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Welcome to Week 10 MA1310 College Math 2 Systems of Equations A set of two equations containing two variables ... A set of three equations containing three variables ... Are you seeing a pattern? Systems of Equations 2x + 5y = 38 x + 7y = 37 Lots of equations come in groups! Systems of Equations If they give you values for “x” and “y” and ask if they are true values, plug them in to both equations – both equations must be true for the values Systems of Equations IN-CLASS PROBLEM Are x = 4 and y = 6 true values for the system: 2x + 5y = 38 x + 7y = 37 Plug in x = 4 and y = 6 Systems of Equations IN-CLASS PROBLEM Plug in x = 4 and y = 6 2(4) + 5(6) =? 38 (4) + 7(6) =? 37 Are the values x = 4 and y = 6 a true solution to this system? Systems of Equations To solve a system, you must have at least as many equations as you have variables Systems of Equations IN-CLASS PROBLEM Suppose we wanted to find “x” and “y” for the equations: 2x + 5y = 38 x + 7y = 37 Systems of Equations IN-CLASS PROBLEM Strategy: try to get one of the variables written in terms of the other variable: 2x + 5y = 38 x + 7y = 37 Look for a “loner” Systems of Equations IN-CLASS PROBLEM x + 7y = 37 Use algebra to solve for “x”: x = 37 – 7y Now, plug in this “x” into the other equation Systems of Equations IN-CLASS PROBLEM 2x + 5y = 38 x = 37 – 7y To get: 2(37-7y) + 5y = 38 Now, solve for “y”! Systems of Equations IN-CLASS PROBLEM 2(37-7y) + 5y = 38 74 – 14y + 5y = 38 -9y = 38 – 74 = -36 y = -36/-9 y = 4 Then we plug in this value for “y” and solve for “x” Systems of Equations IN-CLASS PROBLEM We already have: x = 37 – 7y y = 4 So just plug in y = 4: x = 37 – 7(4) x = 37 – 28 x = 9 Systems of Equations IN-CLASS PROBLEM So, the true solution to the system: 2x + 5y = 38 x + 7y = 37 is x = 9, y = 4 or (x,y) = (9,4) Systems of Equations IN-CLASS PROBLEM To find the true solution to the system: 2x + 5y = 38 3x + 7y = 50 Solve one of the equations for either “x” or “y” Systems of Equations IN-CLASS PROBLEM 2x + 5y = 38 2x = 38 – 5y x = 38 – 5y 2 Then plug in this “x” into the other equation Systems of Equations IN-CLASS PROBLEM 3x + 7y = 50 x = 38 – 5y 2 Systems of Equations IN-CLASS PROBLEM 3(38/2 - 5y/2) + 7y = 50 57 – 7.5y + 7y = 50 Solve for “y”: -7.5y + 7y = 50 – 57 -0.5y = -7 y = -7/-0.5 y = 14 Systems of Equations IN-CLASS PROBLEM You already have: x = 38 – 5y 2 and y = 14 so: x = 38 – 5(14) = 38-70 2 2 x = -32/2 = -16 Systems of Equations IN-CLASS PROBLEM So the true solution to the system: 2x + 5y = 38 3x + 7y = 50 is x = -16, y = 14 or (x,y) = (-16,14) Systems of Equations For more equations – Same stuff, just more of it! Systems of Equations IN-CLASS PROBLEM Solve the following system of equations: x + z = 3 x + 2y – z = 1 2x – y + z = 3 Systems of Equations IN-CLASS PROBLEM x + z = 3 is the easiest – let’s start there: z = 3 – x Plug that in! Systems of Equations IN-CLASS PROBLEM z = 3 – x x + 2y – z = 1 2x – y + z = 3 become: x + 2y – (3 – x) = 1 2x – y + (3 – x) = 3 Now it’s just two equations! Systems of Equations IN-CLASS PROBLEM x + 2y – 3 + x = 1 2x – y + 3 – x = 3 or 2x + 2y = 4 x – y = 0 Use x – y = 0 x = y Systems of Equations IN-CLASS PROBLEM x = y 2x + 2y = 4 2x + 2x = 4 4x = 4 x = 1 Systems of Equations IN-CLASS PROBLEM So, x = x = z = now we have: 1 y 3 – x So, (x,y,z) = (1,1,2) Uses for Systems of Equations Lots of business applications use systems of equations Uses for Systems of Equations The income formula is the total amount generated by selling the product: I(x) = (price per unit sold) * x x is the quantity sold Uses for Systems of Equations The cost formula is the total cost of producing the product: C(x) = fixed cost + (cost per unit produced) * x x is the quantity produced Uses for Systems of Equations The profit formula is the difference between the cost and the revenue formulas: P(x) = I(x) – C(x) x is the quantity of the product Uses for Systems of Equations The x-solution for the two equations Cost and Income is the true solution of the system of the two equations (the Cost and Income equations) Uses for Systems of Equations The solution to the system is called the “break-even point” Uses for Systems of Equations Video Revenue = 25x – 0.01x2 Cost = 1600 + 5x + 10x Systems of Equations IN-CLASS PROBLEM Revenue = 25x – 0.01x2 Cost = 1600 + 5x + 10x Systems of Equations IN-CLASS PROBLEM Example: A widget-maker produces widgets Each widget sells for $300 They sell 20,000 widgets each year What is their annual income? Systems of Equations IN-CLASS PROBLEM I(x) = (price per unit sold) * x I = $300 (20,000) = $6,000,000 Systems of Equations IN-CLASS PROBLEM They have a fixed cost (rent, utilities, salaries, etc.) of $950,000 per year Each widget costs $10 to make They make 20,000 widgets each year What is their annual cost? Systems of Equations IN-CLASS PROBLEM C(x) = fixed cost + (cost per unit produced) * x C = $950,000 + $10(20,000) = $1,150,000 Systems of Equations IN-CLASS PROBLEM What is their annual profit? Systems of Equations IN-CLASS PROBLEM P(x) = I(x) – C(x) P = $6,000,000 - $1,150,000 = $4,850,000 Systems of Equations IN-CLASS PROBLEM Let’s buy Widget stock!!! Systems of Equations IN-CLASS PROBLEM Break-even point: How many widgets you need to make for the cost to equal the income (profit of $0) Systems of Equations IN-CLASS PROBLEM Break-even point: C(x) = I(x) fixed cost + (cost per unit) * x = (price per unit sold) * x Systems of Equations IN-CLASS PROBLEM Break-even point: 950,000 + 10(x) = 300(x) Algebra magic… 950,000 = 300x - 10x 950,000 = 290x 950,000/290 = x So x = 3275.86 units Systems of Equations IN-CLASS PROBLEM They will “break even” when they make and sell 3276 units Since they make and sell a lot more, they are making a huge profit! Uses for Systems of Equations If you graph the income and cost curves, the break-even point is where the two curves cross Uses for Systems of Equations Actually this is true for all systems of equations – the true solution (x,y) is the point where the graph of the two curves cross Uses for Systems of Equations IN-CLASS PROBLEM What is the break-even point? Questions? Matrices Matrix (plural matrices) - a rectangular table of elements (or entries), numbers or abstract quantities Arranged in rows and columns Matrices Matrices in everyday life Matrices The term "matrix" for arrangements of numbers was introduced in 1850 by James Joseph Sylvester Matrices Horizontal lines in a matrix are called rows Vertical lines are called columns Matrices A matrix with m rows and n columns is called an m-by-n matrix (written m × n) m and n are called its dimensions Matrices The dimensions of a matrix are always given with the number of rows first, then the number of columns: r x c Matrices Matrices are usually given capital letter names like “A” Matrices To designate a specific element in a matrix, you use small letters with subscripts r and c: a37 means the element in matrix A in row 3 column 7 Matrices In Math class, matrices are shown by listing the elements inside square braces: [ ] Matrices ┌ │ a11 a12 a13 A = │ a21 a22 a23 │ a31 a32 a33 └ ┐ │ │ │ ┘ Matrices The Matrix Game Questions? Matrices A matrix can be used as a shorthand way of writing a system of equations Matrices Called an augmented matrix: Each row is an equation Each column is the coefficient of a variable or a constant A vertical bar separates coefficients on the left side from constants on the right Matrices For example: 3x + 7y – 4z = 10 2x + 5y + z = 9 x + 6z = 15 Matrices 3x + 7y – 4z = 10 2x + 5y + z = 9 x + 6z = 15 ┌ eq 1 │ eq 2 │ eq 3 │ └ x 3 2 1 y z becomes: k ┐ 7 -4 │ 10│ 5 1 │ 9│ 0 6 │ 15│ ┘ Matrices It is traditional to use “k” to designate the constant rather than “c” (Don’t ask me why!) Matrices IN-CLASS PROBLEM What would be the matrix for the system of equations: 46x – 51y = 709 30x + 88y = -420 Matrices IN-CLASS PROBLEM 46x – 51y = 709 30x + 88y = -420 x y k ┌ ┐ eq 1 │ 46 -51 │ 709│ eq 2 │ 30 88 │-420│ └ ┘ Matrices IN-CLASS PROBLEM What would be the matrix for the system of equations: x = 7y + 8 y = x – 3 Careful! Matrices IN-CLASS PROBLEM x = 7y + 8 y = x – 3 x x – 7y = 8 -x + y = -3 y k ┌ ┐ eq 1 │ 1 -7 │ 8 │ eq 2 │ -1 1 │-3 │ └ ┘ Matrices IN-CLASS PROBLEM You can also recover the original equations from a matrix What would be the equations for the matrix: x y k ┌ ┐ eq 1 │ 10 4 │ 92│ eq 2 │-43 0 │ 87│ └ ┘ Matrices x y k ┌ ┐ eq 1 │ 10 4 │ 92│ eq 2 │-43 0 │ 87│ └ ┘ The equations would be: 10x + 4y = 92 -43x = 87 Matrices The goal of a system of equations is to solve the system for values of the variables that work in all of the equations in the system Matrices Same thing with a matrix! Matrices IN-CLASS PROBLEM How would you solve: ┌ eq 1 │ eq 2 │ eq 3 │ └ x y 1 0 0 0 1 0 z k ┐ 0 │ 10│ 0 │ 9│ 1 │ 15│ ┘ Matrices IN-CLASS PROBLEM How would you solve: ┌ eq 1 │ eq 2 │ eq 3 │ └ x = 10, x 1 0 0 y y z k ┐ 0 0 │ 10│ 1 0 │ 9│ 0 1 │ 15│ ┘ = 9, z = 15 Matrices This pattern is easy to solve: upper triangle ┌ ┐ eq 1 │ 1 0 0 │ 10│ eq 2 │ 0 1 0 │ 9│ eq 3 │ 0 0 1 │ 15│ └ ┘ lower triangle Matrices IN-CLASS PROBLEM What if you had: ┌ eq 1 │ eq 2 │ eq 3 │ └ x y 6 0 0 3 5 0 z k ┐ 2 │ 19│ 4 │ 15│ 1 │ 5│ ┘ Matrices IN-CLASS PROBLEM This is called a “lower triangular matrix” x y z k ┌ ┐ eq 1 │ 6 3 2 │ 19│ eq 2 │ 0 5 4 │ 15│ eq 3 │ 0 0 1 │ 5│ └ ┘ Matrices IN-CLASS PROBLEM Not quite as easy! What value do you know? ┌ eq 1 │ eq 2 │ eq 3 │ └ x y 6 0 0 3 5 0 z k ┐ 2 │ 19│ 4 │ 15│ 1 │ 5│ ┘ Matrices IN-CLASS PROBLEM ┌ eq 1 │ eq 2 │ eq 3 │ └ x y 6 0 0 3 5 0 z k ┐ 2 │ 19│ 4 │ 15│ 1 │ 5│ z = 5 ! ┘ Matrices IN-CLASS PROBLEM We’ll use the same techniques you used in solving systems of equations to solve for y and x ┌ eq 1 │ eq 2 │ eq 3 │ └ x y 6 0 0 3 5 0 z k ┐ 2 │ 19│ 4 │ 15│ 1 │ 5│ ┘ Matrices Remember - each row in an augmented matrix represents an equation So…anything that you can do to an equation can be done to the row of a matrix! Matrices Scalar multiplication: 3x + 7y = 15 2(3x + 7y) = 2(15) You can multiply a row of an augmented matrix by a constant Matrices Scalar multiplication uses a special notation: 2R1 This means multiply all of the elements in row 1 by the constant value 2 Matrices Just like you can add two polynomial equations, you can also add two rows of an augmented matrix: 3x + 7y = 15 5x - 2y = 40 8x + 5y = 55 Matrices You can multiply by a constant and add rows of a matrix at the same time Matrices 3R1 + R2 means multiply the elements in row 1 by three, add it to the elements in row 2, and replace the elements in ROW 2 with the result Matrices IN-CLASS PROBLEM So… back to our problem: ┌ eq 1 │ eq 2 │ eq 3 │ └ x y 6 0 0 3 5 0 z k ┐ 2 │ 19│ 4 │ 15│ 1 │ 5│ ┘ Matrices IN-CLASS PROBLEM If you multiply row 3 by -4 and add it to row 2, you can solve for y easily! ┌ eq 1 │ eq 2 │ eq 3 │ └ x y 6 0 0 3 5 0 z k ┐ 2 │ 19│ 4 │ 15│ 1 │ 5│ ┘ Matrices IN-CLASS PROBLEM ┌ eq 1 │ eq 2 │ eq3*-2 new eq2 eq 3 │ └ x y z 6 0 0 0 0 3 2 5 4 0 -4 5 0 0 1 k ┐ │ 19│ │ 15│ -20 -5 │ 5│ ┘ Matrices IN-CLASS PROBLEM x y ┌ eq 1 │ 6 3 eq 2 │ 0 5 eq 3 │ 0 0 └ Now what? z k ┐ 2 │ 19│ 0 │ -5│ 1 │ 5│ ┘ Matrices IN-CLASS PROBLEM x ┌ eq 1 │ 6 eq 2 │ 0 eq 3 │ 0 └ Divide row y z k ┐ 3 2 │ 19│ 5 0 │ -5│ 0 1 │ 5│ ┘ 2 by 5! Matrices IN-CLASS PROBLEM x y z ┌ eq 1 │ 6 3 2 │ eq 2 │ 0 1 0 │ eq 3 │ 0 0 1 │ └ Now what values do k ┐ 19│ -1│ 5│ ┘ you know? Matrices IN-CLASS PROBLEM So now we know 2 values: ┌ eq 1 │ eq 2 │ eq 3 │ └ x y 6 0 0 3 1 0 z k ┐ 2 │ 19│ 0 │ -1│ y = -1 1 │ 5│ z = 5 ┘ Matrices IN-CLASS PROBLEM Use the same strategy to solve for x! ┌ eq 1 │ eq 2 │ eq 3 │ └ x y 6 0 0 3 1 0 z k ┐ 2 │ 19│ 0 │ -1│ 1 │ 5│ ┘ Matrices IN-CLASS PROBLEM -3R2 + R1: ┌ eq 1 │ x 6 +0 6 eq 2 │ 0 eq 3 │ 0 └ y z k ┐ 3 2 │ 19│ -3 +0 3 0 2 22 1 0 │ -1│ 0 1 │ 5│ ┘ Matrices IN-CLASS PROBLEM Now what? ┌ eq 1 │ eq 2 │ eq 3 │ └ x y 6 0 0 0 1 0 z k ┐ 2 │ 22│ 0 │ -1│ 1 │ 5│ ┘ Matrices IN-CLASS PROBLEM -2R3 + R1: ┌ eq 1 │ x y z 6 0 2 +0 +0 -2 6 0 0 eq 2 │ 0 1 0 eq 3 │ 0 0 1 └ k ┐ │ 22│ -10 12 │ -1│ │ 5│ ┘ Matrices IN-CLASS PROBLEM x y ┌ eq 1 │ 6 0 eq 2 │ 0 1 eq 3 │ 0 0 └ What now? z k ┐ 0 │ 12│ 0 │ -1│ 1 │ 5│ ┘ Matrices IN-CLASS PROBLEM x ┌ eq 1 │ 6 eq 2 │ 0 eq 3 │ 0 └ Divide row y z k ┐ 0 0 │ 12│ 1 0 │ -1│ 0 1 │ 5│ ┘ 1 by 6! Matrices IN-CLASS PROBLEM x y z ┌ eq 1 │ 1 0 0 │ eq 2 │ 0 1 0 │ eq 3 │ 0 0 1 │ └ Now what values do k ┐ 2│ -1│ 5│ ┘ we know? Matrices IN-CLASS PROBLEM Now what do we know? ┌ eq 1 │ eq 2 │ eq 3 │ └ x y 1 0 0 0 1 0 z k ┐ 0 │ 2│ x = 2 0 │ -1│ y = -1 1 │ 5│ z = 5 ┘ Matrices So… if you have an upper and lower triangular matrix, the solution is obvious: upper triangle ┌ ┐ eq 1 │ 1 0 0 │ 10│ eq 2 │ 0 1 0 │ 9│ eq 3 │ 0 0 1 │ 15│ └ ┘ lower triangle Matrices And… if you have only one zero triangle, it’s more work: ┌ ┐ eq 1 │ 1 1 2 │ 19│ eq 2 │ 0 1 0 │ 3│ eq 3 │ 0 0 1 │ 5│ └ ┘ lower triangular matrix Matrices The upper and lower triangular form is called the “GaussJordan” solution, the lower triangular form is called the “Gauss” solution Matrices Carl Friedrich Gauss Wilhelm Jordan Matrices Wilhelm Jordan Camille Jordan Matrices Computers and calculators manipulate the matrix to achieve the Gauss-Jordan solution Matrices Computers and calculators do these processes using a strategy to reduce the matrix to the Gauss-Jordan “Easy-ToSolve” form Excel Solver Demo Suppose you had the system of equations: 7u + 6v + 15w – 44x + 18y + 11z = 70 9u + 25v + 10w + 3x – 14y + 8z = 161 13u + 8v + 2w + 6x + 62y + 15z = 671 –2u + v – 5w + 57x + 9y – 3z = 354 5u – 17v + 18w + 9x – 7y + 5z = 82 4u + 3v + 16w + 14x + 2y + 7z = 258 Solve for u, v, w, x, y, z Questions? Liberation! Be sure to turn in your assignments from last week to me before you leave Don’t forget your homework due next week! Have a great rest of the week!