quad x = \frac{-12}{2} = -6
\][/tex]
5. Check for Extraneous Solutions:
Substitute [tex]\(x = -6\)[/tex] and [tex]\(x = 2\)[/tex] into the original equation to validate if they are solutions:
- For [tex]\(x = 2\)[/tex]:
[tex]\[
\frac{2}{3} + \frac{1}{6} = \frac{5}{6} \quad \text{which simplifies correctly to}\quad \frac{5}{6} = \frac{5}{6}
\][/tex]
[tex]\(x = 2\)[/tex] is a valid solution.
- For [tex]\(x = -6\)[/tex]:
[tex]\[
\frac{-6}{3} + \frac{-7}{-2} = -2 + \frac{7}{-2}
\quad \text{denominator cannot be zero}
\][/tex]
As [tex]\(x = -6\)[/tex] results in an invalid denominator in the given equation.
6. Conclude the Number of Solutions:
Since [tex]\(x = 2\)[/tex] is a valid solution and [tex]\(x = -6\)[/tex] is an extraneous solution:
The correct statement is:
D. The equation has two valid solutions and no extraneous solutions.
With this detailed explanation, we conclude that the solution accurately corresponds to:
D. The equation has two valid solutions and no extraneous solutions.
