Question

which graph shows a system of equations that solves $-\frac{1}{x - 1} = 4$, and the solution itself?

Explanation:

Step1: Rewrite the equation as a system

To solve \(-\frac{1}{x - 1}=4\), we can set up a system of equations: \(y = -\frac{1}{x - 1}\) and \(y = 4\). The solution to the original equation will be the \(x\)-coordinate of the intersection point of these two graphs.

Step2: Analyze the vertical asymptote

The function \(y = -\frac{1}{x - 1}\) has a vertical asymptote at \(x = 1\) (since the denominator is zero when \(x = 1\)). So we can eliminate graphs with vertical asymptotes at other \(x\)-values (like \(x=\frac{1}{2}\) or \(x = \frac{2}{3}\) or \(x=-\frac{2}{3}\)).

Step3: Analyze the horizontal line \(y = 4\)

The line \(y = 4\) is a horizontal line. We need to find the graph where the horizontal line \(y = 4\) intersects the hyperbola \(y=-\frac{1}{x - 1}\), and the hyperbola has a vertical asymptote at \(x = 1\).

Looking at the graphs:

• The first graph has a vertical asymptote at \(x=\frac{1}{2}\), so it's incorrect.
• The second graph has a vertical asymptote at \(x = 1\) (we can check the position of the hyperbola's branches; the vertical asymptote is where the two branches of the hyperbola approach, and here it's at \(x = 1\)) and the horizontal line \(y = 4\).
• The third graph has a vertical asymptote at \(x=-\frac{2}{3}\) (or similar non - 1 value), so it's incorrect.

Answer:

The Middle Graph (the second graph among the three shown)