Simultaneous equations - word problem

[sammmiewong]

The student council is providing lunch and music for the grade 10 class. They have two quotes from Lunch Express and Let’s Do Lunch. The costs for each were given as follow: Lunch Express: If 100 students attend, it will cost $1 000. If 200 students attend, it will cost $1 500. Let’s Do Lunch: If 50 students attend, it will cost $700. If 150 students attend, it will cost $1 350. Solve the system using the three different methods ie. graphing, substitution and elimination.

So first of all I'm trying to to solve by graphing method but got stuck on setting up the equations already! I thought the two equations are 100x + 200y = 2500 and 50x +150y = 2050 but the graph given is from 0 to 300 for x-axis and from 0 to 3000 for the y-axis, so the scale just doesn't seem right to me. May I know if I'm missing something here?

 

[Deleted member 4993]

The student council is providing lunch and music for the grade 10 class. They have two quotes from Lunch Express and Let’s Do Lunch. The costs for each were given as follow: Lunch Express: If 100 students attend, it will cost $1 000. If 200 students attend, it will cost $1 500. Let’s Do Lunch: If 50 students attend, it will cost $700. If 150 students attend, it will cost $1 350. Solve the system using the three different methods ie. graphing, substitution and elimination.

So first of all I'm trying to to solve by graphing method but got stuck on setting up the equations already! I thought the two equations are 100x + 200y = 2500 and 50x +150y = 2050 but the graph given is from 0 to 300 for x-axis and from 0 to 3000 for the y-axis, so the scale just doesn't seem right to me. May I know if I'm missing something here?

Your equations

100x + 200y = 2500

50x +150y = 2050

What does 'x' and 'y' represent? Be totally explicit!!

 

[Romsek]

The problem isn't stated very clearly.

Are we supposed to assume these costs are linear in the number of people attending? What is the cost at Lunch Express if less than 100 students attend? We aren't told. Same with Do Lunch and 50 students. Are we just supposed to extrapolate?

If so then we can figure up linear functions of attendees and solve for the intersection as usual. Let's take a look.

[MATH]LE(100) = 1000, ~LE(200) = 1500[/MATH]
[MATH]LE(n) =1000+ (n-100)\dfrac{1500-1000}{200-100} = 5(n-100)+1000 = 5n + 500[/MATH]
similarly

[MATH]DL(n) = 700 + (n-50)\dfrac{1350-700}{150-50} = 6.5(n-50)+700 = 6.5n +375[/MATH]
Can you solve for the intersection now?[MATH][/MATH]

 

[Steven G]

I too think that there is no reason to assume the price is linear, although I have no idea what else to assume.