Question

Make up a word problem to solve for system of equations. Be creative

Answer 1

Step-by-step 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

`x\text{ +y + z = 24}`
`4x\text{ + 6y + 10z = 148}`
`(1)/(2)x\text{ + }(1)/(4)y+(1)/(3)z\text{ = 9}`

Multiplying the second equation by 1/2 and the third equation by 12, we get:

`x\text{ +y + z = 24}`
`2x\text{ + 3y + 5z = 74}`
`6x\text{ + 3}y+4z\text{ = 108}`

Now, multiply the first equation by -2 and add it to the second equation. In this way we obtain:

`x\text{ +y + z = 24}`
`y+3z\text{ =26}`
`6x\text{ + 3}y+4z\text{ = 108}`

Multiply the first equation by -6 and add it to the last equation. In this way we obtain:

`x\text{ +y + z = 24}`
`y+3z\text{ =26}`
`\text{ -3y -2z = -36}`

Finally, the process is completed by adding the second multiplied by 3 to the third equation.

`x\text{ +y + z = 24}`
`y+3z\text{ =26}`
`7z\text{ = 42}`

then, if we perform back substitution we get the desired solutions:

`x=10`
`z=6`

and

`y=8`

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:

`x\text{ +y + z = 24}`
`4x\text{ + 6y + 10z = 148}`
`(1)/(2)x\text{ + }(1)/(4)y+(1)/(3)z\text{ = 9}`

Solution for this system of equations:

`x=10`
`y=8`

`z=6`