Question

Solve for to determine if the
equation has one, none, or
infinite solutions.
(1 point)
x
5 (2x + 2) − 7x = 3x + 6
The equation has no
solution.
The equation has one
solution: x = 6.
The equation has
infinite solutions.
The equation has one
solution: x = 10.

Answer 1

To determine the number of solutions for the equation

\[
\frac{x}{5} (2x + 2) - 7x = 3x + 6,
\]

let's simplify it step-by-step.

1. Distribute \(\frac{x}{5}\) to \((2x + 2)\):

\[
\frac{x}{5} \cdot 2x + \frac{x}{5} \cdot 2 = \frac{2x^2}{5} + \frac{2x}{5}.
\]

So, the equation can be rewritten as:

\[
\frac{2x^2}{5} + \frac{2x}{5} - 7x = 3x + 6.
\]

2. Combine like terms. First, let's get all terms on one side of the equation:

\[
\frac{2x^2}{5} + \frac{2x}{5} - 7x - 3x - 6 = 0.
\]

3. Combine the \(x\) terms:

\[
\frac{2x^2}{5} + \frac{2x}{5} - 10x - 6 = 0.
\]

4. To eliminate fractions, multiply through by 5:

\[
2x^2 + 2x - 50x - 30 = 0.
\]

5. Combine like terms:

\[
2x^2 - 48x - 30 = 0.
\]

6. Divide the entire equation by 2 for simplification:

\[
x^2 - 24x - 15 = 0.
\]

7. Now, we will use the quadratic formula to find the roots:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
\]

where \(a = 1\), \(b = -24\), and \(c = -15\):

\[
x = \frac{24 \pm \sqrt{(-24)^2 - 4(1)(-15)}}{2(1)} = \frac{24 \pm \sqrt{576 + 60}}{2} = \frac{24 \pm \sqrt{636}}{2} = \frac{24 \pm \sqrt{4 \cdot 159}}{2} = \frac{24 \pm 2\sqrt{159}}{2} = 12 \pm \sqrt{159}.
\]

Since \(\sqrt{159}\) is not a perfect square, there are two distinct real solutions. Hence:

**Conclusion:** The equation has **two distinct solutions**, meaning it has **one solution** in terms of any option. This doesn't match the provided multiple-choice answers exactly. However, since it doesn't have infinite solutions or no solutions, the most appropriate conclusion is that it has one solution (although it has two distinct values of \(x\)).

**Final Answer:** The equation has two distinct solutions (none of the provided statements are accurate).