Question

1. which inequality is shown by the graph below?

a) $-x + y \ge 2$ b) $-x + y < 2$
c) $-x + y > 2$ d) $-x + y \le 2$

Explanation:

Step1: Analyze the boundary line

The equation of the boundary line is \(-x + y = 2\) (rewritten as \(y=x + 2\)). The line is dashed (assuming from the graph's appearance, as the shaded region is on one side and the line style indicates it's a strict inequality or we check the inequality type).

Step2: Test a point in the shaded region

Let's take a point in the shaded region, say \((0,0)\) (if it's in the shaded area; wait, actually, looking at the graph, the shaded region is below or above? Wait, the line \(y=x + 2\): when \(x = 0\), \(y = 2\); when \(y=0\), \(x=-2\). The shaded region—let's pick a point like \((0,0)\). Plug into \(-x + y\): \(-0+0 = 0\). Now check the inequality: \(0>2\)? No. Wait, maybe a point like \((3,0)\). \(-3 + 0=-3\), no. Wait, maybe I got the line wrong. Wait, the inequality options: let's re - express the line as \(y=x + 2\). The shaded region—if the line is dashed (since the options have \(>\) or \(<\), not \(\geq\) or \(\leq\) for some, but wait the options are: a) \(-x + y\geq2\) (solid line if included), b) \(-x + y<2\), c) \(-x + y>2\), d) \(-x + y\leq2\). Wait, the graph's boundary: if the shaded region is where \(-x + y>2\) is false? No, wait let's take a point in the shaded area. Suppose the shaded region is above the line? Wait, no, let's take the line \(y=x + 2\). Let's take a point in the shaded region, say \((0,3)\). Then \(-0 + 3=3>2\), which satisfies \(-x + y>2\)? Wait, no, if the shaded region is where \(-x + y>2\), then plugging \((0,3)\): \(3>2\), yes. Wait, maybe I misread the graph. Wait, the correct approach: the boundary line is \(y=x + 2\). The shaded region: if the line is dashed (so inequality is strict, \(>\) or \(<\)), and we test a point. Let's take a point in the shaded area. Let's assume the shaded region is where \(-x + y>2\) is true. Wait, the option c is \(-x + y>2\), option b is \(-x + y<2\). Let's take a point on the shaded side. Let's say the point \((0,3)\): \(-0 + 3=3>2\), which satisfies \(-x + y>2\). If we take a point on the other side of the line, say \((0,0)\): \(-0+0 = 0<2\), which is the opposite. So the shaded region satisfies \(-x + y>2\)? Wait, no, wait the line is \(y=x + 2\), so \(-x + y=2\). If the shaded region is above the line, then for a point above the line, say \((0,3)\), \(-x + y=3>2\), so the inequality is \(-x + y>2\), which is option c. Wait, but the original answer mark is c? Wait, let's re - check. The options: a) \(-x + y\geq2\) (solid line, includes the line), b) \(-x + y<2\), c) \(-x + y>2\), d) \(-x + y\leq2\). If the boundary line is dashed (indicating the inequality is strict, not including the line), so we can eliminate a and d. Then, test a point in the shaded region. Let's take a point in the shaded area. Suppose the shaded region is where \(-x + y>2\). Let's take \(x = 0\), \(y = 3\) (in shaded area): \(-0+3 = 3>2\), which is true for option c. If we take a point not in the shaded area, say \(x = 0\), \(y = 0\): \(-0 + 0=0<2\), which is true for option b, but since \(0\) is not in the shaded area, the shaded area must satisfy \(-x + y>2\), so the inequality is \(-x + y>2\), which is option c.

Answer:

c) \(-x + y>2\)