Question

4 use the graphed system of inequalities to answer a-c.
a. write the system of inequalities represented on the graph:
(-1, 2)
b. list three ordered pairs that are solutions to the system.
c. list three ordered pairs that are not solutions to the system.
5 jim has over 40 contacts saved in his phone. the contacts are made up of friends and family. the number of contacts that are friends is at least two times the number of contacts that are family. if x represents family and y represents friends, the

Explanation:

Step1: Analyze the dashed line

The dashed line passes through \((-1, 2)\) and has a slope. Let's find its equation. The slope \(m\) between two points (let's assume another point, but since it's a dashed line, let's find the equation. Let's see, the dashed line: let's take two points. Wait, the dashed line seems to have a slope of \(1\) (since from \((-1,2)\), moving 1 right and 1 up gives \((0,3)\), but maybe better to use slope-intercept form \(y = mx + b\). Wait, maybe the dashed line is \(y = x + 3\)? Wait, no, let's check the solid line. The solid line is a straight line, maybe \(y=-x + 1\)? Wait, maybe first, the shaded region is above the dashed line and below the solid line? Wait, the problem is to write the system of inequalities. Let's re-examine.

Wait, the graph has a dashed line and a solid line. Let's find the equations of both lines.

First, the dashed line: let's find two points on it. Let's say the dashed line passes through \((-1, 2)\) and maybe \((0, 3)\)? Wait, no, maybe \((-1,2)\) and \((1,4)\)? Wait, slope \(m=\frac{4 - 2}{1 - (-1)}=\frac{2}{2}=1\). So equation of dashed line: \(y - 2 = 1(x + 1)\) → \(y = x + 3\). Since it's dashed, the inequality is \(y < x + 3\) (because the shaded region is below? Wait, no, the shaded region is to the right? Wait, maybe I got the lines wrong. Wait, the solid line: let's find its equation. Let's take two points on the solid line. Suppose it passes through \((-1, 2)\) and \((0,1)\)? Slope \(m=\frac{1 - 2}{0 - (-1)}=\frac{-1}{1}=-1\). So equation: \(y - 2 = -1(x + 1)\) → \(y = -x + 1\). Since it's solid, the inequality is \(y \leq -x + 1\)? Wait, no, the shaded region is between the two lines? Wait, the problem says "Write the system of inequalities represented on the graph: \((-1, 2)\)". Wait, maybe the two lines are:

Dashed line: let's find its equation. Let's assume the dashed line has a slope of \(1\) (since it's a diagonal line going up). Let's take point \((-1, 2)\), so \(y = x + 3\) (because when \(x=-1\), \(y=-1 + 3 = 2\), correct). So the dashed line is \(y = x + 3\), and since it's dashed, the inequality is \(y < x + 3\) (because the shaded region is below the dashed line? Wait, no, the shaded region is above the dashed line? Wait, maybe I messed up. Wait, the solid line: let's find its equation. Let's take two points on the solid line. Suppose it passes through \((-1, 2)\) and \((0,1)\), slope is \(-1\), so equation \(y = -x + 1\). Since it's solid, the inequality is \(y \leq -x + 1\) (if the shaded region is below the solid line) or \(y \geq -x + 1\). Wait, the shaded region is between the two lines? Wait, the problem's point is \((-1, 2)\), which is on both lines? Wait, no, \((-1, 2)\) is the intersection point. So the two lines intersect at \((-1, 2)\). The dashed line is, say, \(y = x + 3\) (dashed, so inequality is \(y < x + 3\) or \(y > x + 3\)) and the solid line is \(y = -x + 1\) (solid, so \(y \leq -x + 1\) or \(y \geq -x + 1\)). Now, the shaded region: let's pick a test point. Let's take \((0, 0)\). Plug into dashed line: \(0 < 0 + 3\) (true) and into solid line: \(0 \leq -0 + 1\) (true). So the shaded region is below the solid line (\(y \leq -x + 1\)) and above the dashed line (\(y > x + 3\))? Wait, no, when \(x=0\), dashed line is \(y=3\), solid line is \(y=1\). Wait, that can't be. Wait, maybe I got the lines reversed. Let's re-express.

Wait, maybe the dashed line has a slope of \(1\) and the solid line has a slope of \(-1\), intersecting at \((-1, 2)\). Let's check the dashed line: if \(x=-1\), \(y=2\); if \(x=0\), \(y=3\) (so \(y = x + 3\)). Solid line: \(x=-1\), \(y=2\); \(x=0\), \(y=1\) (so \(y = -x + 1\)). Now, the shaded region: let's take a point in the shaded area, say \((0, 2)\). Plug into dashed line: \(2 > 0 + 3\)? No, \(2 < 3\). Plug into solid line: \(2 \leq -0 + 1\)? No, \(2 \leq 1\) is false. Wait, maybe the shaded region is above the solid line and below the dashed line? Wait, \((0, 2)\): solid line at \(x=0\) is \(y=1\), so \(2 \geq 1\) (true for solid line, which is solid, so \(y \geq -x + 1\)). Dashed line at \(x=0\) is \(y=3\), so \(2 < 3\) (true for dashed line, which is dashed, so \(y < x + 3\)). Ah, that works! So the system is:

\(y \geq -x + 1\) (solid line, shaded above)

\(y < x + 3\) (dashed line, shaded below)

Step2: Verify with the point \((-1, 2)\)

Plug \((-1, 2)\) into both inequalities:

For \(y \geq -x + 1\): \(2 \geq -(-1) + 1 = 2\), which is true (equality holds, so solid line is correct).

For \(y < x + 3\): \(2 < -1 + 3 = 2\)? No, \(2 < 2\) is false. Wait, that's a problem. Wait, maybe the dashed line is \(y = x + 3\) but the inequality is \(y > x + 3\)? No, \((-1,2)\) is on the dashed line, so it's not a solution. Wait, maybe the two lines are:

Wait, maybe the dashed line is \(y = x + 3\) (dashed, so \(y < x + 3\)) and the solid line is \(y = -x + 1\) (solid, so \(y \leq -x + 1\)), but the intersection point \((-1,2)\) is on both lines. Wait, maybe I made a mistake in the slope. Let's try another approach. Let's find the equations of the two lines.

First line (dashed): let's take two points. Suppose the dashed line goes through \((-1, 2)\) and \((1, 4)\), slope \(m = \frac{4 - 2}{1 - (-1)} = 1\), so equation \(y = x + 3\) (since \(2 = -1 + 3\)).

Second line (solid): goes through \((-1, 2)\) and \((1, 0)\), slope \(m = \frac{0 - 2}{1 - (-1)} = -1\), so equation \(y = -x + 1\) (since \(2 = -(-1) + 1 = 2\)).

Now, the shaded region: let's pick a point in the shaded area, say \((0, 2)\).

Check \(y \geq -x + 1\): \(2 \geq 0 + 1\) → \(2 \geq 1\), true.

Check \(y < x + 3\): \(2 < 0 + 3\) → \(2 < 3\), true.

Another point: \((1, 2)\).

\(y \geq -x + 1\): \(2 \geq -1 + 1 = 0\), true.

\(y < x + 3\): \(2 < 1 + 3 = 4\), true.

Another point: \((-2, 2)\).

\(y \geq -x + 1\): \(2 \geq 2 + 1 = 3\)? No, \(2 \geq 3\) is false. Wait, so \((-2, 2)\) is not in the shaded region. So the shaded region is to the right of the intersection? Wait, maybe the correct system is:

After re-evaluating, the two lines are:

Dashed line: \(y = x + 3\) (dashed, so inequality \(y < x + 3\))

Solid line: \(y = -x + 1\) (solid, so inequality \(y \leq -x + 1\))? No, that didn't work. Wait, maybe the solid line is \(y = -x + 1\) (solid, so \(y \geq -x + 1\)) and dashed line \(y = x + 3\) (dashed, so \(y < x + 3\)), and the shaded region is between them. Let's check \((0, 2)\):

\(2 \geq -0 + 1\) → \(2 \geq 1\), true.

\(2 < 0 + 3\) → \(2 < 3\), true.

\((1, 2)\):

\(2 \geq -1 + 1 = 0\), true.

\(2 < 1 + 3 = 4\), true.

\((-1, 2)\):

\(2 \geq -(-1) + 1 = 2\), true (equality holds, so solid line includes it).

\(2 < -1 + 3 = 2\)? No, \(2 < 2\) is false, so \((-1, 2)\) is on the dashed line, so it's not a solution (since dashed line means strict inequality). So the system is:

\(y \geq -x + 1\)

\(y < x + 3\)

Answer:

The system of inequalities is \(y \geq -x + 1\) and \(y < x + 3\) (or in boxed form, but as per instruction, no boxed, just the system). Wait, the problem says "Write the system of inequalities represented on the graph: \((-1, 2)\)". So the two inequalities are \(y \geq -x + 1\) (solid line, shaded above) and \(y < x + 3\) (dashed line, shaded below).