Question

exercises within reach® solutions in english & spanish and tutorial videos at collegeprepalgebra.com
matching in exercises 17–20, match the inequality with its graph.
(a) (b) (c) (d)

1. $2x - y \leq 1$ 18. $2x - y < 1$ 19. $2x - y \geq 1$ 20. $2x - y > 1$

sketching the graph of a linear inequality in exercises 21–28, sketch the graph of the linear inequality.

1. $y \leq 3x$ 22. $y \geq 6x$ 23. $y \leq 2x - 3$ 24. $y \geq -x + 2$
2. $x - y < 0$ 26. $x + y > 0$ 27. $y > -2x + 10$ 28. $y < 3x - 7$

Explanation:

Step1: Analyze the inequality \(2x - y \leq 1\)

First, rewrite the inequality in slope - intercept form (\(y=mx + b\)):
Starting with \(2x - y\leq1\), we can solve for \(y\). Subtract \(2x\) from both sides: \(-y\leq - 2x + 1\). Multiply both sides by \(- 1\) (and reverse the inequality sign) to get \(y\geq2x - 1\).

Step2: Determine the line type and shading

The inequality \(y\geq2x - 1\) has a "greater than or equal to" sign. This means the boundary line \(y = 2x-1\) is a solid line (because the inequality includes equality). To determine the shading, we can test a point. Let's test the point \((0,0)\):
Substitute \(x = 0\) and \(y = 0\) into \(y\geq2x - 1\): \(0\geq2(0)-1\), which simplifies to \(0\geq - 1\), a true statement. But we need to check the region relative to the line \(y = 2x - 1\).
Now, let's analyze the graphs:
- Graph (a): Dashed line, so it's for a strict inequality (\(<\) or \(>\)), so not for \(\leq\) or \(\geq\). Eliminate (a).
- Graph (b): Solid line. Let's check the line equation. The line \(y = 2x - 1\) has a slope of \(2\) and a \(y\) - intercept of \(-1\). The line in graph (b) has a slope of \(2\) (since for a change in \(x\) of \(1\), \(y\) changes by \(2\)) and \(y\) - intercept around \(- 1\). Now, the shading: for \(y\geq2x - 1\), we shade above the line. Wait, no, wait. Wait, our inequality is \(y\geq2x - 1\), but let's re - express the original inequality \(2x - y\leq1\) as \(y\geq2x - 1\). Wait, maybe I made a mistake. Let's re - do the slope - intercept form.
Starting with \(2x - y\leq1\), we can also rewrite it as \(y\geq2x - 1\). The line \(y = 2x - 1\): when \(x = 0\), \(y=-1\); when \(x = 1\), \(y = 1\).
Now, graph (b): The line is solid. Let's check the shading. The region in graph (b) is below or above? Wait, maybe I mixed up. Let's take the inequality \(2x - y\leq1\), which is equivalent to \(y\geq2x - 1\). So we shade the region where \(y\) is greater than or equal to \(2x - 1\).
Graph (c): Solid line. Let's check the line. The line in graph (c) has the same slope and intercept as \(y = 2x - 1\). Now, the shading: for \(y\geq2x - 1\), we shade above the line. Wait, no, let's test a point in graph (c). Let's take \((0,0)\): \(0\geq2(0)-1\) is true, but in graph (c), the green region (shaded) at \((0,0)\): Wait, the graph (c) has green on the left and yellow on the right? Wait, maybe the axes are different. Wait, the original problem's graphs:
Graph (b): The line is solid, and the shading is on the right - hand side of the line? Wait, no, let's re - express the inequality \(2x - y\leq1\) as \(y\geq2x - 1\). The line \(y = 2x - 1\) goes through \((0,-1)\) and \((1,1)\).
Graph (b): The line is solid, and the shading is the region that satisfies \(y\geq2x - 1\)? Wait, maybe I got the inequality direction wrong. Let's solve \(2x - y\leq1\) for \(y\) again: \(y\geq2x - 1\). So the boundary line is solid (because of \(\leq\) which includes equality), and we shade above the line (since \(y\) is greater than or equal to the line).
Now, graph (b): The line is solid, and the shading is on the side where \(y\) is greater than the line? Wait, no, let's look at the four graphs:
- Graph (a): Dashed line, so for strict inequality (\(<\) or \(>\)), so not 17.
- Graph (b): Solid line. Let's check the line equation. The line has slope \(2\), \(y\) - intercept \(-1\). The shading: if we take a point above the line, say \((0,2)\): \(2\geq2(0)-1\) (true), and in graph (b), the region with the green color (assuming green is the shaded region) - wait, the graphs have green and yellow. Let's check the original problem's graphs:
Graph (b): The line is solid, and the green region is on the right of the line. Wait, maybe my initial analysis of the inequality direction was wrong. Let's rewrite \(2x - y\leq1\) as \(y\geq2x - 1\). So when \(x = 0\), \(y\geq - 1\), and when \(x = 1\), \(y\geq1\).
Graph (b): The line is solid, and the shading is the region that includes points where \(y\) is greater than or equal to \(2x - 1\). Let's check graph (b) with the line \(y = 2x - 1\). The line passes through \((0,-1)\) and \((1,1)\). In graph (b), the green region is on the side of the line where, for example, when \(x = 0\), \(y\) values are greater than \(-1\) (since the green region at \(x = 0\) is above \(y=-1\)). So graph (b) is for \(2x - y\leq1\)? Wait, no, wait the inequality \(2x - y\leq1\) is equivalent to \(y\geq2x - 1\), so the shading should be above the line.
Wait, maybe I made a mistake in the inequality transformation. Let's do it again:
\(2x - y\leq1\)
Subtract \(2x\) from both sides: \(-y\leq - 2x + 1\)
Multiply both sides by \(-1\) (reverse inequality): \(y\geq2x - 1\)
Yes, that's correct. So the boundary line is solid, and we shade above the line.
Now, graph (b): The line is solid, and the shading is on the side where \(y\) is greater than the line. Let's check graph (c): The line is solid, and the shading is on the left? No, graph (c) has green on the left and yellow on the right. Wait, maybe the correct graph for \(2x - y\leq1\) is graph (b)? Wait, no, let's check the four graphs again:
Wait, the problem is to match \(2x - y\leq1\) with its graph. Let's consider the boundary line: solid (because of \(\leq\)), so eliminate (a) and (d) (which have dashed lines). Now between (b) and (c).
The inequality \(2x - y\leq1\) can be rewritten as \(y\geq2x - 1\). So for a point below the line, say \((0,0)\): \(0\geq - 1\) (true), but wait, \(y = 0\) is above \(y=-1\) (the line at \(x = 0\) is \(y=-1\)). Wait, maybe the line is \(y = 2x - 1\), and the shading is above the line.
Graph (b): The line is solid, and the shading is on the side where \(y\) is greater than the line. Graph (c): The line is solid, and the shading is on the side where \(y\) is less than the line? Wait, no, let's take the inequality \(2x - y\leq1\) and test the point \((0,0)\): \(2(0)-0\leq1\) ( \(0\leq1\) ), which is true. So \((0,0)\) is in the solution set. Now, where is \((0,0)\) relative to the line \(y = 2x - 1\)? At \(x = 0\), the line is at \(y=-1\), so \((0,0)\) is above the line. So we need a graph where the region containing \((0,0)\) is shaded and the line is solid.
Graph (b): The line is solid, and \((0,0)\) is in the green region (assuming green is the shaded region). Graph (c): The line is solid, and \((0,0)\) is in the yellow region (not shaded). So graph (b) is for \(2x - y\leq1\)? Wait, no, wait the original inequality is \(2x - y\leq1\), and when we test \((0,0)\), it satisfies the inequality. Now, let's check the line in graph (b): the line is \(y = 2x - 1\) (since slope is \(2\), \(y\) - intercept \(-1\)). The shaded region in graph (b) includes \((0,0)\) (since \((0,0)\) is in the green area). So graph (b) is the graph for \(2x - y\leq1\). Wait, but let's check the other inequalities.
Wait, maybe I messed up the first step. Let's take the inequality \(2x - y\leq1\), rewrite it as \(y\geq2x - 1\). The boundary line is solid, so options (a) and (d) (dashed lines) are out. Now, between (b) and (c).
For graph (b): The line is solid, and the shading is on the side where \(y\) is greater than the line (since \((0,0)\) is in the shaded region and \(0\geq - 1\)). For graph (c): The line is solid, and the shading is on the side where \(y\) is less than the line (since \((0,0)\) is in the yellow region, not shaded). So the correct graph for \(17.2x - y\leq1\) is graph (b).

Step3: Analyze \(2x - y\lt1\) (Exercise 18)

Rewrite the inequality as \(y\gt2x - 1\). The boundary line is dashed (because of \(\lt\), strict inequality), and we shade above the line (since \(y\) is greater than the line).
- Graph (a): Dashed line. Let's check the line: slope \(2\), \(y\) - intercept \(-1\). The shading: \((0,0)\) is in the green region. \(0\gt2(0)-1\) ( \(0\gt - 1\) ), which is true. So graph (a) is for \(2x - y\lt1\).

Step4: Analyze \(2x - y\geq1\) (Exercise 19)

Rewrite as \(y\leq2x - 1\). Boundary line is solid, shade below the line.
- Graph (c): Solid line. Test \((0,0)\): \(0\leq2(0)-1\) ( \(0\leq - 1\) ), false. So the shaded region should be where \(y\) is less than or equal to the line. In graph (c), the green region is on the left, and \((0,0)\) is in the yellow region. Let's take a point below the line, say \((0,-2)\): \(-2\leq2(0)-1\) ( \(-2\leq - 1\) ), true. So graph (c) is for \(2x - y\geq1\).

Step5: Analyze \(2x - y\gt1\) (Exercise 20)

Rewrite as \(y\lt2x - 1\). Boundary line is dashed, shade below the line.
- Graph (d): Dashed line. Test \((0,0)\): \(0\lt2(0)-1\) ( \(0\lt - 1\) ), false. Take a point below the line, say \((0,-2)\): \(-2\lt2(0)-1\) ( \(-2\lt - 1\) ), true. So graph (d) is for \(2x - y\gt1\).

Answer:

1. B. Graph (b)
2. A. Graph (a)
3. C. Graph (c)
4. D. Graph (d)