We define the following notation:
• z = cost of an item from shelf A,
,• y = cost of an item from shelf B.
From the statement, we know that:
• 3 items from shelf A and 2 from shelf B cost $26, so we have:
[tex]3z+2y=26,[/tex]• 2 items from shelf A and 5 from shelf B for $32, so we have:
[tex]2z+5y=32.[/tex]We have the following system of equations:
[tex]\begin{gathered} 3z+2y=26, \\ 2z+5y=32. \end{gathered}[/tex]1) To solve this system, we multiply the first equation by 2 and the second equation by 3:
[tex]\begin{gathered} 2\cdot(3z+2y)=2\cdot26\rightarrow6z+4y=52, \\ 3\cdot(2z+5y)=3\cdot32\rightarrow6z+15y=96. \end{gathered}[/tex]2) We subtract equation 1 to equation 2, and then we solve for y:
[tex]\begin{gathered} (6z+15y)-(6z+4y)=96-52, \\ 11y=44, \\ y=\frac{44}{11}=4. \end{gathered}[/tex]We found that y = 4.
3) We replace the value y = 4 in the first equation, and then we solve for z:
[tex]\begin{gathered} 3z+2\cdot4=26, \\ 3z+8=26, \\ 3z=26-8, \\ 3z=18, \\ z=\frac{18}{3}=6. \end{gathered}[/tex]Answer
The cost of the items are:
• z = 6, for items from shelf A,
,• y = 4, for items from shelf B.