Andre is buying snacks for his team. He buys pounds of apricots for [tex]$6[/tex] a pound and pounds of dried bananas for [tex]$4[/tex] per pound. He buys a total of 5 pounds and spends a total of [tex]\$24.50[/tex].

Write and solve a system of equations that will show how many pounds of apricots and bananas he bought.

1. Let [tex]a[/tex] be the pounds of apricots.
2. Let [tex]b[/tex] be the pounds of bananas.

Equations:
1. [tex]a + b = 5[/tex]
2. [tex]6a + 4b = 24.50[/tex]

Sure! Let's solve this problem step-by-step:

1. Define your variables:
- Let [tex]$ a $[/tex] represent the number of pounds of apricots.
- Let [tex]$ b $[/tex] represent the number of pounds of bananas.

2. Set up the system of equations:

- The first equation comes from the total weight of the snacks:
[tex]\[
a + b = 5
\][/tex]
This equation represents the fact that Andre buys a total of 5 pounds of snacks.

- The second equation comes from the total cost of the snacks:
[tex]\[
6a + 4b = 24.50
\][/tex]
This equation represents the total cost, where apricots cost [tex]$6 per pound and bananas $[/tex]4 per pound, and the total spending is $24.50.

3. Solve the system of equations:

- Start with the first equation to express one variable in terms of the other. Let's solve for [tex]$ b $[/tex]:
[tex]\[
b = 5 - a
\][/tex]

- Substitute [tex]$ b = 5 - a $[/tex] into the second equation:
[tex]\[
6a + 4(5 - a) = 24.50
\][/tex]

- Distribute and simplify:
[tex]\[
6a + 20 - 4a = 24.50
\][/tex]

- Combine like terms:
[tex]\[
2a + 20 = 24.50
\][/tex]

- Solve for [tex]$ a $[/tex]:
[tex]\[
2a = 24.50 - 20
\][/tex]
[tex]\[
2a = 4.50
\][/tex]
[tex]\[
a = 2.25
\][/tex]

- Now, substitute back to find [tex]$ b $[/tex]:
[tex]\[
b = 5 - a = 5 - 2.25 = 2.75
\][/tex]

4. Conclusion:
- Andre bought 2.25 pounds of apricots and 2.75 pounds of bananas.