Explore IDNLearn.com to discover insightful answers from experts and enthusiasts alike. Whether it's a simple query or a complex problem, our community has the answers you need.
Sagot :
Answer:
12 regular
15 large
10 extra large
Explanation:
Variables:
s = number of coffees with regular size
l = number of coffees with large size
e = number of coffees with extra large size
Equations.
Equation for the quantity of coffees sold:
we add all of the quantities s, l and e:
[tex]s+l+e=37[/tex]this is because the diner sold a total of 37 coffees. This will be equation 1.
Equation for the volume of coffee:
We need to multiply the volume that each side has with the number of coffees with that size:
[tex]300s+500l+800e=19100[/tex]This is because the total volume of coffee sold was 19100mL. This will be equation 2.
Equation for the coffee sales:
We multiply the price of each coffee by the amount of coffes with that price:
[tex]2.25s+3.25l+5.75e=133.25[/tex]This is because the small cost $2.25, the large $3.25 and the extra large $5.75, and the total of coffee sales was $133.25. This will be equation 3.
Solve.
There are many ways to solve a 3 variable system of equations. I will multiply the first equation by -300 to add it to the second equation and eliminate s.
Multiply 1 by -300:
[tex]\begin{gathered} -300(s+l+e=37) \\ -300s-300l-300e=-11,100 \end{gathered}[/tex]Now we add this to equation 2
[tex]\begin{gathered} -300s-300l-300e=-11,100 \\ 300s+500l+800e=19,100 \\ ---------------- \\ 200l+500e=8,000 \end{gathered}[/tex]I will call this equation 4.
Next, we will use equation 2 and equation 3 to eliminate the same variable s, for this we multiply equation 2 by -2.25
[tex]\begin{gathered} -2.25(300s+500l+800e=19,100) \\ -675s-1,125l-1,800=-42,975 \end{gathered}[/tex]and equation 3 by 300:
[tex]\begin{gathered} 300(2.25s+3.25l+5.75e=133.25) \\ 675s+975l+1,725e=39,975 \end{gathered}[/tex]And we add this two equations to elimate s:
[tex]\begin{gathered} -675s-1,125l-1,800=-42,975 \\ 675s+975l+1,725e=39,975 \\ ----------------- \\ -150l-75e=-3,000 \end{gathered}[/tex]This will be equation 5.
Now we use equations 4 and 5 to find l and e. For that, we multiply equation 5 by 4/3:
[tex]\begin{gathered} \frac{4}{3}(-150l-75e=-3,000) \\ -200l-100e=-4,000 \end{gathered}[/tex]And add this to equation 4:
[tex]\begin{gathered} 200l+500e=8,000 \\ -200l-100e=-4,000 \\ ----------- \\ 400e=4,000 \\ e=4,000/400 \\ e=10 \end{gathered}[/tex]We have the value of e. Now we substitute this into equation 5, to solve for l:
[tex]\begin{gathered} -150l-75e=-3,000 \\ -150l-75(10)=-3,000 \\ -150l-750=-3,000 \\ -150l=-3,000+750 \\ -150l=-2,250 \\ l=-2,250/(-150) \\ l=15 \end{gathered}[/tex]We have tha value of l, now we need to substitute e and l into equation 1 to find s:
[tex]\begin{gathered} s+l+e=37 \\ s+15+10=37 \\ s+25=37 \\ s=37-25 \\ s=12 \end{gathered}[/tex]Answer:
12 regular
15 large
10 extra large
Thank you for participating in our discussion. We value every contribution. Keep sharing knowledge and helping others find the answers they need. Let's create a dynamic and informative learning environment together. Find clear answers at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.
