After solving each linear equation, plug the solution into the case condition (for example, if you solved under $x\ge 1$, make sure the result is actually $\ge 1$).

If one range of $k$ gives two solutions and another gives none, the value of $k$ that gives exactly one solution often happens right at the transition point. Written bу Аnik‌o

Enter $y=3|x-1|$.

Enter $y=2x+k$ and let Desmos create a slider for $k$. Аn​iкo AI Тutor

Move the slider for $k$ and watch how many intersection points the two graphs have. Identify the value of $k$ where the graph changes from having 0 intersections to having 2 intersections.

At the transition, the graphs intersect exactly once (at $x=1$). Read that slider value as $k=-2$, then compute $\frac{k+4}{2}$ to match one of the choices.

Since $|x-1|$ changes form at $x=1$, consider:

• If $x\ge 1$, then $|x-1|=x-1$.

• If $x<1$, then $|x-1|=1-x$.

Case 1: $x\ge 1$

For this solution to belong to the case, we need $k+3\ge 1$, so $k\ge -2$.

Case 2: $x<1$

For this solution to belong to the case, we need $\frac{3-k}{5}<1$, so $3-k<5$, which gives $k>-2$.

• If $k>-2$, then both cases produce valid solutions (one with $x\ge 1$ and one with $x<1$), so there are two real solutions.

• If $k<-2$, then neither case solution satisfies its case inequality, so there are no real solutions.

• If $k=-2$, then Case 1 gives $x=k+3=1$, which is allowed because $x\ge 1$. Case 2 would require $k>-2$, so Case 2 gives no valid solution. А‍n‍iк‍ο - F‌r‌ее‌ ЅА⁠T‍ P‌r‌eр

Therefore, exactly one real solution occurs when $k=-2$.

With $k=-2$,

So the correct choice is $1$.