Meha is helping her aunt to plan a birthday party for her cousin at the science center. Her aunt

To represent the situation with a two-variable inequality, we need to define the variables first:

Let:

• ( c ) = number of children attending the party
• ( a ) = number of adults attending the party

The total cost of the party can be expressed as:

[ \text{Total Cost} = 150 + 10.50c + 5.50a ]

Given that Meha's aunt has budgeted a maximum of $300 for the party, we can set up the inequality:

[ 150 + 10.50c + 5.50a \leq 300 ]

To isolate the variable terms, we can subtract 150 from both sides:

[ 10.50c + 5.50a \leq 150 ]

Additionally, since there must be at least one adult and at least one child at the party, we have the following constraints:

[ c \geq 1 \quad \text{(at least one child)} ] [ a \geq 1 \quad \text{(at least one adult)} ]

Putting it all together, the system of inequalities representing the situation is:

1. ( 10.50c + 5.50a \leq 150 )
2. ( c \geq 1 )
3. ( a \geq 1 )

These inequalities collectively represent the conditions that Meha's aunt has for the birthday party.

Is it possible for the party to include 10 children and five adults? Use your inequality from

part A, 10.50c + 5.50a ≤ 150, to check. Explain your answer.

To determine if it is possible for the party to include 10 children and 5 adults, we can use the inequality we derived:

[ 10.50c + 5.50a \leq 150 ]

We substitute ( c = 10 ) (number of children) and ( a = 5 ) (number of adults) into the inequality:

[ 10.50(10) + 5.50(5) \leq 150 ]

Calculating the left-hand side:

[ 10.50(10) = 105 ] [ 5.50(5) = 27.50 ]

Now, adding these two results together:

[ 105 + 27.50 = 132.50 ]

Now we check if the sum ( 132.50 ) is less than or equal to ( 150 ):

[ 132.50 \leq 150 ]

Since this inequality holds true, it is possible for the party to include 10 children and 5 adults while staying within the budget constraint.

Conclusion:

Yes, including 10 children and 5 adults is possible because the total cost of $132.50 is within the budget limit of $150.

Graph the equation, but do not yet consider the constraints.

To graph the equation derived from the inequality (10.50c + 5.50a = 150), we can start by rearranging it into a more familiar linear form, such as (y = mx + b). For this purpose, let's express (a) in terms of (c).

Starting with the equation:

[ 10.50c + 5.50a = 150 ]

First, isolate (a):

[ 5.50a = 150 - 10.50c ]

Now divide everything by (5.50):

[ a = \frac{150 - 10.50c}{5.50} ]

To make calculations easier, we can simplify ( \frac{150}{5.50} ) and ( \frac{10.50}{5.50} ):

1. ( \frac{150}{5.50} \approx 27.27 )
2. ( \frac{10.50}{5.50} \approx 1.91 )

This gives us the equation:

[ a \approx 27.27 - 1.91c ]

Finding Intercepts:

1. Find the y-intercept (where (c = 0)):

• Plug (c = 0) into the equation:

[ a = 27.27 \quad \text{(y-intercept)} ]

2. Find the x-intercept (where (a = 0)):

• Set (a = 0) and solve for (c):

[ 0 = 27.27 - 1.91c ] [ 1.91c = 27.27 \implies c = \frac{27.27}{1.91} \approx 14.27 \quad \text{(x-intercept)} ]

Summary of Intercepts:

• The y-intercept is approximately ( (0, 27.27) )
• The x-intercept is approximately ( (14.27, 0) )

Graphing:

1. Plotting the Intercepts:

   • Mark the point ( (0, 27.27) ) on the y-axis.
   • Mark the point ( (14.27, 0) ) on the x-axis.

2. Drawing the Line:

   • Draw a straight line connecting these two points, which represents the equation (10.50c + 5.50a = 150).

Note:

Since we have not yet considered the constraints ( c \geq 1 ) and ( a \geq 1 ), the entire line and area would be part of the graph until we apply those constraints. If you were to graph it on graph paper or using graphing software, just make sure to extend the line between those intercepts.

Graph the solution set for the inequality you wrote in part A, 10.50c + 5.50a ≤ 150.

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

To graph the solution set for the inequality (10.50c + 5.50a \leq 150), we will follow these steps:

1. Convert to the equation:

First, we'll use the corresponding equation to find the boundary line: [ 10.50c + 5.50a = 150 ]

2. Finding Intercepts:

As calculated earlier, the intercepts for the equation are:

• y-intercept ((c = 0)): [ a = 27.27 \quad \text{(point (0, 27.27))} ]
• x-intercept ((a = 0)): [ c \approx 14.27 \quad \text{(point (14.27, 0))} ]

3. Plotting the Line:

• Plot the points ((0, 27.27)) and ((14.27, 0)) on a coordinate plane (with (c) on the x-axis and (a) on the y-axis).
• Draw a solid line through these points because the inequality is non-strict (≤), meaning the boundary line itself is included in the solution set.

4. Shading the Solution Region:

The inequality (10.50c + 5.50a \leq 150) indicates that we want to find the region below the line (including the line itself), because we are looking for values of (c) and (a) where the total cost is less than or equal to $150.

5. Choosing Test Points:

To confirm that the correct region is shaded, you can choose a test point not on the line, such as ((0,0)):

• Substitute into the inequality: [ 10.50(0) + 5.50(0) \leq 150 \quad \Rightarrow \quad 0 \leq 150 \quad \text{(True)} ] Thus, the point ((0,0)) satisfies the inequality, confirming that this region is included.

Summary:

• The graph should have a solid line running from ((0, 27.27)) to ((14.27, 0)) and the area below the line shaded.

Note:

Since I can't create visual images directly in this text format, you can follow these instructions to graph it on paper or using graphing software. The shaded region would represent all possible combinations of the number of children ((c)) and adults ((a)) that keep the cost of the birthday party at or below $150.

Simplify.

To simplify the inequality (10.50c + 5.50a \leq 150), we can start by eliminating the decimal by multiplying through by 100. This helps to work with whole numbers:

Step 1: Eliminate Decimals

Multiply all terms in the inequality by 100:

[ 100(10.50c) + 100(5.50a) \leq 100(150) ]

This simplifies to:

[ 1050c + 550a \leq 15000 ]

Step 2: Divide by 50

Next, we can simplify the coefficients by dividing the entire inequality by 50:

[ \frac{1050c}{50} + \frac{550a}{50} \leq \frac{15000}{50} ]

This simplifies to:

[ 21c + 11a \leq 300 ]

Final Simplified Inequality:

The simplified form of the inequality is:

[ 21c + 11a \leq 300 ]

This inequality can be used for further analysis or graphing while maintaining the relationships originally described.