2x + 3y < 6, x - y > 1. Describe how you could determine whether a given point is a solution to your system of

Question

linear inequalities. Include an example in your description.

Answer 1

To determine whether a given point is a solution to the system of linear inequalities defined by \(2x + 3y < 6\) and \(x - y > 1\), you need to substitute the coordinates of the point into each inequality and check if the inequalities hold true.

Steps to Determine if a Point is a Solution:

1. Select a Point: Choose a point \((x, y)\). For example, let’s use the point \((2, 1)\).

2. Substitute into the First Inequality:

   • Substitute \(x = 2\) and \(y = 1\) into the first inequality: \[ 2(2) + 3(1) < 6 \] Simplifying this, \[ 4 + 3 < 6 \implies 7 < 6 \] This inequality is false.

3. Substitute into the Second Inequality:

   • Next, substitute the same values into the second inequality: \[ 2 - 1 > 1 \] Simplifying, \[ 1 > 1 \] This inequality is also false.

4. Determine Overall Solution: Since both inequalities must be satisfied for \((2, 1)\) to be a solution to the system, and both inequalities are false, the point \((2, 1)\) is not a solution to the system.

Example of a Valid Point:

Let’s use a different point, say \((0, 1)\).

1. First Inequality: \[ 2(0) + 3(1) < 6 \] Simplifying, \[ 0 + 3 < 6 \implies 3 < 6 \] This is true.

2. Second Inequality: \[ 0 - 1 > 1 \] Simplifying, \[ -1 > 1 \] This is false.

Conclusion:

In this case, since the first inequality is true but the second is false, the point \((0, 1)\) is also not a solution.

Solution Verification:

For a point to be a solution to the system of inequalities, both inequalities need to be satisfied simultaneously.

Final Note: Choose different points and perform the same checks until you find a point that makes both inequalities true. For example, the point \((3, -1)\):

1. Checking \(2(3) + 3(-1) < 6\): \[ 6 - 3 < 6 \implies 3 < 6 \text{ (true)} \]
2. Checking \(3 - (-1) > 1\): \[ 3 + 1 > 1 \implies 4 > 1 \text{ (true)} \] Thus, \((3, -1)\) is a valid solution for this system of inequalities.

Answer 2

Simplify what you said.

Answer 3

To determine whether a given point is a solution to the system of inequalities \(2x + 3y < 6\) and \(x - y > 1\), follow these steps:

1. Choose a Point: Select a point, for example, \((2, 1)\).

2. Check the First Inequality: Substitute \(x\) and \(y\) into the first inequality: \[ 2(2) + 3(1) < 6 \implies 4 + 3 < 6 \implies 7 < 6 \text{ (false)} \]

3. Check the Second Inequality: Now check the second inequality: \[ 2 - 1 > 1 \implies 1 > 1 \text{ (false)} \]

Since both inequalities must be true for the point to be a solution, and in this case both are false, \((2, 1)\) is not a solution.

Example of a Valid Point:

Now let’s try the point \((3, 0)\):

1. First Inequality: \[ 2(3) + 3(0) < 6 \implies 6 < 6 \text{ (false)} \]

2. Second Inequality: \[ 3 - 0 > 1 \implies 3 > 1 \text{ (true)} \]

So \((3, 0)\) is also not a solution.

Finding a Valid Point:

After testing some points, you might find \((2, 0)\):

1. First Inequality: \[ 2(2) + 3(0) < 6 \implies 4 < 6 \text{ (true)} \]

2. Second Inequality: \[ 2 - 0 > 1 \implies 2 > 1 \text{ (true)} \]

Since both inequalities are true for \((2, 0)\), this point is a valid solution to the system.

Answer 4

Graph the solution set for the system of linear inequalities on a piece of graph paper. Make
sure to clearly indicate where the solution is on the graph.

Answer 5

While I cannot directly create a graph on paper for you, I can guide you through the steps to graph the solution set for the system of linear inequalities \(2x + 3y < 6\) and \(x - y > 1\).

Steps to Graph the Inequalities:

1. Graph the First Inequality: \(2x + 3y < 6\)

   • Convert to Equation: Start by converting the inequality to an equation:
     \[ 2x + 3y = 6 \]
   • Find X-Intercept: Set \(y = 0\):
     \[ 2x = 6 \implies x = 3 \quad \text{(Point: (3, 0))} \]
   • Find Y-Intercept: Set \(x = 0\):
     \[ 3y = 6 \implies y = 2 \quad \text{(Point: (0, 2))} \]
   • Draw the Line: Plot the points (3, 0) and (0, 2). Connect them with a dashed line because the inequality is strict (<).
   • Shade the Region: Since it’s \(<\), shade below the line.

2. Graph the Second Inequality: \(x - y > 1\)

   • Convert to Equation: Convert to the equation:
     \[ x - y = 1 \]
   • Find X-Intercept: Set \(y = 0\):
     \[ x = 1 \quad \text{(Point: (1, 0))} \]
   • Find Y-Intercept: Set \(x = 0\):
     \[ -y = 1 \implies y = -1 \quad \text{(Point: (0, -1))} \]
   • Draw the Line: Plot the points (1, 0) and (0, -1). Connect them with a dashed line because the inequality is strict (>).
   • Shade the Region: Since it’s \(>\), shade above the line.

Finding the Solution Set:

The solution set for the system of inequalities is the region where the shaded areas of both inequalities overlap.

Example Points in the Solution Set:

To confirm that a certain point is in the solution set, you can test points within the overlapping shaded area, such as (2, 0) or (2, 1).

Finalizing the Graph:

• Make sure to label the two inequalities as \(2x + 3y < 6\) and \(x - y > 1\).
• Clearly indicate the shaded regions and where the solution set is located in the overlap of the two regions.

This graphical representation will help visualize the solution set of the given inequalities! If you proceed with creating it on graph paper, ensure proper labeling and use of dashed lines for strict inequalities.

Answer 6

Could a system of linear inequalities ever have no solutions? What would this look like on
a graph? Explain in a short-form.

Answer 7

Yes, a system of linear inequalities can have no solutions. This occurs when the shaded regions of the inequalities do not overlap at all.

Example:

Consider the system:

1. \(y > 2\) (a horizontal line at \(y = 2\) with shading above)
2. \(y < 1\) (a horizontal line at \(y = 1\) with shading below)

Graph Representation:

• On the graph, the line \(y = 2\) is dashed (indicating it is not included), and the region above it is shaded.
• The line \(y = 1\) is also dashed, and the region below it is shaded.

Visualization:

The two shaded areas do not intersect; one is above \(y = 2\) and the other is below \(y = 1\). Thus, there are no points that satisfy both inequalities, indicating that the system has no solutions.