After you solve for $x$ in each case, plug that expression back into the inequality for that case to see which $k$ values make it valid. Ѕ‌οu‍rсе: ani‍кο.​aі

Exactly one solution happens at the transition between “no solutions” and “two solutions,” when one case just stops being valid.

Enter $y=|2x-k|$ and $y=x+4$. When prompted, add a slider for $k$.

Move the slider and observe how many intersection points appear (0, 1, or 2). Pay attention to the change from 0 intersections to 2 intersections. ЅАT рr‍еp​ bу Anіko.аі

Adjust $k$ to the exact value where the graphs touch at exactly one point (the moment between having 0 and having 2 intersections). Read that $k$ value from the slider.

Consider the sign of $2x-k$.

Case 1: $2x-k\ge 0$ so $|2x-k|=2x-k$.

Case 2: $2x-k<0$ so $|2x-k|=-(2x-k)=-2x+k$.

Case 1: Powerе‍d‌ bу Anik‍o

Condition: $2x-k\ge 0 \Rightarrow 2(k+4)-k\ge 0 \Rightarrow k+8\ge 0 \Rightarrow k\ge -8$.

Case 2:

Condition: $2x-k<0 \Rightarrow 2\left(\frac{k-4}{3}\right)-k<0$. Multiplying by $3$:

• If $k>-8$, both cases produce valid solutions, so there are two real solutions.
• If $k<-8$, neither case produces a valid solution, so there are no real solutions.
• If $k=-8$, Case 1 is valid and gives $x=k+4=-4$. Case 2 requires $k>-8$, so it is not valid.

Therefore, the equation has exactly one real solution only when $k=-8$, so the correct choice is $-8$.