After you collect the $x$ terms, isolate $x$ by moving the constant terms (the ones without $x$) to the other side and then dividing by the coefficient of $x$, which involves $n+2$.

Your expression for $x$ will involve division by $n+2$. For what value of $n$ does $n+2$ become $0$, and what does that mean for the equation?

Once you find the value of $n$ that makes $n+2=0$, substitute it back into the original equation and see whether you get a true statement (like $5=5$), a false statement (like $5=10$), or something that still has $x$.

For each answer choice, rewrite the equation as a single expression equal to $0$: $(n+2)x + 5 - \big(2(n+2)x + 5n\big)$. In Desmos, replace $n$ with the specific value from a choice (for example, use $-1$, $0$, $2$, or $-2$) and enter that as $y_1=$.

For each value of $n$ you test, you will see a line for $y_1$. Check whether this line ever crosses the $x$-axis (where $y=0$). If it crosses once, the equation has one solution for that $n$. If the line is horizontal and never touches the $x$-axis (always above or always below), then there is no value of $x$ that makes the original equation true for that $n$—that $n$ is the correct choice.

Start with the equation:

Subtract $(n+2)x$ from both sides to get all $x$ terms together:

This simplifies to:

From

subtract $5n$ from both sides:

Now, as long as $n+2 \neq 0$, you can divide both sides by $n+2$ to solve for $x$:

This shows that for most values of $n$, the equation has exactly one solution for $x$. The only time we cannot do this is when $n+2=0$.

The fraction $\dfrac{5-5n}{n+2}$ is not defined when the denominator $n+2$ is $0$ because division by $0$ is impossible.

So we must find the value of $n$ that makes

This value of $n$ is a special case that we have to check separately by substituting it back into the original equation to see what happens.

Solve the small equation from the previous step:

Now substitute $n=-2$ into the original equation:

becomes

which simplifies to

This is impossible, so there is no value of $x$ that makes the equation true when $n=-2$. Therefore, the equation has no solution when $n=-2$.