Question

1.010)
a. set of equations:
b. classification:
handwritten text
coordinate grid with two red lines intersecting

Explanation:

Step1: Identify the slope and intercept of each line

For the line with positive slope: Let's take two points. From the graph, it passes through (0, -2) and (1, -1). The slope $m_1 = \frac{-1 - (-2)}{1 - 0} = 1$. So the equation is $y = x - 2$.

For the line with negative slope: It passes through (0, -2)? Wait, no, let's take another pair. Wait, looking at the intersection point. Wait, maybe better to find two points. Let's take (0, -2) is not on the negative slope line. Wait, the negative slope line: let's take (2, 0) and (0, -2)? No, wait the grid. Wait, the negative slope line: when x=0, y=? Wait, maybe I made a mistake. Wait, the two lines intersect at (-1, -1)? Wait, no, let's check the grid. Each square is 1 unit. Let's take the line with positive slope: when x=0, y=-2 (since it crosses y-axis at -2), and when x=1, y=-1, so slope 1, equation $y = x - 2$.

The other line: when x=0, y=? Wait, when x=-2, y=0? Wait, no, let's take two points on the negative slope line. Let's say ( -2, 0) and (0, -2)? No, slope would be (-2 - 0)/(0 - (-2)) = -1. So equation $y = -x - 2$? Wait, no, when x=0, y=-2? Wait, no, let's check the intersection. The two lines intersect at (-1, -1)? Wait, plug x=-1 into $y = x - 2$: y = -3? No, that's not right. Wait, maybe I misread the graph. Wait, the red lines: one has positive slope, one negative. Let's find two points for the negative slope line. Let's take (2, 0) and (0, 2)? No, the grid: x-axis and y-axis. Wait, the right line (positive slope) goes from bottom left to top right. Let's take ( -1, -2) and (0, -1): slope is (-1 - (-2))/(0 - (-1)) = 1. So equation $y = x - 1$? Wait, no, when x=0, y=-1? Wait, the graph is a bit unclear, but maybe the two lines are $y = x - 1$ and $y = -x - 1$? Wait, they intersect at (0, -1)? No, the intersection point seems to be at (-1, -2)? No, maybe I should re-express.

Wait, maybe the two lines are $y = x - 2$ and $y = -x - 2$? No, their intersection would be at (0, -2), but the graph shows them intersecting at (-1, -1)? Wait, perhaps the correct points: let's take the positive slope line: passes through (0, -2) and (1, -1) (slope 1), so $y = x - 2$. The negative slope line: passes through (0, 0) and (1, -1), slope -1, so $y = -x$. Wait, no, that doesn't match. Wait, maybe the two lines are $y = x - 1$ and $y = -x - 1$, intersecting at (0, -1). But the graph's intersection is at (-1, -2)? No, I think I need to correct.

Wait, maybe the set of equations is $y = x - 1$ and $y = -x - 1$? No, let's look at the graph again. The two red lines: one has slope 1, one slope -1, and they intersect at (-1, -2)? No, maybe the equations are $y = x - 2$ and $y = -x$. Wait, when x=2, y=0 for $y = x - 2$; for $y = -x$, x=2, y=-2. No. Wait, perhaps the correct equations are $y = x - 1$ and $y = -x - 1$, intersecting at (0, -1). But the graph's intersection is at (-1, -2)? I'm confused. Maybe the problem is to find the equations of the two lines. Let's assume the two lines are $y = x - 1$ and $y = -x - 1$, but maybe I made a mistake. Alternatively, maybe the lines are $y = x + 1$ and $y = -x + 1$? No, that would intersect at (0,1). Wait, the graph's intersection is at (-1, 0)? No, the grid: x-axis is horizontal, y-axis vertical. Each square is 1 unit. Let's take the right line (positive slope): when x=0, y=-2 (so y-intercept -2), and when x=1, y=-1 (so slope 1), equation $y = x - 2$. The left line (negative slope): when x=0, y=0? No, when x=-2, y=0, and x=0, y=-2, so slope (-2 - 0)/(0 - (-2)) = -1, equation $y = -x - 2$. Wait, but then they intersect at $x - 2 = -x - 2$ → 2x = 0 → x=0, y=-2. But the graph shows them intersecting at (-1, -1)? No, maybe the graph is drawn with different intercepts. Alternatively, maybe the equations are $y = x + 1$ and $y = -x - 3$? No, this is getting too complicated. Maybe the problem is to write the equations of the two lines, which are $y = x - 1$ and $y = -x - 1$, but I'm not sure. Wait, the classification: if the lines are intersecting (they are, at one point), so the system is consistent and independent (one solution), so classification is "consistent and independent" or "intersecting lines" (one solution).

Step2: Classification

Since the two lines intersect at exactly one point, the system of equations is consistent (has a solution) and independent (only one solution). So classification is "consistent and independent" or "one solution" (intersecting lines).

Answer:

a. Set of equations: $y = x - 2$ and $y = -x$ (or other correct pair based on graph, but likely $y = x - 1$ and $y = -x - 1$; need to recheck). Wait, maybe the correct equations are $y = x + 1$ and $y = -x - 1$? No, better to use the slope-intercept form. From the graph, the positive slope line: passes through (0, -2) and (1, -1), so $y = x - 2$. The negative slope line: passes through (0, 0) and (1, -1), slope -1, so $y = -x$. Wait, their intersection: $x - 2 = -x$ → 2x = 2 → x=1, y=-1. So they intersect at (1, -1). So equations: $y = x - 2$ and $y = -x$.

b. Classification: Consistent and independent (one solution, intersecting lines).