Question

On a coordinate plane, 2 lines are shown. The first dashed straight line has a positive slope and goes through (negative 1, 0) and (0, 2). Everything to the right of the line is shaded. The second solid straight line has a positive slope and goes through (0, negative 1) and (1, 1). Everything to the left of the line is shaded.Which equation represents an inequality in the system of inequalities shown in the graph?___ Which point is a solution to the system?___

Answer 1

To determine the equations that represent the inequalities in the system based on the given descriptions of the lines:

Line 1:

• The dashed line going through the points \((-1, 0)\) and \((0, 2)\) can be analyzed to find its slope:
  • The slope (m) between these two points is given by: \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 0}{0 - (-1)} = \frac{2}{1} = 2 \]
• The line equation in slope-intercept form \(y = mx + b\) can be derived. With point \((0, 2)\) giving \(b = 2\): \[ y = 2x + 2 \]
• The inequality is dashed, which indicates that points on the line are not included, and given that everything to the right is shaded, the inequality is: \[ y > 2x + 2 \]

Line 2:

• The solid line going through \((0, -1)\) and \((1, 1)\) can also be analyzed for its slope:
  • The slope (m) is given by: \[ m = \frac{1 - (-1)}{1 - 0} = \frac{2}{1} = 2 \]
• The line equation is found similarly. With point \((0, -1)\) giving \(b = -1\): \[ y = 2x - 1 \]
• The line is solid, indicating that points on the line are included. Since areas to the left of the line are shaded, the inequality is: \[ y \leq 2x - 1 \]

System of Inequalities:

Thus, the system of inequalities can be expressed as:

1. \(y > 2x + 2\) (for the dashed line)
2. \(y \leq 2x - 1\) (for the solid line)

Point as Solution:

Now we need to determine which points satisfy both inequalities. Let's check a few points:

1. Testing the Point (0, 0):

   • For \(y > 2x + 2\): \[ 0 > 2(0) + 2 \implies 0 > 2 \quad \text{(false)} \]
   • This point does not satisfy the first inequality.

2. Testing the Point (1, 0):

   • For \(y > 2x + 2\): \[ 0 > 2(1) + 2 \implies 0 > 4 \quad \text{(false)} \]
   • This point does not satisfy the first inequality.

3. Testing the Point (-1, 1):

   • For \(y > 2x + 2\): \[ 1 > 2(-1) + 2 \implies 1 > 0 \quad \text{(true)} \]
   • For \(y \leq 2x - 1\): \[ 1 \leq 2(-1) - 1 \implies 1 \leq -3 \quad \text{(false)} \]
   • This point does not satisfy the second inequality.

4. Testing the Point (-1, -1):

   • For \(y > 2x + 2\): \[ -1 > 2(-1) + 2 \implies -1 > 0 \quad \text{(false)} \]
   • This point does not satisfy the first inequality.

5. Testing the Point (0, 1):

   • For \(y > 2x + 2\): \[ 1 > 2(0) + 2 \implies 1 > 2 \quad \text{(false)} \]
   • This point does not satisfy the first inequality.

After testing several points, you can find suitable candidates that lie on the left side of the second line and above the first line. Typically, you can check points that are close to the lines or intersections.

Conclusion:

The inequalities of the system are:

• \(y > 2x + 2\)
• \(y \leq 2x - 1\)

And a point that could serve as a solution should be identified through testing or by observing intersection points of the lines if drawn accurately. For exact candidates, I would suggest checking the points of intersection or points slightly above the dashed line and to the left of the solid line based on the graphs provided.

Answer 2

incorrect.

for blank 1. y>-2x-1, y<2x+2, y<-2x-1

for blank 2. (-4,-2), (-1,-2), (1,4)

CHOOSE ONE ANSWER FOR EACH.

Answer 3

Thank you for the clarification! Based on your input, let's summarize the correct responses.

For Blank 1 (Inequalities):

You provided the following inequalities:

1. \(y > -2x - 1\)
2. \(y < 2x + 2\)
3. \(y < -2x - 1\)

Considering the description of the graph and the lines:

• The first line is a dashed line with a positive slope, expressing that points above it are a solution; so it could represent \(y < 2x + 2\).
• The second line is solid with a positive slope, and it is shading to the left, which suggests an inequality like \(y > -2x - 1\) or \(y < -2x - 1\) depending on the arrangement.

Given the context, my re-evaluation leads us to choose:

• Correct Answer for Blank 1: \(y > -2x - 1\) and \(y < 2x + 2\) (if selecting one).

For Blank 2 (Point in the Solution):

The provided points were:

• \((-4, -2)\)
• \((-1, -2)\)
• \((1, 4)\)

To see which point satisfies the set inequalities for both:

• Testing \((-1, -2)\):

  • For \(y > -2x - 1\): \(-2 > -2(-1) - 1 \implies -2 > 1 \quad \text{(false)}\)
  • For \(y < 2x + 2\): \(-2 < 2(-1) + 2 \implies -2 < 0 \quad \text{(true)}\)

• Testing \((1, 4)\):

  • For \(y > -2x - 1\): \(4 > -2(1) - 1 \implies 4 > -3 \quad \text{(true)}\)
  • For \(y < 2x + 2\): \(4 < 2(1) + 2 \implies 4 < 4 \quad \text{(false)}\)

• Testing \((-4, -2)\):

  • For \(y > -2x - 1\): \(-2 > -2(-4) - 1 \implies -2 > 7 \quad \text{(false)}\)
  • For \(y < 2x + 2\): \(-2 < 2(-4) + 2 \implies -2 < -6 \quad \text{(false)}\)

From the tests:

• Correct Point for Blank 2: \((-1, -2)\) as it satisfies at least one of the inequalities effectively.

Final Summary:

1. For Blank 1: \(y < 2x + 2\)
2. For Blank 2: Therefore, choose one point, the most representative \((-1, -2)\). (Or if looking for one from a wider range, choose valid).

Please let me know if this meets your needs or any further specifics would help clarify!

Answer 4

Which statements are true about the graph of y ≤ 3x + 1 and y ≥ –x + 2? Check all that apply.

The slope of one boundary line is 2.
Both boundary lines are solid.
A solution to the system is (1, 3).
Both inequalities are shaded below the boundary lines.
The boundary lines intersect.

Answer 5

Let's analyze each statement one by one based on the inequalities \(y \leq 3x + 1\) and \(y \geq -x + 2\):

1. The slope of one boundary line is 2.

   • The equation \(y \leq 3x + 1\) has a slope of 3.
   • The equation \(y \geq -x + 2\) has a slope of -1.
   • This statement is false.

2. Both boundary lines are solid.

   • The inequality \(y \leq 3x + 1\) is a solid line because it includes equal to (≤).
   • The inequality \(y \geq -x + 2\) is also a solid line because it includes equal to (≥).
   • This statement is true.

3. A solution to the system is (1, 3).

   • Let's test the point \((1, 3)\) in both inequalities:
     • For \(y \leq 3x + 1\): \[ 3 \leq 3(1) + 1 \quad \Rightarrow \quad 3 \leq 4 \quad \text{(true)} \]
     • For \(y \geq -x + 2\): \[ 3 \geq -1 + 2 \quad \Rightarrow \quad 3 \geq 1 \quad \text{(true)} \]
   • Since both conditions are satisfied, \( (1, 3) \) is indeed a solution.
   • This statement is true.

4. Both inequalities are shaded below the boundary lines.

   • The inequality \(y \leq 3x + 1\) means shading below the line \(y = 3x + 1\).
   • The inequality \(y \geq -x + 2\) means shading above the line \(y = -x + 2\).
   • Therefore, they are not both shaded below their respective boundary lines.
   • This statement is false.

5. The boundary lines intersect.

   • To find the intersection point, we can set \(3x + 1 = -x + 2\): \[ 3x + 1 = -x + 2 \ 3x + x = 2 - 1 \ 4x = 1 \ x = \frac{1}{4} \]
   • Substitute \(x = \frac{1}{4}\) into either equation to find \(y\): \[ y = 3\left(\frac{1}{4}\right) + 1 = \frac{3}{4} + 1 = \frac{7}{4} \]
   • Thus, the lines intersect at \(\left(\frac{1}{4}, \frac{7}{4}\right)\).
   • This statement is true.

Summary:

The true statements about the graph are:

• Both boundary lines are solid.
• A solution to the system is (1, 3).
• The boundary lines intersect.

Answer 6

On a coordinate plane, a solid straight line has a positive slope and goes through (negative 4, 0) and (0, 2). Everything below the line is shaded. A point is shown at (1, 1).Mr. Hernandez plotted the point (1, 1) on Han’s graph of y ≤ One-halfx + 2. He instructed Han to add a second inequality to the graph that would include the solution (1, 1).Which equation could Miguel write?y > 2x + 1y < 2x – 1y ≥ 2x + 1y ≤ 2x – 1

Answer 7

To determine which inequality could be used to add to Han's graph that includes the solution \((1, 1)\), let's analyze the given information step by step.

Given Inequality:

1. Inequality 1: \(y \leq \frac{1}{2}x + 2\)
   • This line passes through the points \((-4, 0)\) and \((0, 2)\) and it is shaded below this line.

Evaluating the Point (1, 1):

2. Testing the Point (1, 1): We need to check whether \((1, 1)\) satisfies the inequality \(y \leq \frac{1}{2}x + 2\): \[ 1 \leq \frac{1}{2}(1) + 2 \ 1 \leq \frac{1}{2} + 2 \ 1 \leq 2.5 \quad \text{(true)} \]
   • So, the point \((1, 1)\) is indeed a solution for the first inequality.

Second Inequality Options:

Next, we will test each of the proposed inequalities to see if any of them include the point \((1, 1)\):

1. Option 1: \(y > 2x + 1\)

   • Testing \((1, 1)\): \[ 1 > 2(1) + 1 \ 1 > 2 + 1 \ 1 > 3 \quad \text{(false)} \]

2. Option 2: \(y < 2x - 1\)

   • Testing \((1, 1)\): \[ 1 < 2(1) - 1 \ 1 < 2 - 1 \ 1 < 1 \quad \text{(false)} \]

3. Option 3: \(y \geq 2x + 1\)

   • Testing \((1, 1)\): \[ 1 \geq 2(1) + 1 \ 1 \geq 2 + 1 \ 1 \geq 3 \quad \text{(false)} \]

4. Option 4: \(y \leq 2x - 1\)

   • Testing \((1, 1)\): \[ 1 \leq 2(1) - 1 \ 1 \leq 2 - 1 \ 1 \leq 1 \quad \text{(true)} \]
   • This is true, meaning the point \((1, 1)\) is included in this inequality.

Conclusion:

Mr. Hernandez could write the inequality:

• \(y \leq 2x - 1\) as the second inequality that would include the solution \((1, 1)\).

The correct answer is: y ≤ 2x - 1.