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Describe how you could determine whether a given point is a solution to your system of

linear inequalities. Include an example in your description.
x+y=5
x-y=2

5 answers

To determine whether a given point is a solution to the system of linear inequalities, you first need to clarify that the equations you've provided are equalities, not inequalities. However, we can work with those equations to see if a specific point satisfies both.

Given the equations:

  1. \( x + y = 5 \)
  2. \( x - y = 2 \)

To check if a point (let's call it \( (x_0, y_0) \)) is a solution, follow these steps:

  1. Substitute the coordinates of the point into both equations.
  2. Evaluate each equation:
    • For the first equation, check if substituting \( (x_0, y_0) \) results in a true statement.
    • For the second equation, repeat the process.
  3. Conclusion: If both equations are true with the given point, then that point is a solution to the system.

Example

Let’s determine if the point \( (3, 2) \) is a solution to the system.

  1. Substituting into the first equation: \[ x + y = 5 \ 3 + 2 = 5 \quad \text{(True)} \]

  2. Substituting into the second equation: \[ x - y = 2 \ 3 - 2 = 2 \quad \text{(True)} \]

Since both equations are satisfied, the point \( (3, 2) \) is a solution to the system of equations \( x + y = 5 \) and \( x - y = 2 \).

Additional Example with an Inequality

If you were working with a system of inequalities, say:

  1. \( x + y > 5 \)
  2. \( x - y < 2 \)

You would proceed similarly but would check for inequalities:

For the point \( (4, 2) \):

  1. Check the first inequality: \[ x + y > 5 \ 4 + 2 > 5 \quad \text{(True, since 6 > 5)} \]

  2. Check the second inequality: \[ x - y < 2 \ 4 - 2 < 2 \quad \text{(False, since 2 is not less than 2)} \]

In this case, since the second condition is not satisfied, \( (4, 2) \) would not be a solution to the system of inequalities.

The method is consistent across equations and inequalities: substitute the point into the given expressions and verify the conditions to see if they are satisfied.

could you make it more simple

Sure! Let's simplify how to determine if a point is a solution to a set of equations or inequalities.

Steps to Check a Point

  1. Choose a point \((x_0, y_0)\) you want to check.
  2. Plug the point into each equation.
  3. Check if both equations are true:
    • If both are true, the point is a solution.
    • If one or both are false, the point is not a solution.

Example with Equations

Given equations:

  1. \(x + y = 5\)
  2. \(x - y = 2\)

Let’s check the point \( (3, 2) \):

  1. Plug the point into the first equation: \[ 3 + 2 = 5 \quad \text{(True)} \]

  2. Plug the point into the second equation: \[ 3 - 2 = 2 \quad \text{(True)} \]

Since both are true, the point (3, 2) is a solution.

Example with Inequalities

Given inequalities:

  1. \(x + y > 5\)
  2. \(x - y < 2\)

Let’s check the point \( (4, 2) \):

  1. Plug the point into the first inequality: \[ 4 + 2 > 5 \quad \text{(True, since 6 > 5)} \]

  2. Plug the point into the second inequality: \[ 4 - 2 < 2 \quad \text{(False, since 2 is not less than 2)} \]

Since the second part is false, the point (4, 2) is not a solution.

Summary

  1. Choose a point.
  2. Substitute the point into each equation/inequality.
  3. Check if all conditions are satisfied to determine if it's a solution.
Could a system of linear inequalities ever have no solutions? What would this look like on
a graph? Explain.

Yes, a system of linear inequalities can indeed have no solutions. This situation occurs when the inequalities describe regions on a graph that do not overlap at all.

Example of a System with No Solutions

Let's look at an example:

  1. \( y \leq x + 1 \) (This represents the area below or on the line \( y = x + 1 \))
  2. \( y \geq x + 3 \) (This represents the area above or on the line \( y = x + 3 \))

Graphical Representation

  1. Graph the first inequality:
    • The line \( y = x + 1 \) has a slope of 1 and a y-intercept of 1. The area below this line is included in the solution (since it is "less than or equal to").
  2. Graph the second inequality:
    • The line \( y = x + 3 \) also has a slope of 1 but a y-intercept of 3. The area above this line is included in the solution (since it is "greater than or equal to").

Visualization

On a graph, you would draw both lines:

  • The line for \( y = x + 1 \) would extend upwards and to the right starting from the point (0, 1).
  • The line for \( y = x + 3 \) would extend upwards and to the right starting from the point (0, 3).

When you shade the regions created by these inequalities:

  • The area below the first line \( y = x + 1 \) includes all points where \( y \) values are less than or equal to \( x + 1 \).
  • The area above the second line \( y = x + 3 \) includes all points where \( y \) values are greater than or equal to \( x + 3 \).

Conclusion

Because the region below \( y = x + 1 \) and the region above \( y = x + 3 \) do not overlap at any point on the graph, there are no points that satisfy both inequalities simultaneously. Thus, this system has no solutions.

In summary, a system of linear inequalities can have no solutions if the shaded areas of the inequalities do not intersect at all on the graph, meaning there are no points that meet all conditions at once.