Explore \[ \begin{array}{l}\text { The following lines are hidden in the graph: } \\ y=3 x+6 \\ y=3(x-5) \\ y=-\frac{1}{2} x-15 \\ -x+y=12\end{array} \] Capture all of the lines by writing coordinates for points on the lines.

To explore the hidden lines in the graph, let's start by identifying specific points on each line based on their equations. 1. For the equation \(y = 3x + 6\), we can find points by substituting values for \(x\): - If \(x = 0\), then \(y = 3(0) + 6 = 6\) → Point: \((0, 6)\) - If \(x = 2\), then \(y = 3(2) + 6 = 12\) → Point: \((2, 12)\) 2. For the equation \(y = 3(x - 5)\), simplifying this gives us \(y = 3x - 15\): - If \(x = 5\), then \(y = 3(5) - 15 = 0\) → Point: \((5, 0)\) - If \(x = 7\), then \(y = 3(7) - 15 = 6\) → Point: \((7, 6)\) 3. For the equation \(y = -\frac{1}{2}x - 15\): - If \(x = 0\), then \(y = -\frac{1}{2}(0) - 15 = -15\) → Point: \((0, -15)\) - If \(x = -10\), then \(y = -\frac{1}{2}(-10) - 15 = -10\) → Point: \((-10, -10)\) 4. For the equation \(-x + y = 12\), we can rearrange it to \(y = x + 12\): - If \(x = 0\), then \(y = 0 + 12 = 12\) → Point: \((0, 12)\) - If \(x = 3\), then \(y = 3 + 12 = 15\) → Point: \((3, 15)\) Collectively, the coordinates for points on these lines are: 1. \((0, 6)\), \((2, 12)\) 2. \((5, 0)\), \((7, 6)\) 3. \((0, -15)\), \((-10, -10)\) 4. \((0, 12)\), \((3, 15)\) Explore these points on a graph to visualize the lines!