Andre is buying snacks for his team. He buys apricots for [tex]\$6[/tex] a pound and 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 to show how many pounds of apricots and bananas he bought.

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

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

Sure! Let's solve the problem step-by-step to find out how many pounds of apricots and bananas Andre bought.

1. Understand the Problem:
- Andre buys two types of snacks: apricots and dried bananas.
- Apricots cost \[tex]$6 per pound.
- Dried bananas cost \$[/tex]4 per pound.
- He buys a total of 5 pounds of snacks.
- He spends a total of \$24.50.

2. Set Up the Equations:
- Let [tex]$ a $[/tex] be the pounds of apricots.
- Let [tex]$ b $[/tex] be the pounds of dried bananas.

Based on the information:
- Equation for the total pounds:
[tex]\[
a + b = 5
\][/tex]

- Equation for the total cost:
[tex]\[
6a + 4b = 24.50
\][/tex]

3. Solve the System of Equations:
We have two equations:
[tex]\[
\begin{align*}
1) & \quad a + b = 5 \\
2) & \quad 6a + 4b = 24.50
\end{align*}
\][/tex]

- From equation (1), express [tex]$ b $[/tex] in terms of [tex]$ a $[/tex]:
[tex]\[
b = 5 - a
\][/tex]

- Substitute [tex]$ b = 5 - a $[/tex] into equation (2):
[tex]\[
6a + 4(5 - a) = 24.50
\][/tex]

- Simplify and solve for [tex]$ a $[/tex]:
[tex]\[
6a + 20 - 4a = 24.50
\][/tex]
[tex]\[
2a + 20 = 24.50
\][/tex]
[tex]\[
2a = 4.50
\][/tex]
[tex]\[
a = 2.25
\][/tex]

- Now, use the value of [tex]$ a $[/tex] to find [tex]$ 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 dried bananas.

By following these steps, we can determine the number of pounds Andre bought for each snack.