# Section 6.4 Applications of Linear System

## Presentation on theme: "Section 6.4 Applications of Linear System"— Presentation transcript:

Section 6.4 Applications of Linear System

Break even point

Answer? Income: y = 12.50x or y = 12.5x Expense: y = 5.5x + 1400
A fashion designer makes and sells hats. The material for each hat costs \$ The hats sell for \$12.50 each. The designer spends \$1400 on advertising. How many hats must the designer sell to break even? Define Variables: x = number of hats sold y = the # of dollars of expense or income Write a system of equations. Income: y = 12.50x or y = 12.5x Expense: y = 5.5x Choose a method: graph, substitution, elimination. y = 5.5x ( ) = 5.5x 12.5x 12.5x = 5.5x 7x = 1400 Answer? x = 200 200 hats

Never: it is impossible to have time be -3 hours. -7h = 21 g = 20
The local zoo is filling two water tanks for the elephant exhibit. One water tank contains 50 gal of water and is filled at a constant rate of 10 gal/h. The second water tank contains 29 gal of water and is filled at a constant rate of 3 gal/h. When will the two tanks have the same amount of water? Explain. The local zoo is filling two water tanks for the elephant exhibit. One water tank contains 50 gal of water and is filled at a constant rate of 10 gal/h. The second water tank contains 29 gal of water and is filled at a constant rate of 3 gal/h. When will the two tanks have the same amount of water? Explain. Write a system of equations. Let h = the number of hours the tanks are filling. Let g = the number of gallons in the tank. Tank 1: g = 10h + 50 Tank 2: g = 3h + 29 Solve the system g = 10h + 50 solve for g: Hours = -3 ( ) = 10h + 50 3h + 29 g = 10h + 50 Gallons = 20 3h = 10h + 50 g = 10( ) + 50 -3 Answer? -7h+ 29 = 50 g = Never: it is impossible to have time be -3 hours. -7h = 21 g = 20 h = -3

Assignment: Pg 390: 7-10, 13-16, 19-21