Systems of Linear Equations in Two Variables. Break Even. Example. 240x x This is when total cost equals total revenue.

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1 Systems of Linear Equations in Two Variables 1 Break Even This is when total cost equals total revenue C(x) = R(x) A company breaks even when the profit is zero P(x) = R(x) C(x) = 0 2 R x 565x C x x R x C x 565x x 240x x

2 Market Equilibrium q Number of units produced p Price per unit Supply - How much of something is available Demand - How much of something people want Surplus More units are produced than are demanded Shortfall Fewer units are produced than are demanded 4 Law of Demand If all other factors remain equal, the higher the price of a good, the less people will demand that good. In other words, the higher the price, the lower the quantity demanded. p Demand Function q 5 Law of Supply Producers supply more of a higher quantity when the price increases This means that the higher the price, the higher the quantity supplied p Supply Function q 6 2

3 Market Equilibrium Supply and demand are equal Suppliers are selling all the goods that they have make Consumers are getting all the goods that they want Demand Supply p Function Function Equilibrium Point q 7 Remarks The Revenue and Demand functions are also related q is the number of units sold p is the price per unit p(q) is the demand function, the price per unit as a function of the number of units Revenue is the number of units sold times the price per unit R(q) = q p(q) 8 Demand 3q Demand 3q p 340 p 340 3q If the quantity increases by 1, then the price the consumer is willing to pay decreases by $3 If the quantity decreases by 1, then the price the consumer is willing to pay increases by $3 9 3

4 Demand 3q Supply p 4q 220 p 4q We find q( p) p 4q 220 q p 55 4 If the price increases by $1, then the producer will manufacture ¼ more units 10 System of Linear Equations A system of equations is a grouping of two or more linear equations where each equation contains one or more variables 3q p 340 p 4q 220 A brace is used to remind us that we are dealing with a system of equations 11 Solve By Graphing Solve both equations for y Enter both equations in the Y= Editor Find the point of intersection Solution y x 12 4

5 Demand 3q Demand 3q p 340 p 340 3q Supply p 4q 220 p 4q Demand 3q Set window This step often requires trial and error 14 Demand 3q Find point of intersection q = 80, p =

6 Demand 3q We can also use the table to confirm our result The table can help us find an appropriate viewing window 16 2x y 2 x 3y 2 If using a graphing tool, we sometimes must find two integer points that satisfy and equation. Note Do this with one equation at a time. Graph each equation. To solve find the point of intersection. 2x y 2 x 3y 2 If we let x=0 we have y=2 If we take y=0 we have x=-1 If we let x=1 we have y=1 If we let y=0 we have x=-2 17 Solve Using Elimination 1: Simplify if needed 2: Multiply one or both equations by a number that will create opposite coefficients for like terms if needed. 3: Add equations 4: Solve for remaining variable 5: Solve for second variable 6: Check the proposed ordered pair solution(s) in both original equations 18 6

7 Number the equations 3q p 340 p 4q 220 Demand 3q 1 2 Prepare for elimination (multiply equation 2 by negative one and rearrange) 3q p 340 4q p Demand 3q Eliminate 3q (Add) 4q p220 7 q 560 Solve for q q 80 Solve for p using equation 1 3q Solve Using Substitution 1: Simplify if needed. 2: Solve one equation for either variable. 3: Substitute what you get for step 2 into the other equation. 4: Solve for the remaining variable. 5: Solve for second variable. 6: Check the proposed ordered pair solution(s) in both original equations. 21 7

8 Demand 3q Number the equations Solve the 2 nd equation for p, call equation 3 3q p 340 p 4q p 4q220 (3) Substitute into the 1 st equation 3q 4q Demand 3q Solve for q q q q 4q q q 560 q 80 Solve for p using equation 3 p Solve Using Matrices Write each equation in standard form Ax + By = C We enter an augmented matrix which represents the system of equations into our calculator We use the calculator to find the row reduced equivalent matrix We write the solution 24 8

9 Demand 3q We enter the augmented matrix Demand 3q p 340 p 3q 340 Supply p 4q 220 p 4q Demand 3q We enter the augmented matrix Demand 3q p 340 p 3q 340 Supply p 4q 220 p 4q Demand 3q The row reduced echelon form (rref) p = 80, q =

10 Solve Using TI-89 We use the solve feature of the TI-89 from the home screen Solve( Equation 1 AND Equation 2, x) Under F2 2 nd Math 8 8 or y 28 Demand 3q 29 Possible Solutions The solution set is exactly one of the following: 1) One Solution 2) No Solution 3) Infinite Number of Solutions x y 10 x y 0 2x3y 8 x 3y 4 x y 3 2x2y

11 Consistent Independent Solution There is one solution x y 10 x y 0 One point of intersection 31 Consistent Dependent Solution There are an infinite number of solutions x y 3 2x2y 6 Same line There is no solution Inconsistent Solution 2x3y 8 2x 3y 4 Parallel Lines

12 Infinite number of solutions x y 3 2x2y 6 The solution is {( x, y) x y 3} 34 No solution 2x3y 8 2x 3y 4 The last row say 0 1 which is not true Hence there is no solution 35 12