After you multiply the second equation by $2$, compare the coefficients of $y$ in the two equations. What happens if you subtract one equation from the other?

When you eliminate one variable, think about what the resulting equation tells you. For infinitely many solutions, should you get a specific value for $x$, a contradiction like $0=5$, or an identity like $0=0$?

In Desmos, type k= and create a slider. This will let you change the value of $k$ easily.

Type the two equations using $k$:

• (k+2)x + (k-1)y = k
• 3x + (k-1)/2*y = 2 Desmos will graph two lines that depend on the current value of $k$.

Move the $k$ slider to each of the answer choices ($1$, $2$, $4$, $6$) in turn. For most values, the two lines will intersect at one point or be parallel. Note the value of $k$ where the two graphs completely overlap (they are the exact same line); that is the value where the system has infinitely many solutions.

Multiply the second equation by $2$ to remove the fraction:

Original system:

After multiplying the second equation by $2$:

Now both equations have the same $y$-coefficient, $(k-1)$, which makes elimination easier.

Subtract the second (new) equation from the first to eliminate $y$:

The $y$-terms cancel:

So any solution of the system must satisfy $(k-4)x = k-4$.

Rewrite

as

This product is zero if either $x-1=0$ or $k-4=0$. If $x-1=0$, then $x$ is fixed at a single value, which corresponds to one solution (after finding $y$). For the system to have infinitely many solutions, the elimination step must produce an equation that is true for all $x$, not one that fixes $x$ to a single number — that happens when the factor on $x$ is zero so the equation becomes $0=0$.

From $(k-4)(x-1)=0$, getting an identity (true for all $x$) requires $k-4=0$, so $k=4$.

Substitute $k=4$ back into the system:

First equation:

Second equation:

Multiply this by $2$:

Both equations are exactly the same line, so the system has infinitely many solutions when $k=4$.

Therefore, the correct answer is $4$.