Question

a city determines that a planned community must have at least 4 acres of developed and open space, and the difference between the number of developed acres, y, and the number of open acres, x, can be no more than 1. which graph represents the system of inequalities for this scenario?
x + y ≥ 4
y - x ≤ 1

Explanation:

Step1: Analyze \( x + y \geq 4 \)

Rewrite as \( y \geq -x + 4 \). The boundary line is \( y=-x + 4 \) (slope -1, y - intercept 4), and we shade above the line (since \( y\geq \)).

Step2: Analyze \( y - x \leq 1 \)

Rewrite as \( y \leq x + 1 \). The boundary line is \( y = x+1 \) (slope 1, y - intercept 1), and we shade below the line (since \( y\leq \)).

Step3: Check Intersection

Find the region that is above \( y=-x + 4 \) and below \( y = x + 1 \). Also, \( x\geq0,y\geq0 \) (acres can't be negative). The correct graph should have the overlapping region of these two shadings. From the options, the graph (need to check the visual, but based on the inequalities, the region where \( y\geq -x + 4 \) (upper part of that line) and \( y\leq x + 1 \) (lower part of that line) with \( x,y\geq0 \) would match the second graph (middle one) when considering the shading directions. Wait, actually, let's re - check:

For \( x + y\geq4 \), when \( x = 0,y = 4 \); \( y = 0,x = 4 \). The line goes from (0,4) to (4,0), shaded above. For \( y - x\leq1\) or \( y\leq x + 1 \), when \( x = 0,y = 1 \); \( y = 0,x=-1 \) (but x can't be negative, so in positive x - y, the line goes from (0,1) with slope 1. Shaded below. The overlapping region should be where both shadings meet, and also x and y are non - negative. Looking at the options, the middle graph (the second one in the row) has the correct shading: above \( y=-x + 4 \) and below \( y = x + 1 \), with x and y non - negative.

Answer:

The Middle Graph (assuming the middle one in the set of graphs provided)