Question
if lydia buys 1/2 pound of the breakfast tea and 2 pounds of the dark roast coffee, hiw ay 1-pound bags of pumpkin spice coffee she can buy? graph the solution set on the number line
-1 0 1 2 3 4 5 6 7 8 9
-1 0 1 2 3 4 5 6 7 8 9
Answers
Answered by
GPT-4o mini
To determine how many 1-pound bags of pumpkin spice coffee Lydia can buy, we first need to establish her budget for purchasing the breakfast tea and dark roast coffee.
Given:
- Price per 1/2 pound of breakfast tea = \( P_T \)
- Price per pound of dark roast coffee = \( P_C \)
- She buys \( \frac{1}{2} \) pound of breakfast tea.
- She buys 2 pounds of dark roast coffee.
Calculating the total cost of the breakfast tea:
- Cost of breakfast tea = \( \frac{1}{2} P_T \)
Calculating the total cost of the dark roast coffee:
- Cost of dark roast coffee = \( 2 P_C \)
Let’s assume Lydia has a budget \( B \) for these purchases, then we can set up the inequality:
\[
\text{Total cost} + \text{Cost of pumpkin spice coffee} \leq B
\]
Where the cost of pumpkin spice coffee is \( P_S \times x \) (where \( x \) is the number of 1-pound bags).
Thus, the inequality can be written as:
\[
\frac{1}{2} P_T + 2 P_C + x P_S \leq B
\]
To find the maximum number of bags of pumpkin spice coffee (\( x \)):
\[
x P_S \leq B - \left( \frac{1}{2} P_T + 2 P_C \right)
\]
\[
x \leq \frac{B - \left( \frac{1}{2} P_T + 2 P_C \right)}{P_S}
\]
Since you did not provide specific prices or a budget, let's assume:
- \( P_T = 4 \) (for 1/2 pound of breakfast tea, the full pound is $8)
- \( P_C = 10 \) (for 1 pound of dark roast coffee)
- \( P_S = 6 \) (for 1 pound of pumpkin spice coffee)
- \( B = 30 \)
Now we can plug these values into the formula:
1. Calculate cost of breakfast tea:
\[
\text{Cost of breakfast tea} = \frac{1}{2} \times 8 = 4
\]
2. Calculate cost of dark roast coffee:
\[
\text{Cost of dark roast coffee} = 2 \times 10 = 20
\]
3. Now, plug these into the inequality:
\[
x \leq \frac{30 - (4 + 20)}{6}
\]
\[
x \leq \frac{30 - 24}{6}
\]
\[
x \leq \frac{6}{6}
\]
\[
x \leq 1
\]
So, Lydia can buy a maximum of **1 pound bag** of pumpkin spice coffee.
### Graphing the Solution Set on the Number Line
The solution \( x \leq 1 \) indicates that she can buy 0 or 1 bag. To graph this on the provided number line (-1 to 9):
- Draw a solid point at **1**, indicating that she can buy 1 bag.
- Shade to the left of this point (including 0), to indicate that she can also buy 0 bags.
```
-1 0 [---------•====|==========|===========|===========|===========|===========|===========|===========|===========|====] 9
1
```
Here, the solid point represents that she can buy exactly 1 bag, and the shaded area to the left includes the possibility of buying 0 bags.
Given:
- Price per 1/2 pound of breakfast tea = \( P_T \)
- Price per pound of dark roast coffee = \( P_C \)
- She buys \( \frac{1}{2} \) pound of breakfast tea.
- She buys 2 pounds of dark roast coffee.
Calculating the total cost of the breakfast tea:
- Cost of breakfast tea = \( \frac{1}{2} P_T \)
Calculating the total cost of the dark roast coffee:
- Cost of dark roast coffee = \( 2 P_C \)
Let’s assume Lydia has a budget \( B \) for these purchases, then we can set up the inequality:
\[
\text{Total cost} + \text{Cost of pumpkin spice coffee} \leq B
\]
Where the cost of pumpkin spice coffee is \( P_S \times x \) (where \( x \) is the number of 1-pound bags).
Thus, the inequality can be written as:
\[
\frac{1}{2} P_T + 2 P_C + x P_S \leq B
\]
To find the maximum number of bags of pumpkin spice coffee (\( x \)):
\[
x P_S \leq B - \left( \frac{1}{2} P_T + 2 P_C \right)
\]
\[
x \leq \frac{B - \left( \frac{1}{2} P_T + 2 P_C \right)}{P_S}
\]
Since you did not provide specific prices or a budget, let's assume:
- \( P_T = 4 \) (for 1/2 pound of breakfast tea, the full pound is $8)
- \( P_C = 10 \) (for 1 pound of dark roast coffee)
- \( P_S = 6 \) (for 1 pound of pumpkin spice coffee)
- \( B = 30 \)
Now we can plug these values into the formula:
1. Calculate cost of breakfast tea:
\[
\text{Cost of breakfast tea} = \frac{1}{2} \times 8 = 4
\]
2. Calculate cost of dark roast coffee:
\[
\text{Cost of dark roast coffee} = 2 \times 10 = 20
\]
3. Now, plug these into the inequality:
\[
x \leq \frac{30 - (4 + 20)}{6}
\]
\[
x \leq \frac{30 - 24}{6}
\]
\[
x \leq \frac{6}{6}
\]
\[
x \leq 1
\]
So, Lydia can buy a maximum of **1 pound bag** of pumpkin spice coffee.
### Graphing the Solution Set on the Number Line
The solution \( x \leq 1 \) indicates that she can buy 0 or 1 bag. To graph this on the provided number line (-1 to 9):
- Draw a solid point at **1**, indicating that she can buy 1 bag.
- Shade to the left of this point (including 0), to indicate that she can also buy 0 bags.
```
-1 0 [---------•====|==========|===========|===========|===========|===========|===========|===========|===========|====] 9
1
```
Here, the solid point represents that she can buy exactly 1 bag, and the shaded area to the left includes the possibility of buying 0 bags.