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Graph the solution set for the inequality — 1.50x + 0.50y = 20 — on a piece of graph paper.

Note: At this point, do not yet consider any constraints given by the situation.

11 answers

To graph the inequality \(-1.50x + 0.50y = 20\), we first need to convert it into the slope-intercept form \(y = mx + b\).

  1. Rearranging the equation: Start with the original equation: \[ -1.50x + 0.50y = 20 \]

    First, isolate \(y\): \[ 0.50y = 1.50x + 20 \]

    Now, divide every term by \(0.50\): \[ y = 3x + 40 \]

  2. Finding points for graphing: This equation tells us that the y-intercept (where the line crosses the y-axis) is \(40\), and the slope \(m\) is \(3\). You can find two points to draw the line.

    • Point 1 (y-intercept): When \(x = 0\): \[ y = 3(0) + 40 = 40 \] So, point 1 is \((0, 40)\).

    • Point 2: When \(x = 5\): \[ y = 3(5) + 40 = 15 + 40 = 55 \] So, point 2 is \((5, 55)\).

  3. Drawing the line:

    • Plot the points \((0, 40)\) and \((5, 55)\) on the graph paper.
    • Draw the line that passes through these points. This will be a straight line with a slope of \(3\).

  4. Determining the type of line: Since the original problem was an equation (not an inequality), you would normally draw a solid line to indicate that all points on this line satisfy the equation. However, since you've mentioned the inequality but didn't specify it, you might consider it as inequality like \(y \leq 3x + 40\) or \(y > 3x + 40\) later.

  5. Shading the region: If you want to illustrate an inequality, you would typically shade above the line for \(y > 3x + 40\) or below for \(y \leq 3x + 40\).

  6. Final graph: On your graph paper, the line will slope upward from left to right, starting from \((0, 40)\) and going through \((5, 55)\). The shading (if needed) would depend on the inequality direction, which you may define in the next steps.

This completed process will yield the graph you need on paper based on the given equation. If you eventually have a specific inequality, this approach can be adjusted accordingly!

So what is the answer?

To summarize the steps clearly:

  1. Equation: You are working with the equation \(-1.50x + 0.50y = 20\).

  2. Convert to slope-intercept form:

    • Rearranged to: \[ y = 3x + 40 \]

  3. Key points for graphing:

    • y-intercept: \((0, 40)\)
    • When \(x = 5\): \((5, 55)\)

  4. Graphing:

    • Plot the points \((0, 40)\) and \((5, 55)\) on your graph paper.
    • Draw a solid line through these points, indicating \(y = 3x + 40\).

  5. Line type:

    • Since the equation provided is a straight equation, the line is solid.

  6. Shading (Inequality context):

    • If you treat the equation as part of an inequality (e.g., if it were \(y \leq 3x + 40\) or \(y > 3x + 40\)), you would shade the appropriate area:
      • For \(y \leq 3x + 40\), shade below the line.
      • For \(y > 3x + 40\), shade above the line.

Answer:

Graph the line represented by the equation \(y = 3x + 40\) by plotting points \((0, 40)\) and \((5, 55)\) on graph paper. Draw a solid line connecting these points.

If you later incorporate an inequality (e.g., \(y > 3x + 40\)), add shading above the line; for \(y \leq 3x + 40\), shade below the line.

What are the constraints of the situation?

To determine the constraints of a situation, we generally need context regarding what is being modeled by the inequality \(-1.50x + 0.50y = 20\). However, since the context isn't provided, I can offer some common constraints that might apply, especially if this relates to a real-world problem (like budgeting or production).

If we consider this equation in the context of a real-world scenario, such as:

  1. Budgeting:

    • Assume \(x\) represents the number of items (e.g., units produced or sold), and \(y\) represents total revenue.
    • A constraint could be that both \(x\) and \(y\) must be non-negative: \( x \geq 0 \) and \( y \geq 0 \), as negative quantities typically do not make sense in most practical situations.

  2. Production Limits:

    • If there is a limited availability of resources, you might have a constraint on \( x \) such as a maximum number of items that can be produced: \( x \leq \text{maximum units} \).
    • You might have a constraint on total profits or revenues, such as \( y \leq \text{maximum revenue} \).

  3. Raw Material or Time Constraints:

    • If \( x \) represents time spent on production and \( y \) represents output, constraints could exist based on available working hours or raw materials available.

Example Constraints:

  • Non-negativity: \( x \geq 0 \) and \( y \geq 0 \)
  • Production/Cost constraints: May have limits based on sales forecasts or budgets.

Without specific context, it’s difficult to provide exact constraints. If you can give a specific situation where the inequality is being applied, I could offer more tailored constraints relevant to that case.

Conclusion:

In general, constraints will depend on the specifics of the problem and may involve limits on inputs, outputs, resources, or costs.

What are the constraints of the inequality 10.50c + 5.50a ≤ 150

The inequality \(10.50c + 5.50a \leq 150\) likely represents a situation involving budgeting or resource constraints, where:

  • \(c\) could represent the number of a certain item or category of expenditure (for example, categories of products or costs).
  • \(a\) might represent another category of expenditures or items.

Given that context, we can infer some constraints:

General Constraints:

  1. Non-Negativity Constraints:

    • Typically, in problems involving quantities of items or resources, the quantities cannot be negative. Therefore, the following constraints would be typical: \[ c \geq 0 \] \[ a \geq 0 \]

  2. Budget Constraint:

    • The inequality \(10.50c + 5.50a \leq 150\) suggests that the total cost represented by \(10.50c\) and \(5.50a\) should not exceed $150. This is your primary constraint, indicating that the amounts spent (or allocated) on \(c\) and \(a\) together must stay within the bounds of your budget or resources.

Possibly Further Constraints:

Depending on the specific context, additional constraints may apply, including:

  1. Tool or Resource Limits:

    • There may be upper limits on how many of \(c\) and \(a\) can be produced or purchased, based on availability, production capacity, or market demand.

  2. Ratio or Relationship Constraints:

    • If there are other relationships between \(c\) and \(a\) based on operational requirements, you might have constraints like: \[ c \leq k \cdot a \quad (k \text{ being some proportion}) \] (This would depend on the specific relationship in your scenario.)

Example Context:

If you were purchasing items for an event, each type \(c\) and \(a\) might represent different kinds of supplies (e.g., catering and decorations), and the constraints ensure that the total expenditures do not exceed the available budget of $150.

Conclusion:

The primary constraints based on the inequality \(10.50c + 5.50a \leq 150\) would be:

  • \(c \geq 0\)
  • \(a \geq 0\)
  • \(10.50c + 5.50a \leq 150\)

Additional constraints may arise based on the specifics of the situation you are modeling.

Show all viable graph solutions.

To graph the inequality \(10.50c + 5.50a \leq 150\) and find all viable solutions, we will follow these steps:

Step 1: Rewrite the Equation

First, we'll convert the inequality to the equation to find the line: \[ 10.50c + 5.50a = 150 \]

Step 2: Finding Intercepts

To graph this line, we can find the intercepts:

  1. Find the y-intercept (set \(c = 0\)): \[ 10.50(0) + 5.50a = 150 \implies 5.50a = 150 \implies a = \frac{150}{5.50} \approx 27.27 \] So, the y-intercept is \((0, 27.27)\).

  2. Find the x-intercept (set \(a = 0\)): \[ 10.50c + 5.50(0) = 150 \implies 10.50c = 150 \implies c = \frac{150}{10.50} \approx 14.29 \] So, the x-intercept is \((14.29, 0)\).

Step 3: Graphing the Line

Now, plot the intercepts:

  • Plot the point \((0, 27.27)\) on the vertical axis (y-axis).
  • Plot the point \((14.29, 0)\) on the horizontal axis (c-axis).

Draw a solid line connecting these two points because the inequality is "less than or equal to."

Step 4: Shading the Area

The inequality \(10.50c + 5.50a \leq 150\) means that we shade the area below the line (since we want values where the total is less than or equal to 150).

Step 5: Including Non-negativity Constraints

Since we also have the constraints \(c \geq 0\) and \(a \geq 0\), we only consider the first quadrant of the coordinate system (where both coordinates are non-negative).

Summary of the Graph

  1. Solid Line: The boundary line represented by the equation \(10.50c + 5.50a = 150\).
  2. Shaded Region: The area under and including the line (where \(a\) and \(c\) satisfy the inequality).
  3. First Quadrant Only: Only include the region in the first quadrant since both \(c\) and \(a\) should be non-negative.

Resulting Graph Visualization

Here is a description of the final graph:

  • The x-axis (c-axis) will range from \(0\) to about \(14.29\) (on the right).
  • The y-axis (a-axis) will range from \(0\) to about \(27.27\) (upwards).
  • The line drawn between \((0, 27.27)\) and \((14.29, 0)\) is solid, indicating that points on this line satisfy the equality, while points below it satisfy the inequality.
  • The shading will cover all points in the first quadrant below the line.

Conclusion

The graph will show all viable solutions for the given inequality in the feasible region of non-negative \(c\) and \(a\) within the bounds indicated by your inequality \(10.50c + 5.50a \leq 150\).

If you need an actual visual representation, please use graphing software or graphing paper to plot the coordinates and shade the viable area.

What are the solutions?

To find the solutions for the inequality \(10.50c + 5.50a \leq 150\), we can identify the feasible set of \((c, a)\) pairs that satisfy the inequality based on our previously discussed graph.

Steps to Identify Solutions:

  1. Identifying the boundary: The boundary line of the inequality is given by the equation: \[ 10.50c + 5.50a = 150. \] We've already found the intercepts:

    • \(c\)-intercept: \((14.29, 0)\)
    • \(a\)-intercept: \((0, 27.27)\)

  2. Shaded Region: Since the inequality is \( \leq \), we shade the region below and including the line in the first quadrant (\(c \geq 0\) and \(a \geq 0\)).

Potential Solutions:

The solutions to the inequality are all the points \((c, a)\) such that:

\[ 10.50c + 5.50a \leq 150, \] along with the constraints: \[ c \geq 0, \] \[ a \geq 0. \]

Points within the Feasible Region:

To give you an idea of specific solutions, here are examples of points that satisfy the inequality:

  1. Boundary Points (on the line):

    • \(c = 0\), \(a = 27.27\) (when \(c = 0\))
    • \(c = 14.29\), \(a = 0\) (when \(a = 0\))
    • \(c = 5\),
      \[ 10.50(5) + 5.50a = 150 \implies 52.5 + 5.50a = 150 \implies 5.50a = 97.5 \implies a = \frac{97.5}{5.50} \approx 17.73. \] So, one point is \((5, 17.73)\).

  2. Any Point Below the Line: Here are some example points below the line:

    • \( (0, 0) \): Substitute into the inequality to get \(10.50(0) + 5.50(0) = 0 \leq 150\) (valid).
    • \( (5, 10) \): \[ 10.50(5) + 5.50(10) = 52.5 + 55 = 107.5 \leq 150 \quad (\text{valid}). \]
    • \( (10, 5) \): \[ 10.50(10) + 5.50(5) = 105 + 27.5 = 132.5 \leq 150 \quad (\text{valid}). \]
    • \( (14, 0) \): \[ 10.50(14) + 5.50(0) = 147 \leq 150. \]
    • \( (1, 25) \): \[ 10.50(1) + 5.50(25) = 10.50 + 137.5 = 148 \leq 150 \quad (\text{valid}). \]

Conclusion

The solution set is any point \((c, a)\) in the area below the line \(10.50c + 5.50a = 150\), including points like:

  • \((0, 0)\)
  • \((5, 10)\)
  • \((10, 5)\)
  • \((14, 0)\)
  • \((1, 25)\)

And this extends to other points within the region defined by non-negative values of \(c\) and \(a\).

If you need a specific list or a greater number of solutions, any combinations of \(c\) and \(a\) that satisfy \(10.50c + 5.50a \leq 150\) are valid. You could calculate more specific values using the equation based on sets of \(c\) or \(a\) and checking the resulting outputs.