Transcript 6.4 Application of Linear Systems

6.4 Application of Linear
Systems
I can choose the best method for solving a system of
linear equations
When to Use
 Graphing: When you want a visual display of the equations,
or when you want to estimate a solution.
 Substitution: When one equation is already solved for one of
the variables, or when it is easy to solve for one of the
variables.
 Elimination: When the coefficients of one variable are the
same or opposites, or when it is not convenient to use
graphing or substitution.
Modeling Problems
 A break-even point is where a business’ expenses equal its
income
 Ex:
Practice
 A fashion designer makes and sells hats. The material for
each hat costs $5.50 and the hats sell for $12.50 each. If the
designer spends $1400 on advertising, how many hats must
be sold for the designer to break even?
 Let x = # of hats and y = amount of money
 Expenses: y = 5.50x + 1400
 Income:
y = 12.50x
 Use substitution: 12.50x = 5.50x +1400
 7x = 1400
 x = 200
 After selling 200 hats
Constraints and Viable Solutions
 The local zoo is filling two water tanks for the elephant exhibit. One has
50 gallons in it and is being filled at a rate of 10 gal/h while the other
has 29 gallons and is being filled at a rate of 3 gal/h when will they
have the same amount?
 Let x = # of hours and y = # of gallons
 1st tank: y = 10x + 50
 2nd tank: y = 3x + 29
 Substitute: 10x + 50 = 3x + 29
 7x = -21
 x = -3
 Since x is the number of hours and the solution is negative, we know
the tanks will never have the same amount.
Wind or Current
 Tailwind: air speed + wind speed = ground speed
 Headwind: air speed – wind speed = ground speed
 Ex:
 LA to Charlotte: a + w = 550
 Charlotte to LA: a – w = 495
 Use Elimination: 2a = 1045
 a = 522.5 mi/h
 Plug a in to find w = 27.5 mi/h
Assignment
 P.390 odds #7-25