Question 251·Hard·Algebra

On the left side, try factoring $4x+12$. Can you cancel anything with the denominator $x+3$?

After simplifying the left side, rewrite the equation. What happens if you subtract 4 from both sides? What kind of equation do you get for the remaining fraction?

For a fraction of the form $\dfrac{a}{b}$, under what condition does it equal 0? Does $\dfrac{2}{x+3}$ ever meet that condition for any real $x$?

In Desmos, enter the left side as y = (4x + 12)/(x + 3) and the right side as y = 4 + 2/(x + 3). Make sure both graphs are turned on.

Visually inspect the graph and, if needed, zoom in and out. Use the tap/click feature on Desmos to find any intersection points between the two curves. The $x$-coordinates of intersection points (excluding $x=-3$, where there is a vertical asymptote) are the real solutions to the equation.

Both denominators are $x+3$, so $x$ cannot make a denominator zero.

Set the denominator not equal to zero:

Any solution we find must satisfy $x \neq -3$.

Look at the left-hand side:

Factor the numerator:

So

for all $x$ such that $x \neq -3$ (we already excluded $x=-3$).

Using the simplification from Step 2, the original equation becomes

Subtract 4 from both sides:

Now think about what this means: a fraction $\dfrac{2}{x+3}$ has numerator 2 and denominator $x+3$.

For $\dfrac{2}{x+3}$ to equal 0, its numerator would have to be 0. But the numerator here is 2, which is never 0, no matter what real value $x$ has.

So $\dfrac{2}{x+3} = 0$ has no real solution, and $x = -3$ was already excluded because it makes the denominators zero.

Therefore, the original equation has zero real solutions, which corresponds to choice D.