Let x be the number of pounds of the first type ($4.40) of nut, and y be the number of pounds of the other one, then we can set the following system of equations:
![\begin{gathered} x+y=8(1)/(2)=8.5, \\ 4.40x+6.10y=5.80\cdot8.50=49.30. \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ufus3bo8a2b6aobvqv4cs0d4zeltup7a3d.png)
Solving the first equation for x we get:
![x=8.5-y\text{.}](https://img.qammunity.org/2023/formulas/mathematics/college/sq2p77wmzs4ntadzdqjr0dnza9attzv8ko.png)
Substituting the above result in the second equation we get:
![\begin{gathered} 4.40(8.5-y)+6.10y=49.30, \\ 37.40-4.40y+6.10y=49.30, \\ 1.7y=11.90, \\ y=7. \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qdc1zwax8ice7hodihusadqzzb0dxkm5to.png)
Substituting y=7 in the third equation on the board we get:
![x=8.5-7=1.5.](https://img.qammunity.org/2023/formulas/mathematics/college/xofksynnhil28hgkyvb21sv5y3zmbeo6dp.png)
Answer: You will need to add 1.5 pounds of the $4.40 nut and 7 pounds of the other one.