Respuesta :
Explanation:
Consider the following problem.
A student library has 24 tables, X tables with 4 seats each, Y tables with 6 seats each, and Z tables with 10 seats each. The total seating capacity of the cafeteria is 148. For a special student academic meeting, half of the X tables, 1/4 of the Y tables, and 1/3 of the Z tables will be used, for a total of 9 tables. Determine X, Y, and Z.
The conditions of this problem give rise to the following system of equations
[tex]x\text{ +y + z = 24}[/tex][tex]4x\text{ + 6y + 10z = 148}[/tex][tex]\frac{1}{2}x\text{ + }\frac{1}{4}y+\frac{1}{3}z\text{ = 9}[/tex]Multiplying the second equation by 1/2 and the third equation by 12, we get:
[tex]x\text{ +y + z = 24}[/tex][tex]2x\text{ + 3y + 5z = 74}[/tex][tex]6x\text{ + 3}y+4z\text{ = 108}[/tex]Now, multiply the first equation by -2 and add it to the second equation. In this way we obtain:
[tex]x\text{ +y + z = 24}[/tex][tex]y+3z\text{ =26}[/tex][tex]6x\text{ + 3}y+4z\text{ = 108}[/tex]Multiply the first equation by -6 and add it to the last equation. In this way we obtain:
[tex]x\text{ +y + z = 24}[/tex][tex]y+3z\text{ =26}[/tex][tex]\text{ -3y -2z = -36}[/tex]Finally, the process is completed by adding the second multiplied by 3 to the third equation.
[tex]x\text{ +y + z = 24}[/tex][tex]y+3z\text{ =26}[/tex][tex]7z\text{ = 42}[/tex]then, if we perform back substitution we get the desired solutions:
[tex]x=10[/tex][tex]z=6[/tex]and
[tex]y=8[/tex]We can conclude that the correct answer is:
Answer:Problem:
A student cafeteria has 24 tables, X tables with 4 seats each, Y tables with 6 seats each, and Z tables with 10 seats each. The total seating capacity of the cafeteria is 148. For a special student meeting, half of the X tables, 1/4 of the Y tables, and 1/3 of the Z tables will be used, for a total of 9 tables. Determine X, Y, and Z.
System of equations:
[tex]x\text{ +y + z = 24}[/tex][tex]4x\text{ + 6y + 10z = 148}[/tex][tex]\frac{1}{2}x\text{ + }\frac{1}{4}y+\frac{1}{3}z\text{ = 9}[/tex]Solution for this system of equations:
[tex]x=10[/tex][tex]y=8[/tex][tex]z=6[/tex]