First, let's calculate the amount of money required to purchase two of each type of ball:
![\begin{gathered} C=2(8+9+7+5)\\ \\ C=2\cdot29\\ \\ C=58 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2qyjzzgv610f2k302zl8grnggejkma0n4e.png)
Now, to complete the remaining amount of money, we can choose more balls until the total cost reaches $150.
For example, let's add two more of each type of ball. So the total cost will be 116.
Then, to complete the remaining $34, let's create two possibilities of adding more balls:
- 2 baseballs and 4 softballs
- 3 basketballs and 1 baseball
Therefore two possibilities are:
4 footballs, 4 basketballs, 6 baseballs and 8 softballs;
4 footballs, 7 basketballs, 5 baseballs and 4 softballs.
Let's check if the total cost of each case is $150:
![\begin{gathered} C_1=4\cdot8+4\cdot9+6\cdot7+8\cdot5=32+36+42+40=150\\ \\ C_2=4\cdot8+7\cdot9+5\cdot7+4\cdot5=32+63+35+20=150 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/amq56zq8r081la54uvr80fu80sg4q9ynb0.png)