Answer :
0. Building the equations for each friend.
We will use p to represent the pencils, m to represent the markers, and n to represent the notebooks.
• Stan
[tex]n+3p+2m=7.50[/tex]• Jan
[tex]2n+6p+5m=15.50[/tex]• Fran
[tex]n+2p+2m=6.25[/tex]To solve this system, we can do it by substitution: isolating the variable in one equation, replacing the expression in a second one, and doing the same until we have one equation with one variable.
2. Isolating n from Stan's equation:
[tex]n=7.50-3p-2m[/tex]3. Replacing this equation in Fran's equation and simplifying
[tex](7.50-3p-2m)+2p+2m=6.25[/tex][tex]7.50-3p-2m+2p+2m=6.25[/tex][tex]7.50-3p+2p=6.25[/tex]As in this step, we have one equation with one letter, then we can get the value of p.
4. Getting the value of p
[tex]3p-2p=7.50-6.25[/tex][tex]p=1.25[/tex]Now that we have the value of p we can use the isolated equation of n and the value of p in Jan's equation to solve for m.
5. Replacing the isolated expression of n and the value of p in Jan's equation and simplifying.
[tex]2\cdot(7.50-3\cdot(1.25)-2m)+6\cdot(1.25)+5m=15.50[/tex][tex]15-7.50-4m+7.50+5m=15.50[/tex][tex]15+m=15.50[/tex][tex]m=15.50-15[/tex][tex]m=0.50[/tex]6. Finally, we get n from the isolated expression.
[tex]n=7.50-3\cdot(1.25)-2\cdot(0.5)[/tex][tex]n=2.75[/tex]Answer:
Based on the information obtained, we can say that...
• An individual notebook costs $2.75
,• A package of pencils cost $1.25
,• An individual marker costs $0.50