Presentation on theme: "Systems of Linear Equations"— Presentation transcript:
1 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:x and yIf there are two unknowns (variables) you need two different facts (or equations) to find the unknownsMA30SA_C_SysLinEq.pptRevised:
2 How much does the cow weigh? 6Diagram Ayx7Diagram Bxy?Cxx 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?
3 ‘Systems of Linear Equations’ SolutionDid 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’
4 Definitions Systems of Linear Equations A set of two or more linear equations with the same variables.Solution to a System of Linear EquationsThe 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 variablesSystems of Linear Equations are also sometimes called ‘Simultaneous Linear Equations’ if you google the subject
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 JasonThe sum of two numbers equals 10, their difference equals 4. What are the two numbers?Two unknowns: Number A and Number BIt 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
6 Graphing ToolsThis unit uses the TI-83 Graphing tool primarily for graphingThere are lots of other graphing tools also, many are on-lineAll 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
7 Graph the two equations Graph the 2 equations:Line 1: x + y = 4Line 2: 4x – 4y = 8Notice the tables have one point, and only one, in common.In the graph, it is where the lines meet
8 Graph two more equations Graph the 2 equations:Line 1: y = 3x + 2Line 2: y = (1/2)x – 3Notice the tables have one point, and only one, in common.In the graph, it is where the lines meet
9 Graph even more equations Graph the 2 equations:Line 1: 2x + 2y = 4Line 2: x + y = 2Notice the table has all points in common for both lines.They are really the same lines.
10 Graph a final two equations Graph the 2 equations:Line 1: y = 3x + 2Line 2: y = 3x – 2Notice the table has no points in commonThe lines never meet, they are parallel!
11 Types of Linear Systems ConsistentHave solution(s)Independent and consistent has one solutionLines have different slopes. Lines meet at one pointDependent and consistent has all the points on the line as a solution.The lines have same slopes and same interceptsThey are the same line!InconsistentThe lines are parallel, lines never meet, they have no points in common.
12 Consistent (Dependent or Independent) or Inconsistent?
13 Solving Systems of Equations Using Algebra There are two ways to solve Systems of Linear Equations using AlgebraUsing Algebra is best because you don’t need to graph linesThe two methods:Elimination by Addition or SubtractionElimination by Substitution
14 Elimination by Addition or Subtraction Given two equations2x + y = 82x – y = 0Label the equations2x + y = 82x – y = 0Eliminate one variable by adding or subtracting the equations. (1)+(2)2x + y = 8+(2x – y = 0)4x + 0y = 8 x = 2We know that x =2 nowSo just substitute it back into either equation (1) or (2) to find y:2(2)+y=8y = 4The solution to the System is when x = 2 and y = 4. It is the point (2, 4)
15 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
16 Elimination by Addition or Subtraction 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 = 3Given two equations4x + 2y = 162x – y = 4Label the equations4x + 2y = 162x – y = 4We know that x =3 nowSo just substitute it back into either equation (1) or (2) to find y:4(3)+2y=16y = 2The solution to the System is when x = 3 and y = 2. It is the point (3, 2)
17 More examples Elimination Solve:(1) 5x + 2y = 3(2) 2x + 3y = –13*(1) x + 6y = 9-2*(2) – (4x + 6y = –2)Subtract: 11x – 0y = 1111x = x = 1Substitute value x = 1 into either equation to find y5(1) + 2y = 3 y = –1Solution: (1, –1)Check the answer!!Solve:(1) 2x + 3y = 3(2) –6x + 6y = 63*(1) x + 9y = 9Add (2) +(–6x + 6y = 6)Add: x + 15y = 15y = 1Substitute value y = 1 into either equation to find x2x + 3(1) = 3 2x = 0; x = 0Solution: (0, 1)Check the answer!!
18 Review of Steps to Solve by Elimination Put equations into a ‘standard’ order(ax + by = c)Label the equationsMultiply either, or both, equations by some number so that one variable will be eliminated when the equations are added or subtractedAdd or subtract the equations to eliminate one variable and find the remaining variableFind the eliminated variable by substituting the the value of the solved variable back into either equation of the systemCheck your answer (x, y) in both equations
19 Solving by Substitution The steps are:Label the equations (1), (2)Isolate one variable and express it in terms of the otherSubstitute that expression into the other equationSolve for the now single variableSubstitute the value of the found variable back into either equation to find the other unknownCheck the answer in both equationsGiven:(1): 2x – y = 4(2): x + 2y = – 3From (2): x = – 3 – 2ySubstitute into (1)2(– 3 – 2y) – y = 4– 6 – 4y – y = 4– 6 – 5y = 4– 5y = 10So: y = – 2Substitute into either eqn2x – (– 2) = 42x + 2 = 4 ; 2x = 2So x = 1Now Check the Answer!
20 Sample Problem - Digits Two numbers add to 12The same two numbers have a difference of 4Find the two numbersGive the unknowns labels: like a and bWrite the two facts or equations:(1) a + b = 12(2) a – b = 4Add eqns (1) and (2)2a + 0 =16 2a = 16a = 8. So if a = 8 then b = 4 from either equationSo the numbers are 8 and 4Check to see if these numbers work in both equations
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: speed of airplane;b: speed of winda + b=(1500/3) = 500 mph with the winda – b = 1500/3.333= 450 mph against the windSolve from there!??
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 agesLet: f = father’s current age. s = son’s current age.f – s = 24(f + 8) = 2*(s + 8)f + 8 = 2s + 16f – 2s = 8Eqn (1) – Eqn (2)– (f – 2s = 8)s = 16 f = 40Check: (1): (40) – (16) = 24(2): (40 + 8) = 2*(16 + 8)
23 Coin ProblemsEdgar 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 = 20Use substitution(2): n = 20 – dSubstitute eqn (2) into eqn (1)(20 - d)*(0.05) + d*(0.10)= 1.40d = 1.4 d=8, n=12Check the answer in both equationsIt works!!
24 Word Problems1000 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 soldLet a = nbr of adult tickets sold; let c = nbr of children tickets sold(1): a + c = 1000(2): 8.5a + 4.5c = 7300Substitute: a = 1000 – c8.5(1000 – c) + 4.5c = 73008500 – 4c = 7300c = and a = 700700 adult tickets were sold, 300 children’s tickets were soldNow check answer in both equations!
25 The EndNow you know how to solve Systems of Linear Equations with two unknownsIf you take Pre-Calculus, you will learn how to solve for 3 unknowns!Just a few extra steps for three unknowns