On a coordinate plane, 2 lines are shown. The first solid straight line has an equation of y greater-than-or-equal-to negative one-fifth x + 1, has a negative slope, and goes through (negative 5, 2) and (0, 1). Everything above the line is shaded. The second dashed solid line has equation y less-than 2 x + 1, has a positive slope, and goes through (negative 2, negative 3) and (0, 1). Everything to the right of the line is shaded. Which ordered pairs make both inequalities true? Check all that apply.

Question

(–2, 2)
(0, 0)
(1,1)
(1, 3)
(2, 2)

Answer 1

Let's analyze the inequalities represented by the lines and determine which ordered pairs satisfy both inequalities.

1. The first inequality is: $$y\ge -\frac{1}{5}x+1$$ This line has a negative slope, meaning it decreases as $x$ increases. The area above the line is shaded.

2. The second inequality is: $$y<2x+1$$ This line has a positive slope, meaning it increases as $x$ increases. The area to the right of the line is shaded.

Now we will check each ordered pair to see if it satisfies both inequalities.

Checking the Ordered Pairs:

1. (–2, 2):

   • For the first inequality: $$2\ge -\frac{1}{5}(-2)+1\Rightarrow 2\ge \frac{2}{5}+1\Rightarrow 2\ge \frac{7}{5}\text{ (true)}$$
   • For the second inequality: $$2<2(-2)+1\Rightarrow 2<-4+1\Rightarrow 2<-3\text{ (false)}$$
   • Not a solution.

2. (0, 0):

   • For the first inequality: $$0\ge -\frac{1}{5}(0)+1\Rightarrow 0\ge 1\text{ (false)}$$
   • Not a solution.

3. (1, 1):

   • For the first inequality: $$1\ge -\frac{1}{5}(1)+1\Rightarrow 1\ge -\frac{1}{5}+1\Rightarrow 1\ge \frac{4}{5}\text{ (true)}$$
   • For the second inequality: $$1<2(1)+1\Rightarrow 1<2+1\Rightarrow 1<3\text{ (true)}$$
   • Is a solution.

4. (1, 3):

   • For the first inequality: $$3\ge -\frac{1}{5}(1)+1\Rightarrow 3\ge -\frac{1}{5}+1\Rightarrow 3\ge \frac{4}{5}\text{ (true)}$$
   • For the second inequality: $$3<2(1)+1\Rightarrow 3<2+1\Rightarrow 3<3\text{ (false)}$$
   • Not a solution.

5. (2, 2):

   • For the first inequality: $$2\ge -\frac{1}{5}(2)+1\Rightarrow 2\ge -\frac{2}{5}+1\Rightarrow 2\ge \frac{3}{5}\text{ (true)}$$
   • For the second inequality: $$2<2(2)+1\Rightarrow 2<4+1\Rightarrow 2<5\text{ (true)}$$
   • Is a solution.

Summary:

The ordered pairs that make both inequalities true are:

• (1, 1)
• (2, 2)

Thus, the final answers are: (1, 1) and (2, 2).

Answer 2

Graphically, a point is a solution to a system of two inequalities if and only if the point

lies in the shaded region of the top inequality, but not in the shaded region of the bottom inequality.
lies in the shaded region of the bottom inequality, but not in the shaded region of the top inequality.
lies in the shaded regions of both the top and bottom inequalities.
does not lie in the shaded region of the top or bottom inequalities.

Answer 3

To determine which option accurately describes when a point is a solution to a system of two inequalities, let's clarify the nature of the inequalities:

1. The first inequality represents a region shaded above or on the line (typically denoted as "greater than or equal to").
2. The second inequality represents a region shaded below or to the left of the line (typically denoted as "less than").

In the context of two inequalities:

• For a point to be a solution to both inequalities simultaneously, it needs to satisfy both conditions:
  • It must lie within the region defined by the first inequality (above or on that line).
  • It must also fulfill the condition of the second inequality (below or to the left of that line).

Given this information, the best way to describe when a point is a solution to the system of two inequalities is:

It lies in the shaded regions of both the top and bottom inequalities.

Thus, the correct answer is:

• lies in the shaded regions of both the top and bottom inequalities.

Answer 4

On a coordinate plane, a dashed solid line has an equation of y less-than five-thirds x + 1. It has a positive slope and goes through (negative 3, negative 4) and (0, 1). Everything to the right of the line is shaded.Which linear inequality will not have a shared solution set with the graphed linear inequality?y < Five-thirdsx – 2y < Negative five-thirdsx + 1y > Five-thirdsx + 2y > Negative five-thirdsx + 2

Answer 5

To determine which linear inequality will not have a shared solution set with the given graphed linear inequality, we need to analyze the original inequality and how each proposed inequality relates to it.

The original inequality is:

$$y<\frac{5}{3}x+1$$

This inequality represents the region below the dashed line that has a positive slope ($\frac{5}{3}$). The area shaded represents all points for which $y$ is less than the value given by the equation of the line for each corresponding $x$.

Now let's analyze each option to see if it has any overlap with the original inequality:

1. $y<\frac{5}{3}x-2$:

   • This inequality represents a line that is also below another line ($\frac{5}{3}x-2$ is lower than $\frac{5}{3}x+1$). Since it is on the same side (below) as the original inequality, there will be a shared solution set.

2. $2y<-\frac{5}{3}x+1$:

   • Rearranging gives $y<-\frac{5}{6}x+\frac{1}{2}$, which has a negative slope (-$\frac{5}{6}$) and will shade below that line. This line intersects the y-axis above and to the left of the original line. Therefore, there will be some points that are beneath both lines, meaning there will be a shared solution set.

3. $y>\frac{5}{3}x+2$:

   • This represents a region above the line $\frac{5}{3}x+2$, and since this line is above the original line (which is described by the first inequality), it will not intersect or share any solution set with the original inequality.

4. $y>-\frac{5}{3}x+2$:

   • Similar to option 3, this inequality represents a region above the line $y=-\frac{5}{3}x+2$. The slope is negative, meaning it’s below the regions shaded by the original inequality. Hence, there will be a shared solution set.

Conclusion:

The only inequality that does not have a shared solution set with the graphed inequality $y<\frac{5}{3}x+1$ is:

$y>\frac{5}{3}x+2$.

Answer 6

Which is the graph of the system x + 3y > –3 and y < One-halfx + 1? 1. On a coordinate plane, 2 dashed lines are shown. The first straight line has a positive slope and goes through (negative 2, negative 2) and (2, 0). Everything below the line is shaded. The second straight line has a negative slope and goes through (negative 3, 0) and (0, negative 1). Everything above and to the right of the line is shaded. 2. On a coordinate plane, 2 dashed lines are shown. The first straight line has a positive slope and goes through (negative 2, 0) and (0, 1). Everything below the line is shaded. The second straight line has a negative slope and goes through (0, 1) and (2, 0). Everything above the line is shaded. 3. On a coordinate plane, 2 dashed lines are shown. The first straight line has a positive slope and goes through (negative 2, 0) and (0, 1). Everything below the line is shaded. The second straight line has a negative slope and goes through (0, 1) and (3, 0). Everything above the line is shaded. 4. On a coordinate plane, 2 dashed lines are shown. The first straight line has a positive slope and goes through (0, 1) and (2, 2). Everything below the line is shaded. The second straight line has a negative slope and goes through (negative 3, 0) and (0, negative 1). Everything above the line is shaded.

Answer 7

Let's analyze the two inequalities in the system and identify their graphs to determine which option corresponds to their representations.

Inequalities:

1. First Inequality: $$x+3y>-3$$ To express it in slope-intercept form, we can manipulate it as follows: $$3y>-x-3⟹y>-\frac{1}{3}x-1$$ This line has a negative slope of $-\frac{1}{3}$ and a y-intercept at $-1$. Since it's a "greater than" inequality, the region above this line will be shaded.

2. Second Inequality: $$y<\frac{1}{2}x+1$$ This line has a positive slope of $\frac{1}{2}$ and a y-intercept of $1$. Since it's a "less than" inequality, the region below this line will be shaded.

Analyzing the Options:

Now, we will compare the characteristics of each graph option to see which one matches our derived inequalities.

1. Option 1:

   • Line 1: Has positive slope (through $(-2,-2)$ and $(2,0)$), shading below → This represents $y<\frac{1}{2}x$, not matching.
   • Line 2: Has negative slope (shaded above → does not match $x+3y>-3$).

2. Option 2:

   • Line 1: Has positive slope (through $(-2,0)$ and $(0,1)$), shading below → This represents $y<\frac{1}{2}x+1$, matches.
   • Line 2: Negative slope; shading above, matches for the other inequality, indicating possible matches, but needs confirmation with values.

3. Option 3:

   • Line 1: Has positive slope (through $(-2,0)$ and $(0,1)$), shading below → This supports $y<\frac{1}{2}x+1$, matches.
   • Line 2: Negative slope (through Points that do not correlate to desired coordinates).

4. Option 4:

   • Line 1: Positive slope (through $(0,1)$ and $(2,2)$); shading below → This indicates a different relationship.
   • Line 2: Negative slope (through $(-3,0)$ and $(0,-1)$) shading above this line does not match $y<\frac{1}{2}x+1$.

Conclusion:

The best match for the system of inequalities $x+3y>-3$ and $y<\frac{1}{2}x+1$ is:

Option 3, because it correctly reflects both the conditions of the inequalities:

• The first line corresponds to the correct slope for $y<\frac{1}{2}x+1$ and the second line corresponds to a configuration where shading above indicates solution to $x+3y>-3$ beyond the correct positive-negative alignments.

Answer 8

On a coordinate plane, 2 dashed lines are shown. The first straight line has a positive slope and goes through (0, 1) and (2, 2). Everything below the line is shaded. The second straight line has a negative slope and goes through (negative 3, 0) and (0, negative 1). Everything above the line is shaded. OPTION 4.

its the x that has the left side with the smaller triangle.

On a coordinate plane, 2 dashed lines are shown. The first straight line has a negative slope and goes through (0, 1) and (2, 0). Everything below and to the left of the line is shaded. The second straight line has a positive slope and goes through (negative 1, 0) and (0, 2). Everything above and to the left of the line is shaded. A point is shown at (negative 3, 1).Which system of inequalities with a solution point is represented by the graph?y > 2x – 2 and y < Negative one-halfx – 1; (3, 1)y > 2x – 2 and y < Negative one-halfx + 1; (–3, 1)y > 2x + 2 and y < Negative one-halfx – 1; (3, 1)y > 2x + 2 and y < Negative one-halfx + 1; (–3, 1)

Answer 9

To identify the correct system of inequalities represented by the described graph, let’s first understand the lines and the regions that are shaded based on the given information about the slope and intercepts.

Analyzing the Graph Based on Your Description

1. First Line:

   • Positive slope and goes through (0, 1) and (2, 2).
   • This implies the line has the equation in slope-intercept form (from points):
   • Slope = $\frac{2-1}{2-0}=\frac{1}{2}$
   • Equation: $$y=\frac{1}{2}x+1$$
   • Since everything below the line is shaded, we have the inequality: $$y<\frac{1}{2}x+1$$

2. Second Line:

   • Negative slope and goes through (–3, 0) and (0, –1).
   • Slope = $\frac{-1-0}{0+3}=-\frac{1}{3}$
   • Equation: $$y=-\frac{1}{3}x-1$$
   • Since everything above the line is shaded, we have the inequality: $$y>-\frac{1}{3}x-1$$

Evaluating Given Options

Now we are to evaluate the options provided to see which one contains a solution point at $(-3,1)$.

Checking $(-3,1)$ Against Each Proposed System

1. First Option:

   • $y>2x-2$
     • $1>2(-3)-2\to 1>-6-2\to 1>-8\text{ (true)}$
   • $y<-\frac{1}{2}x-1$
     • $1<-\frac{1}{2}(-3)-1\to 1<\frac{3}{2}-1\to 1<\frac{1}{2}\text{ (false)}$

2. Second Option:

   • $y>2x-2$
     • $1>2(-3)-2\to 1>-8\text{ (true)}$
   • $y<-\frac{1}{2}x+1$
     • $1<-\frac{1}{2}(-3)+1\to 1<\frac{3}{2}+1\to 1<\frac{5}{2}\text{ (true)}$

3. Third Option:

   • $y>2x+2$
     • $1>2(-3)+2\to 1>-6+2\to 1>-4\text{ (true)}$
   • $y<-\frac{1}{2}x-1$
     • $1<-\frac{1}{2}(-3)-1\to 1<\frac{3}{2}-1\to 1<\frac{1}{2}\text{ (false)}$

4. Fourth Option:

   • $y>2x+2$
     • $1>2(-3)+2\to 1>-6+2\to 1>-4\text{ (true)}$
   • $y<-\frac{1}{2}x+1$
     • $1<-\frac{1}{2}(-3)+1\to 1<\frac{3}{2}+1\to 1<\frac{5}{2}\text{ (true)}$

Summary of Evaluation:

• The Second Option and the Fourth Option both have $(-3,1)$ as a solution.
• The second system is valid as it offers the correct evaluations for $y>2x-2$ and $y<-\frac{1}{2}x+1$, confirming $(-3,1)$ as a solution.

Conclusion:

The correct answer that represents the system of inequalities with the solution point at $(-3,1)$ is:

• y > 2x – 2 and y < -\frac{1}{2}x + 1; (-3, 1).

Answer 10

wrong answer y > 2x – 2 and y < -\frac{1}{2}x + 1; (-3, 1). choose another.

Answer 11

Let's carefully analyze the provided options again to find which system corresponds to the graph and has the solution point at $(-3,1)$.

Given Options:

1. Option A:

   • $y>2x-2$
   • $y<-\frac{1}{2}x-1$

2. Option B:

   • $y>2x-2$
   • $y<-\frac{1}{2}x+1$

3. Option C:

   • $y>2x+2$
   • $y<-\frac{1}{2}x-1$

4. Option D:

   • $y>2x+2$
   • $y<-\frac{1}{2}x+1$

Testing Each Option with the Point $(-3,1)$

Inequalities tending toward checking if $(-3,1)$ is in the respective solution sets.

1. Option A: $y>2x-2$ and $y<-\frac{1}{2}x-1$

   • For $y>2x-2$: $$1>2(-3)-2\Rightarrow 1>-6-2\Rightarrow 1>-8\text{(True)}$$
   • For $y<-\frac{1}{2}x-1$: $$1<-\frac{1}{2}(-3)-1\Rightarrow 1<\frac{3}{2}-1\Rightarrow 1<\frac{1}{2}\text{(False)}$$
   • This option does not work.

2. Option B: $y>2x-2$ and $y<-\frac{1}{2}x+1$

   • For $y>2x-2$: $$1>2(-3)-2\Rightarrow 1>-8\text{(True)}$$
   • For $y<-\frac{1}{2}x+1$: $$1<-\frac{1}{2}(-3)+1\Rightarrow 1<\frac{3}{2}+1\Rightarrow 1<2.5\text{(True)}$$
   • This option works.

3. Option C: $y>2x+2$ and $y<-\frac{1}{2}x-1$

   • For $y>2x+2$: $$1>2(-3)+2\Rightarrow 1>-6+2\Rightarrow 1>-4\text{(True)}$$

   • For $y<-\frac{1}{2}x-1$: $$1<-\frac{1}{2}(-3)-1\Rightarrow 1<\frac{3}{2}-1\Rightarrow 1<\frac{1}{2}\text{(False)}$$

   • This option does not work.

4. Option D: $y>2x+2$ and $y<-\frac{1}{2}x+1$

   • For $y>2x+2$: $$1>2(-3)+2\Rightarrow 1>-6+2\Rightarrow 1>-4\text{(True)}$$
   • For $y<-\frac{1}{2}x+1$: $$1<-\frac{1}{2}(-3)+1\Rightarrow 1<\frac{3}{2}+1\Rightarrow 1<2.5\text{(True)}$$
   • This option works.

Conclusion

Both Option B and Option D allow the solution point (-3, 1). However, since you have said Option B does not represent your desired intention, thus it leaves Option D as the identified representation.

So, the correct system of inequalities that contains the point (-3, 1) is:

• Option D: $y>2x+2$ and $y<-\frac{1}{2}x+1;(-3,1)$.

If there are further clarifications needed, let me know!