I interpret the problem as asking the following. Let $f(x)=-x^3+3x^2+9x-11$ and let $g(x)=9x+b$. For what values of $b$ does the equation $f(x)=g(x)$ have three different solutions? Equivalently, for what values of $b$ does the curve $y=f(x)$ meet the line $y=g(x)$ at three different points?
We arrive at the equation $-x^3+3x^2+9x-11=9x+b$. This simplifies to
$$x^3-3x^2=-(11+b).$$
We now draw the graph of $y=x^3-3x^2$. Note that $\frac{dy}{dx}=3x^2-6x$. This is $0$ at $x=0$ and $x=2$.
So $x^3-3x^2$ is steadily increasing until $x=0$, where it reaches $0$. Then it steadily decreases until $x=2$, where it reaches the value $-4$. After that, the curve steadily increases.
By looking at the graph, we can see that the horizontal line $y=-(11+b)$ meets the curve $y=x^3-3x^2$ at $3$ distinct places precisely if $-(11+b)$ lies between $0$ and $-4$ (but not including these points). So the $b$'s that work are all $b$ in the open interval $(-11, -7)$.
There are in a sense three points of intersection when $b=-11$, but two happen to coincide. The same is true when $b=-7$. For values of $b$ not in $[-11,-7]$, there is only one point of intersection.
Remark: An equivalent solution that is not as evident geometrically is that we will have three meeting points between the values of $b$ for which the line $y=9x+b$ is tangent to the curve $y=-x^3+3x^2+9x-11$. For tangency, slopes must match, so we want
$$-3x^2+6x+9=9,$$
from which we get $x=0$ and $x=2$.
In order for $y=9x+b$ to be tangent to $y=f(x)$ at $x=0$, values must match, meaning that $-11=b$. For tangency at $x=2$, values must match, giving $-8+12+18-11=18+b$, meaning that $b=-7$.
The answer is already in the comments, but I'll try explain it a bit more verbosely. If you solve an equation, what you're actually doing is writing a proof that the equation is logically equivalent to some statement $x = ...$, where the dots stand for some number. You do that by writing down a series of equivalent statements, one below the other. In your case, you start out with $$
5x + 3 = 8x + 3
$$
Then you subtract $3$ from both sides. You may do that, because two numbers are equal exactly if those numbers minus 3 are equal. So the statement $$
5x = 8x
$$
is logically equivalent to the equation you started with. Now you divide by $x$. Are you allowed to do that? Is it true that two numbers are equal exactly if they're equal if divided by the same divisor? Turns out they are, except if the divisor is zero. In that case, the division isn't allowed, and so the statement you get - even if it formally looks valid - has no more logical connection to the original equation. In particular, the original equation may have a solution, but the statement you get after the division may be simply untrue. So, by dividing by $x$, you implicitly assume that $x\neq 0$! Everything that follows will be depend on that assumption. In your case, you do the division and end up with $$
5 = 8 \text{,}
$$
an obvious contradiction. But that statement is only equivalent to the original equation if $x \neq 0$ - after all, we had to assume that to get this far. So we now know that, indeed, $x$ has to be zero for the equation to hold, since assuming that it isn't zero got us into trouble. We don't yet know if it actually will hold for $x=0$, thought, so the last step is to set $x=0$ in the original equation, and verify that it checks out.
Best Answer
If we assume that all linear equations have the form:
$$ ax + b= 0 $$
(which is completely valid and should be how we view linear equations)
then linear equations have either 1, 0, or infinite solutions. It's quite simple if $a \neq 0$ then they have exactly one solution: $x = -\frac{b}{a}$.
On the other hand, if $a = 0$ then if $b = 0$ we have infinite solutions (any value of $x$ solves $0x + 0 = 0$) and if $a = 0$ and $b \neq 0$ then there are no solutions (there is no value of $x$ that makes $0x + 1 = 0$, for example).