Question

choose the graph that matches the system of inequalities.
$-x + y \leq -1$
$x + 2y \geq 4$
\\(\bigcirc\\) a. graph a
\\(\bigcirc\\) b. graph b

Explanation:

Step1: Rewrite the first inequality

Rewrite \(-x + y \leq -1\) in slope - intercept form (\(y=mx + b\), where \(m\) is the slope and \(b\) is the y - intercept).
Add \(x\) to both sides: \(y\leq x - 1\). The boundary line is \(y=x - 1\) (a straight line with slope \(m = 1\) and y - intercept \(b=-1\)), and since the inequality is \(\leq\), the line is solid and we shade below the line.

Step2: Rewrite the second inequality

Rewrite \(x + 2y\geq4\) in slope - intercept form.
Subtract \(x\) from both sides: \(2y\geq - x + 4\).
Divide both sides by 2: \(y\geq-\frac{1}{2}x + 2\). The boundary line is \(y =-\frac{1}{2}x+2\) (a straight line with slope \(m =-\frac{1}{2}\) and y - intercept \(b = 2\)), and since the inequality is \(\geq\), the line is solid and we shade above the line.

Step3: Analyze the graphs

• For the first inequality \(y\leq x - 1\), the line \(y=x - 1\) passes through \((0,-1)\) and \((1,0)\). The region \(y\leq x - 1\) is below this line.
• For the second inequality \(y\geq-\frac{1}{2}x + 2\), the line \(y =-\frac{1}{2}x + 2\) passes through \((0,2)\) and \((4,0)\). The region \(y\geq-\frac{1}{2}x + 2\) is above this line.

Now, let's analyze the two options:

• Option A: The shaded region for \(y\leq x - 1\) (below \(y=x - 1\)) and \(y\geq-\frac{1}{2}x + 2\) (above \(y =-\frac{1}{2}x+2\)) is correctly represented.
• Option B: The shading for the inequalities does not match the required regions (the region for \(y\geq-\frac{1}{2}x + 2\) and \(y\leq x - 1\) is not correctly shaded).

Answer:

A. The graph with the shaded region that satisfies \(y\leq x - 1\) (below the line \(y = x-1\)) and \(y\geq-\frac{1}{2}x + 2\) (above the line \(y=-\frac{1}{2}x + 2\))