In the xy-plane, region R consists of all the points \(\mathrm{(x,y)}\) such that 2x + 3y leq 6. Is the...

Understanding the Question

Let's break down what we're being asked. Region R consists of all points \(\mathrm{(x,y)}\) where \(\mathrm{2x + 3y \leq 6}\). We need to determine if point \(\mathrm{(r,s)}\) is in this region.

What We Need to Determine: Is \(\mathrm{2r + 3s \leq 6}\)?

This is a yes/no question. For sufficiency, we need to definitively answer either YES (the point is in region R) or NO (the point is not in region R).

Key Insight: Region R includes everything on or below the line \(\mathrm{2x + 3y = 6}\). Think of this line as a boundary—points below it are in the region, points above it are not.

Analyzing Statement 1

Statement 1 tells us: The point \(\mathrm{(r,s)}\) lies on the line \(\mathrm{3r + 2s = 6}\).

Here's the critical observation: We have a point that must lie on one specific line \(\mathrm{(3r + 2s = 6)}\), and we need to know if it's in a region defined by a different line \(\mathrm{(2r + 3s \leq 6)}\).

Since these two lines have different slopes, they must intersect at exactly one point. This means the line \(\mathrm{3r + 2s = 6}\) must cross through region R's boundary—it will have:

• Some points where \(\mathrm{2r + 3s \leq 6}\) (inside region R)
• Some points where \(\mathrm{2r + 3s > 6}\) (outside region R)

Let's verify with two quick examples:

• If \(\mathrm{r = 0}\): then \(\mathrm{s = 3}\) → Check: \(\mathrm{2(0) + 3(3) = 9 > 6}\) → NOT in region R
• If \(\mathrm{r = 2}\): then \(\mathrm{s = 0}\) → Check: \(\mathrm{2(2) + 3(0) = 4 < 6}\) → IS in region R

Since different points on the line give different answers, we cannot determine whether \(\mathrm{(r,s)}\) is in region R.

Statement 1 is NOT sufficient. [STOP - Not Sufficient!]

This eliminates answer choices A and D.

Analyzing Statement 2

Now let's forget Statement 1 completely and analyze Statement 2 by itself.

Statement 2 tells us: The point must satisfy \(\mathrm{r \leq 3}\) and \(\mathrm{s \leq 2}\).

This creates a rectangular region in the plane. To determine if this entire rectangle is inside region R, we can use a strategic approach: check the corners.

Why corners? Because with linear inequalities like \(\mathrm{2r + 3s \leq 6}\), the maximum and minimum values occur at the vertices of the region.

Testing the extreme points:

• Origin \(\mathrm{(0,0)}\): \(\mathrm{2(0) + 3(0) = 0 < 6}\) → IS in region R
• Corner \(\mathrm{(3,2)}\): \(\mathrm{2(3) + 3(2) = 12 > 6}\) → NOT in region R

Since the rectangular region contains points both inside and outside region R, we cannot determine whether \(\mathrm{(r,s)}\) is in region R.

Statement 2 is NOT sufficient. [STOP - Not Sufficient!]

This eliminates answer choices B and D (already eliminated).

Combining Statements

Combined Information: The point \(\mathrm{(r,s)}\) must:

1. Lie on the line \(\mathrm{3r + 2s = 6}\) (from Statement 1)

2. Satisfy \(\mathrm{r \leq 3}\) and \(\mathrm{s \leq 2}\) (from Statement 2)

This means we're looking at the intersection of a line with a rectangle—which gives us a line segment.

From Statement 1, we can express: \(\mathrm{s = (6 - 3r)/2 = 3 - 1.5r}\)

For this to also satisfy \(\mathrm{s \leq 2}\):

• \(\mathrm{3 - 1.5r \leq 2}\)
• \(\mathrm{1 \leq 1.5r}\)
• \(\mathrm{r \geq 2/3}\)

Combined with \(\mathrm{r \leq 3}\) from Statement 2, we get: \(\mathrm{2/3 \leq r \leq 3}\)

Now let's check the endpoints of this line segment:

• At \(\mathrm{r = 2/3}\): \(\mathrm{s = 2}\) → Check: \(\mathrm{2(2/3) + 3(2) = 4/3 + 6 = 22/3 \approx 7.33 > 6}\) → NOT in region R
• At \(\mathrm{r = 3}\): \(\mathrm{s = -3/2}\) → But wait! This violates \(\mathrm{s \leq 2}\) from Statement 2, so we need to find where the line exits the rectangle.

Actually, when \(\mathrm{s = 0}\) (the bottom boundary of our rectangle): \(\mathrm{3r = 6}\), so \(\mathrm{r = 2}\)

• At \(\mathrm{r = 2}\): \(\mathrm{s = 0}\) → Check: \(\mathrm{2(2) + 3(0) = 4 < 6}\) → IS in region R

So even with both statements combined, we have:

• Some points on our line segment that ARE in region R (like \(\mathrm{(2,0)}\))
• Some points on our line segment that are NOT in region R (like \(\mathrm{(2/3,2)}\))

The statements together are NOT sufficient. [STOP - Not Sufficient!]

This eliminates answer choices A, B, C, and D.

The Answer: E

Since we can find points satisfying both statements that are inside region R AND other points satisfying both statements that are outside region R, we cannot definitively answer whether \(\mathrm{(r,s)}\) is in region R.

Answer Choice E: "The statements together are not sufficient."