Once you rewrite the right-hand side, notice that both sides of the equation have a factor involving $(x+3)$. Try moving everything to one side and factoring out $(x+3)$.

If you end up with an equation like $A(x+3)=0$, when does this hold for every real value of $x$, not just one specific value? What must be true about $A$?

Set the coefficient that multiplies $(x+3)$ equal to $0$ and solve that simple equation for $k$.

In Desmos, type y = 6(k - 2)(x + 3). When prompted, add a slider for k so you can change its value.

On a new line, type y = 3(4x + 12). Desmos will graph this as a straight line.

Move the k slider and watch the graph of y = 6(k - 2)(x + 3). Find the value of k for which this graph lies exactly on top of the graph of y = 3(4x + 12) for all visible $x$-values. The k value at that moment is the one that makes the original equation have infinitely many solutions (the two sides are the same for every $x$).

Start by simplifying the expression on the right:

• Factor $4x+12$ as $4(x+3)$.
• Then

So the original equation becomes

Subtract $12(x+3)$ from both sides so that one side is $0$:

Now factor out the common factor $(x+3)$:

The equation

will be true for all real $x$ (infinitely many solutions) only if the coefficient in brackets is $0$.

• If $x+3=0$, then $x=-3$, which is just one solution.
• To have every $x$ work, the factor multiplying $(x+3)$ must be $0$, so the equation becomes $0 \cdot (x+3)=0$, which is always true.

So we need to solve:

Now solve the equation

Simplify step by step:

So the value of $k$ that makes the equation have infinitely many real solutions is $4$.