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1 Systems of Linear Equations Systems of Linear Equations’This is a Grade 11 Applied Math presentation for ‘Systems of Linear Equations’ The equation for a line has two variables: xy x and y If there are two unknowns (variables) you need two different facts (or equations) to find the unknowns MA30SA_C_SysLinEq.pptRevised:

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2 How much does the cow weigh? 6 Diagram A y x 7 Diagram B x y y ? C x x is the weight of the Cow. y is the weight of the Beaver. 1 Cow and 1 Beaver weigh 6, 1 Cow and 2 Beavers weigh 7. How much does 1 Cow weigh by itself? How much does a beaver weigh?

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3 Solution Did you say the cow weighs 5? So how much does a beaver weigh? –Did you say 1? How did you get that answer?? That is all that this unit is about: ‘Systems of Linear Equations’

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4 Definitions Systems of Linear Equations –A set of two or more linear equations with the same variables. Solution to a System of Linear Equations –The set of all ordered pairs that satisfy all equations (make them true) In Grade 11 Applied we will only look at two equations with two variables Systems of Linear Equations are also sometimes called ‘Simultaneous Linear Equations’ if you google the subject

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5 Other examples of typical problems for Systems of Linear Equations Jason is twice as old as Scott. Scott is 3 years younger than Jason. How old is Jason? –Two unknowns; ages of Scott and Jason The sum of two numbers equals 10, their difference equals 4. What are the two numbers? –Two unknowns: Number A and Number B It takes you two hours to paddle upstream a distance of two km, and one hour to come back to the start. What is the speed of the canoe and of the current? –Two unknowns: your paddling speed and the speed of the current

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6 Graphing Tools This unit uses the TI-83 Graphing tool primarily for graphing There are lots of other graphing tools also, many are on-line All of the problems can be done manually using graph paper if necessary, but you will find graphing tools much more useful. This presentation will demonstrate the manual method

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7 Graph the two equations Graph the 2 equations: Line 1: x + y = 4 Line 2: 4x – 4y = 8 Notice the tables have one point, and only one, in common. In the graph, it is where the lines meet

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8 Graph two more equations Graph the 2 equations: Line 1: y = 3x + 2 Line 2: y = (1/2)x – 3 Notice the tables have one point, and only one, in common. In the graph, it is where the lines meet

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9 Graph even more equations Graph the 2 equations: Line 1: 2x + 2y = 4 Line 2: x + y = 2 all points in common Notice the table has all points in common for both lines. They are really the same lines.

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10 Graph a final two equations Graph the 2 equations: Line 1: y = 3x + 2 Line 2: y = 3x – 2 no points in common Notice the table has no points in common The lines never meet, they are parallel!

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11 Types of Linear Systems Consistent –Have solution(s) –Independent and consistent has one solution Lines have different slopes. Lines meet at one point –Dependent and consistent has all the points on the line as a solution. The lines have same slopes and same intercepts They are the same line! Inconsistent noThe lines are parallel, lines never meet, they have no points in common.

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12 Consistent (Dependent or Independent) or Inconsistent?

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13 Solving Systems of Equations Using Algebra There are two ways to solve Systems of Linear Equations using Algebra –Using Algebra is best because you don’t need to graph lines The two methods: –Elimination by Addition or Subtraction –Elimination by Substitution

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14 Elimination Elimination by Addition or Subtraction Given two equations 2x + y = 8 2x – y = 0 Label the equations (1)2x + y = 8 (2)2x – y = 0 Eliminate Eliminate one variable by adding or subtracting the equations. (1)+(2) (1) 2x + y = 8 (2) +(2x – y = 0) 4x + 0y = 8 x = 2 x = 2 x =2 We know that x =2 now So just substitute it back into either equation (1) or (2) to find y: 2(2)+y=8 y = 4 (2, 4) The solution to the System is when x = 2 and y = 4. It is the point (2, 4)

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15 Elimination Elimination by Addition or Subtraction Sometimes you need to change one of the equations, or both, so that when you add or subtract them one of the variables will be eliminated

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16 Elimination Elimination by Addition or Subtraction Given two equations 4x + 2y = 16 2x – y = 4 Label the equations (1)4x + 2y = 16 (2)2x – y = 4 Eliminate after multiplying one equation by a suitable constant. Eliminate one variable by adding or subtracting the equations after multiplying one equation by a suitable constant. (1)+2*(2) (1) 4x + 2y = 16 (2) +(4x – 2y = 8) 8x + 0y = 24 x = 3 x = 3 x =3 We know that x =3 now So just substitute it back into either equation (1) or (2) to find y: 4(3)+2y=16 y = 2 (3, 2) The solution to the System is when x = 3 and y = 2. It is the point (3, 2)

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17 More examples Elimination Solve: (1)5x + 2y = 3 (2)2x + 3y = –1 3*(1) 15x + 6y = 9 -2*(2) – (4x + 6y = –2) Subtract: 11x – 0y = 11 x = 1 11x = 11 x = 1 Substitute value x = 1 into either equation to find y 1 5(1) + 2y = 3 y = –1 Solution: (1, –1) Check the answer!! Solve: (1) 2x + 3y = 3 (2)–6x + 6y = 6 3*(1) 6x + 9y = 9 Add (2) +(–6x + 6y = 6) Add: 0x + 15y = 15 y = 1 y = 1 Substitute value y = 1 into either equation to find x 1 2x + 3(1) = 3 2x = 0; x = 0 Solution: (0, 1) Check the answer!!

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18 Review of Steps to Solve by Elimination Put equations into a ‘standard’ order (ax + by = c) (ax + by = c) Label the equations Multiply either, or both, equations by some number so that one variable will be eliminated when the equations are added or subtracted Add or subtract the equations to eliminate one variable and find the remaining variable Find the eliminated variable by substituting the the value of the solved variable back into either equation of the system bothCheck your answer (x, y) in both equations

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19 Solving by Substitution The steps are: –Label the equations (1), (2) –Isolate one variable and express it in terms of the other other equation –Substitute that expression into the other equation –Solve for the now single variable –Substitute the value of the found variable back into either equation to find the other unknown –Check the answer in both equations Given: (1): x – y = 4 (2): x + 2y = – 3 From (2): = – 3 – 2y Substitute into (1) 2(– 3 – 2y) – y = 4 – 6 – 4y – y = 4 – 6 – 5y = 4 – 5y = 10 So: y = – 2 Substitute into either eqn 2x – (– 2) = 4 2x + 2 = 4 ; 2x = 2 So x = 1 Now C CC Check the Answer!

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20 Sample Problem - Digits Two numbers add to 12 The same two numbers have a difference of 4 Find the two numbers Give the unknowns labels: like a and b Write the two facts or equations: (1)a + b = 12 (2)a – b = 4 Add eqns (1) and (2) 2a + 0 =16 2a = 16 a = 8. So if a = 8 then b = 4 from either equation 84 So the numbers are 8 and 4 Check Check to see if these numbers work in both equations

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21 Sample Problems - Motion An airplane covers a distance of 1500 miles in 3 hours when it flies with the wind, and in 3 and 1/3 hours when it flies against the wind. What is the speed of the plane in still air. What is the speed of the wind? Unknowns; a a: speed of airplane; b b: speed of wind a + b=500 mph a + b=(1500/3) = 500 mph with the wind a b=450 mph a – b = 1500/3.333= 450 mph against the wind Solve from there! ??

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22 Sample Problems - Age A father is 24 years older than his son. In 8 years he will be twice as old as his son. Determine their current ages Let: f = father’s current age. s = son’s current age. (1)f – s = 24 (2)(f + 8) = 2*(s + 8) f + 8 = 2s + 16 f – 2s = 8 Eqn (1) – Eqn (2) f – s = 24 – (f – 2s = 8) s = 16 f = 40 Check Check: (1): (40) – (16) = 24 (2): (40 + 8) = 2*(16 + 8)

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23 Coin Problems Edgar has 20 coins in dimes and nickels, which together total $1.40. How many of each does he have? Let n = nbr of nickels, let d = nbr of dimes (1): n(0.05)+d(0.10) = 1.40 (2): n + d = 20 Use substitution (2): n = 20 – d Substitute eqn (2) into eqn (1) (20 - d)*(0.05) + d*(0.10)= 1.40 d=8, n= d = 1.4 d=8, n=12 Check Check the answer in both equations It works!!

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24 Word Problems 1000 tickets were sold. Adult tickets cost $8.50, children's cost $4.50, and a total of $7300 was collected. How many tickets of each kind were sold Let a = nbr of adult tickets sold; let c = nbr of children tickets sold (1):a + c = 1000 (2):8.5a + 4.5c = 7300 Substitute Substitute: a = 1000 – c 8.5(1000 – c) + 4.5c = – 4c = 7300 c = 300 and a = adult tickets were sold, 300 children’s tickets were sold check Now check answer in both equations!

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25 The End Now you know how to solve Systems of Linear Equations with two unknowns If you take Pre-Calculus, you will learn how to solve for 3 unknowns! –Just a few extra steps for three unknowns

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