Start from $(k+2)x + 6 = 6$ and try to isolate the term with $x$. What equation involving $x$ and $k$ do you get after subtracting $6$ from both sides?

After you simplify, you will get an equation of the form $(k+2)x = 0$. For which value of $k$ does this equation no longer restrict $x$ to just one value?

You want the equation to become a true statement like $6=6$ for all $x$. What must be true about $k+2$ for the $x$ term to disappear completely?

Think of the equation as $(k+2)x+6-6=0$, which simplifies to $(k+2)x=0$. You want to see for which choice of $k$ this is true for all $x$.

For each option, replace $k$ with that number and type the expression into Desmos as y = (k+2)x with $k$ replaced (for example, y = (0+2)x, y = (2+2)x, etc.).

For each tested value, look at the graph:

• If the graph is a slanted line through the origin, the equation only holds for one $x$.
• The correct choice will make the graph collapse to the horizontal line $y=0$ for all $x$, meaning the equation is always true and has infinitely many solutions.

A linear equation in $x$ has infinitely many solutions only if both sides of the equation are identical for all $x$.

That means, after simplifying, the equation should turn into something like $6=6$, with no $x$ term left.

Start from the given equation:

Subtract $6$ from both sides:

So the equation becomes $(k+2)x = 0$.

Consider the equation $(k+2)x = 0$.

• If $k+2 \neq 0$, then you can divide both sides by $k+2$ and get $x=0$, which is one solution.
• To have infinitely many solutions, the $x$ term must disappear, meaning its coefficient must be $0$.

So we need:

Solve the equation from the previous step:

So the value of $k$ that makes the equation have infinitely many solutions is $-2$.