Question

y = -2x - 2
x - y = 8
plot two lines by clicking the graph.
click a line to delete it.
graph with x and y axes from -10 to 10

Explanation:

Step1: Analyze \( y = -2x - 2 \)

This is in slope - intercept form \( y=mx + b \), where slope \( m=-2 \) and y - intercept \( b = - 2 \). To plot, start at the y - intercept \((0,-2)\). Then, use the slope: from \((0,-2)\), move 1 unit right (since the run is 1) and 2 units down (because the slope is - 2) to get the next point \((1,-4)\), or 1 unit left and 2 units up to get \((-1,0)\).

Step2: Analyze \( x - y=8 \)

Rewrite it in slope - intercept form. Solve for \( y \): \( y=x - 8 \). Here, slope \( m = 1 \) and y - intercept \( b=-8 \). Start at the y - intercept \((0,-8)\). Then, use the slope: move 1 unit right and 1 unit up to get \((1,-7)\), or 1 unit left and 1 unit down to get \((-1,-9)\).

To plot the lines:

• For \( y=-2x - 2 \), plot the points \((0,-2)\) and \((-1,0)\) (or other points found using the slope) and draw the line through them.
• For \( y=x - 8 \) (from \( x - y = 8 \)), plot the points \((0,-8)\) and \((1,-7)\) (or other points found using the slope) and draw the line through them.

(Note: Since the problem is about plotting lines on a graph, the key is to identify the key points for each linear equation using their slope - intercept forms and then plot those points to draw the lines. If we were to find the intersection (the solution of the system), we can set the two equations equal: \(-2x-2=x - 8\). Solving for \( x \): \(-2x-x=-8 + 2\), \(-3x=-6\), \(x = 2\). Then \( y=-2(2)-2=-6 \). So the intersection point is \((2,-6)\), but the main task here is plotting the lines.)

Answer:

To plot \( y=-2x - 2 \): Plot \((0,-2)\) and \((-1,0)\) (or other valid points) and draw the line. To plot \( x - y = 8\) (or \( y=x - 8\)): Plot \((0,-8)\) and \((1,-7)\) (or other valid points) and draw the line. (If the goal was to find the intersection, the solution is \( x = 2,y=-6 \), but based on the plotting task, the lines are drawn through the identified points.)