W&C - Chapter 2

1. W&C - Chapter 2 Systems of Linear Equations(essentially a review of what we did with breakeven analysis, but better and more detailed in every way)

2. Deja vu R(x) = 1.95xC(x) = .03x + 9000 We went through the graphical and table solutions.

3. solving algebraicly • The major change will be to write the functions this cool way: ax + by = c • The number a is called the coefficient of x. • The number b is called the coefficient of y. • x and y are the variables that represent the unknowns. • Example: • Revenue on drinks y = 1.95 x + 0 becomes -1.95x + 1y = 0 • Cost of drinks y = .03 x + 9000becomes-.03x + 1y = 9000 where – 1.95 and -.03 are considered coefficients ofx and 1 and 1 are considered coefficients ofy.

4. A System of Two Linear Equations and Two Unknowns • Because we worked with two functions at a time (Revenue and Cost functions), we get to claim we were working with a “system of two linear equations” ax + by = c px + qy = r where x and y are variables that represent the two unknowns. • The values for x and y that satisfy both equations will be called the “solution for the system.”  • When calculating a breakeven point (or when working with any system of linear equations), we get to call the breakeven point the “solution for the system.”

5. Solution to the system of two linear equations • Revenue on drinks -1.95x + 1y = 0 • Cost of drinks -.03x + 1y = 9000 • Multiply by -1 and add the equations to eliminate y -1.95x + 1y = 0 .03x - 1y = -9000  multiplied equation by -1 -1.92x = -9000 x = 4687.5, the breakeven point that’s half the “solution to the system” • The other half is to calculate y y = 1.95x y = 1.95(4687.5) y = 9140.625 • Breakeven is therefore 4687.5 drinks, when revenue and cost are both $9,140.625. . . Right?

6. Psychology problem • Suppose you have $3 to spend on snacks and a drink. • If x represents the amount you’ll spend on snacks and y the amount you’ll spend on a drink, you can say that x + y = 3 • If you have obsessive-compulsive disorder and must spend exactly $1 more on snacks than on your drink, you can also say that: x - y = 1 • These two equations represent a system of equations. • The values for x and y that satisfy both equations are called the solution for the system. • The solution can be obtained graphically or algebraically.

7. Graphical Solution • One way to solve the system of equations is to graph the lines representing each equation and find the coordinates of the point at which the lines cross: • Thus, the optimal solution is: x = 2 ($2 spent on snacks) y = 1 ($1 spent on a drink) y x - y = 1 (2, 1) x x + y = 3

8. Algebraic Solution: Adding Equations • In the algebraic approach, we try to combine the equations in such a way as to eliminate one variable. • One way of doing this is by adding the equations or multiplesof the equations to find the value of one variable. • Then substituting in one of the equations to find the value of the other variable. x + y = 3 x - y = 1 (2) + y = 3 y = 3 - 2 y = 1 2x + 0 = 4 x = 2

9. Solving a System: Algebraically versus Graphically • The algebraic method always allows us to determine an exact solution, whereas the graphic method often does not.

10. Graphical solution problems • Solve the system: 3x + 5y = 0 2x + 7y = 1 Graphic Method: y (-.5, -.3)? x 2x + 7y = 1 3x + 5y = 0

11. Try it in pairs #1! • Solve the system: x – 3y = 5 – 2x + 6y = 8 Algebraic Method:

12. Try it in pairs #1! x 2 • Solve the system: x – 3y = 5 – 2x + 6y = 8 • Multiply equation one by 2 and add: 2x –6y = 10 – 2x + 6y = 8 0 + 0 = 18 0 = 18 Algebraic Method:

13. Try it in pairs #1! x 2 • Solve the system: x – 3y = 5 – 2x + 6y = 8 • Multiply equation one by 2 and add: 2x –6y = 10 – 2x + 6y = 8 0 + 0 = 18 0 = 18 • The system has no solution and is called inconsistent. (The two lines are parallel, so they never meet) Algebraic Method:

14. Try it in pairs #2! • Solve the system: x + y = 2 2x + 2y = 4 Algebraic Method:

15. Try it in pairs #2! x (– 2) • Solve the system: x + y = 2 2x + 2y = 4 • Multiply equation one by (– 2) and add: – 2x –2y = – 4 2x + 2y = 4 0 + 0 = 0 0 = 0 Algebraic Method:

16. Try it in pairs #2! x (– 2) • Solve the system: x + y = 2 2x + 2y = 4 • Multiply equation one by (– 2) and add: – 2x –2y = – 4 2x + 2y = 4 0 + 0 = 0 0 = 0 • This is called a redundant system: the two equations are in fact two versions of the same equation. • When we have a single equation with two unknowns, there are infinitely many solutions (all the points along the line). Algebraic Method:

17. Example: Equilibrium Price • The demand for refrigerators in West Podunk is given by where q is the number of refrigerators that the citizens will buy each year if they are priced at p dollars each.

18. Example: Equilibrium Price • The demand for refrigerators in West Podunk is given by where q is the number of refrigerators that the citizens will buy each year if they are priced at p dollars each. • The supply is where now q is the number of refrigerators the manufacturers will be willing to ship into town each year if they are priced at p dollars each. • Find the equilibrium price and quantity.

19. Example: Equilibrium Price • Solve the system: Graphic Method: Solution: • The equilibrium price is $500. • The equilibrium quantity is 50 refrigerators per year. q (500, 50) 125 100 75 50 25 p Now solve the system using the algebraic method 0 100 300 500 700

20. Example: Equilibrium Price Algebraic Method: • Write both equations in standard form:

21. Example: Equilibrium Price Algebraic Method: • Write both equations in standard form: • Multiply the first equation by 10 and the second by 20 to clear fractions, then add and solve: 1p + 10q = 1000 – 1p + 20q = 500 0p + 30q = 1500 q = 50 x 10 x 20 • Then substituteq = 50 in equation one to find p: p + 10(50) = 1000 p + 500 = 1000 p = 1000 – 500 p = 500

22. 1p + 10q = 1000 – 1p + 20q = 500 0p + 30q = 1500 q = 50 Something interesting here. We almost don’t even need p and q or the + or the = to do the math. 1 10 1000 – 1 20 500 0 30 1500 I’m lazy. Tell me more!