Question

w x y z
y = -2x - 3
y = 2x - 1

y = x + 4
y = -x + 2

y = -2x - 3
y = x + 4

y = 2x - 1
y = -x + 2

Answer:

To solve this, we analyze the slopes and y - intercepts of each pair of linear equations and match them to the graph.

Pair 1: \(y = - 2x-3\) and \(y = 2x - 1\)

• For \(y=-2x - 3\), the slope \(m_1=-2\) (negative, steep) and y - intercept \(b_1=-3\).
• For \(y = 2x-1\), the slope \(m_2 = 2\) (positive, steep) and y - intercept \(b_2=-1\).
• These two lines have steep slopes (magnitude 2) and y - intercepts - 3 and - 1.

Pair 2: \(y=x + 4\) and \(y=-x + 2\)

• For \(y=x + 4\), slope \(m_3 = 1\) (positive, moderate) and y - intercept \(b_3 = 4\).
• For \(y=-x + 2\), slope \(m_4=-1\) (negative, moderate) and y - intercept \(b_4 = 2\).
• These two lines have moderate slopes (magnitude 1) and y - intercepts 4 and 2.

Pair 3: \(y=-2x-3\) and \(y=x + 4\)

• \(y=-2x-3\) has slope - 2 (steep, negative) and \(y=x + 4\) has slope 1 (moderate, positive).

Pair 4: \(y = 2x-1\) and \(y=-x + 2\)

• \(y = 2x-1\) has slope 2 (steep, positive) and \(y=-x + 2\) has slope - 1 (moderate, negative).

Now, looking at the graph (even with partial view), we can match:

• The pair \(y=-2x-3\) and \(y = 2x - 1\) (steep slopes) will correspond to one of the "steep" looking lines. Let's assume the first pair (steep) is matched to a box (say W, but we need to check the graph's visual cues). But since we know the slopes:

1. \(y=-2x-3\) and \(y = 2x - 1\) → Let's say this is matched to a box (e.g., if we consider the steep lines, this pair has slopes with magnitude 2).
2. \(y=x + 4\) and \(y=-x + 2\) → These have slopes with magnitude 1, so they are the "less steep" lines.
3. \(y=-2x-3\) and \(y=x + 4\) → One steep negative, one moderate positive.
4. \(y = 2x-1\) and \(y=-x + 2\) → One steep positive, one moderate negative.

If we assume the graph has two sets of lines: one set with steep slopes (slope magnitude 2) and one set with moderate slopes (slope magnitude 1).

• The pair \(y=-2x-3\) and \(y = 2x - 1\) (steep slopes) will be matched to a box (let's say the first box, but we can also use the fact that in typical such problems, the pair \(y=x + 4\) and \(y=-x + 2\) have y - intercepts 4 and 2 which are positive, so they are the upper - most lines.

Let's do the matching properly:

• Pair \(y=-2x-3\) and \(y = 2x - 1\): These two lines have steep slopes. Let's say this pair is matched to a box (e.g., if we label the boxes from left to right as W, X, Y, Z, and the steep - sloped lines are on the left - ish part, but we can also use the slope and intercept values. The correct matching (assuming standard problems) is:
• \(y=-2x-3\) and \(y = 2x - 1\) → Let's say W (or the first box)
• \(y=x + 4\) and \(y=-x + 2\) → Let's say X (or the second box)
• \(y=-2x-3\) and \(y=x + 4\) → Let's say Y (or the third box)
• \(y = 2x-1\) and \(y=-x + 2\) → Let's say Z (or the fourth box)

But to get the exact match, we can also solve for intersection points (optional, but helps):

For \(y=-2x-3\) and \(y = 2x - 1\), set \(-2x-3=2x - 1\)
\(-2x-2x=-1 + 3\)
\(-4x=2\)
\(x=-\frac{1}{2}\), \(y=-2\times(-\frac{1}{2})-3=1 - 3=-2\)

For \(y=x + 4\) and \(y=-x + 2\), set \(x + 4=-x + 2\)
\(x+x=2 - 4\)
\(2x=-2\)
\(x=-1\), \(y=-1 + 4 = 3\)

If we assume the graph has two intersection points: one at \((-\frac{1}{2},-2)\) (from the steep lines) and one at \((-1,3)\) (from the moderate lines).

So the correct matching (assuming the boxes are labeled W, X, Y, Z from left to right) is:

• \(y=-2x-3\) and \(y = 2x - 1\) → W
• \(y=x + 4\) and \(y=-x + 2\) → X
• \(y=-2x-3\) and \(y=x + 4\) → Y
• \(y = 2x-1\) and \(y=-x + 2\) → Z

(Note: The exact box - to - pair match may depend on the full graph's visual, but the above is the logical way to match based on slope and intercept analysis)

Final Answer (assuming standard matching):

• \(y=-2x-3\), \(y = 2x - 1\) → W
• \(y=x + 4\), \(y=-x + 2\) → X
• \(y=-2x-3\), \(y=x + 4\) → Y
• \(y = 2x-1\), \(y=-x + 2\) → Z