The graph shows the system y = 2/(x - 2) - 1 and y = 2|x - 2| - 4. Which set of points is the complete solution: (4,0); (0,0) and (4,0); (0,0), (0,-2) and (4,0); or (0,-2) and (4,0)?

Step-by-step solution

1. 1

   Substitute u = x - 2 to exploit the shared shift. Both curves are built around $x=2$, so the substitution removes it from both:

   $\frac{2}{u}-1=2|u|-4\quad\Longrightarrow\quad \frac{2}{u}+3=2|u|,\qquad u\neq 0.$

   The restriction $u\neq 0$ (that is, $x\neq 2$) comes from the vertical asymptote of the first curve.

2. 2

   Split on the sign of u — case u > 0. Then $|u|=u$, and multiplying through by $u>0$ keeps the equation intact:

   $2+3u=2u^{2}\ \Longrightarrow\ 2u^{2}-3u-2=0\ \Longrightarrow\ (2u+1)(u-2)=0.$

   The roots are $u=2$ and $u=-\tfrac12$; only $u=2$ satisfies $u>0$, giving $x=4$.

3. 3

   Case u < 0. Then $|u|=-u$, and multiplying by the negative $u$ flips nothing in an equation:

   $2+3u=-2u^{2}\ \Longrightarrow\ 2u^{2}+3u+2=0.$

   Its discriminant is $9-16=-7<0$, so there is no real root and this case contributes nothing.

4. 4

   Find the y-value of the surviving solution. At $x=4$:

   $y=\frac{2}{4-2}-1=1-1=0,\qquad y=2|4-2|-4=4-4=0.$

   Both curves give $y=0$, so $(4,0)$ is genuinely on both.

5. 5

   Test the distractor points. At $x=0$: the first curve gives $y=\tfrac{2}{-2}-1=-2$, while the second gives $y=2(2)-4=0$. So $(0,0)$ lies only on the absolute-value graph and $(0,-2)$ only on the rational graph — neither is a solution of the system. That is exactly the trap: a point on one curve is not a solution unless it is on both.

6. 6

   State the complete solution. The system has exactly one solution, $(4,0)$.