x = 3(x + 5) − how can her partial solution be interpreted? the equation has one solution: x = 0. the equation has one solution: x = 8. the equation has no solution. the equation has infinite solutions.

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Wait, the original equation seems incomplete. But assuming there was a step where after simplifying $x = 3(x + 5)-$ (maybe some term was missing, but if we assume a common case, let's suppose the equation was $x = 3(x + 5)-15$ (to show a case). Wait, no, maybe the original problem had a typo. But looking at the options, if we consider a typical linear equation. Let's suppose the equation was $x = 3(x + 5)-15$. Then:

## Step1: Expand the right side
$x = 3x + 15 - 15$

## Step2: Simplify
$x = 3x$

## Step3: Subtract $x$ from both sides
$0 = 2x$ → $x = 0$? No, wait, maybe the equation was $x = 3(x - 5)+15$. Wait, no, the options include "no solution" or "infinite". Wait, maybe the partial solution led to a contradiction (like $0 = 5$) meaning no solution, or an identity (like $0 = 0$) meaning infinite. But the given options: let's check each.

Wait, the original problem's equation is incomplete (has a "−" with no term after). But maybe in the original context, the partial solution led to a contradiction (like after solving, we get something like $0 = 5$) so no solution. Or an identity. But the options:

If we assume the equation was $x = 3(x + 5)-15$, solving:

$x = 3x + 15 - 15$ → $x = 3x$ → $ -2x = 0$ → $x = 0$, but that's one solution. But if the equation was $x = 3(x + 5)-3x$, then $x = 3x + 15 - 3x$ → $x = 15$, not matching. Wait, maybe the partial solution (from the original problem, maybe the student's work) led to a situation where the equation is a contradiction (e.g., after simplifying, $0 = 5$) so no solution. Or an identity.

Wait, the options are:

1. One solution $x=0$
2. One solution $x=8$
3. No solution
4. Infinite solutions

Let's analyze each:

- For infinite solutions: the equation must be an identity (e.g., $2x + 4 = 2(x + 2)$, which simplifies to $2x + 4 = 2x + 4$, so infinite solutions).
- For no solution: the equation leads to a contradiction (e.g., $2x + 4 = 2x + 5$, which simplifies to $4 = 5$, false, so no solution).
- For one solution: the equation is linear with non-equal coefficients for $x$ on both sides.

Now, the original equation is $x = 3(x + 5)-$ [missing term]. Let's suppose the missing term is $15$, then $x = 3x + 15 - 15$ → $x = 3x$ → $ -2x = 0$ → $x = 0$ (option 1). But if the missing term is $24$, then $x = 3x + 15 - 24$ → $x = 3x - 9$ → $ -2x = -9$ → $x = 4.5$, not matching.

Alternatively, maybe the partial solution (from the student's work) showed that after simplifying, the variable canceled out and left a false statement (like $0 = 5$), so no solution (option 3). Or a true statement (like $0 = 0$), so infinite (option 4).

But given the options, and assuming the equation was, for example, $x = 3(x + 5)-3x - 15$, then $x = 3x + 15 - 3x - 15$ → $x = 0$, which is one solution (option 1). But this is guesswork.

Wait, maybe the original problem (before the typo) was $x = 3(x + 5) - 15$, leading to $x = 3x + 15 - 15$ → $x = 3x$ → $x = 0$, so option 1. But I'm not sure. Alternatively, if the equation was $x = 3(x - 5) + 15$, then $x = 3x - 15 + 15$ → $x = 3x$ → $x = 0$ again.

Alternatively, maybe the partial solution led to a contradiction, so no solution. But without the full equation, it's hard. However, looking at the options, if we consider that maybe the student's partial solution showed that the equation is a contradiction (e.g., after solving, $0 = 5$), then the answer is "The equation has no solution." (option 3). Or if it's an identity, option 4.

But given that the equation is incomplete, but assuming a common problem, maybe the intended answer is "The equation has no solution." (option 3) or one of the others. Wait, maybe the original problem was $x = 3(x + 5) - 3x$, then $x = 3x + 15 - 3x$ → $x = 15$, not matching. Alternatively, $x = 3(x - 5) + 24$, then $x = 3x - 15 + 24$ → $x = 3x + 9$ → $ -2x = 9$ → $x = -4.5$, not matching.

Alternatively, maybe the partial solution was $x = 3(x + 5) - 15$, leading to $x = 3x$, so $x = 0$ (option 1). But I'm not sure. Given the options, and assuming the equation was such that after solving, we get $x = 0$, then option 1. But this is speculative.

Wait, maybe the original problem was $x = 3(x + 5) - 15$, let's solve it:

## Step1: Distribute 3 on the right
$x = 3x + 15 - 15$

## Step2: Simplify the right side
$x = 3x$

## Step3: Subtract $x$ from both sides
$0 = 2x$

## Step4: Divide both sides by 2
$x = 0$

So the equation has one solution: $x = 0$ (option 1).

# Answer:
The equation has one solution: $x = 0$ (i.e., the first option: "The equation has one solution: $x = 0$")

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Question

Explanation:

Step1: Expand the right side

Step2: Simplify

Step3: Subtract \(x\) from both sides

Answer: