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Unit 6 Notes Day 5

Unit 6 Notes Day 5. EQ: How do I use a two-way table to answer questions about data?. Complete the following table Students are asked if they prefer to go swimming or to the gym. Some people are asked about their favorite outdoor sport. How many boys liked mash?

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Unit 6 Notes Day 5

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  1. Unit 6 Notes Day 5 EQ: How do I use a two-way table to answer questions about data?

  2. Complete the following table Students are asked if they prefer to go swimming or to the gym.

  3. Some people are asked about their favorite outdoor sport.

  4. How many boys liked mash? • How many teachers preferred new potatoes? • How many girls were asked? • Out of the people who liked chips, how many were boys? What is the probability the person picked is a boy? • What is the probability the person liked mash? • What is the probability the person was a teacher who preferred new potatoes? • What is the probability that, out of the girls, the person liked chips? • Out of the people who liked chips, what is the probability the person was a boy?

  5. 100g 200g 300g Total Ground 50 120 Powder 80 35 26 Granules 40 45 Total 135 135 400 For the following table, copy and complete it, then answer the questions about types of coffee bought. • Coffee is sold in three types and in three weights. • How many people bought 100g of powdered coffee? • How many people bought ground coffee? • Out of the packets weighing 200g, what is the probability the packet bought contained granules?

  6. Unit 7 Notes Day 1 EQ: How do I solve a system of equations?

  7. Profit In business, the point at which income equals expenses is called the break-evenpoint.When starting a business, people want to know the point a which their income equals their expenses, that’s the point where they start to make a profit. In the example above the values of y on the blue line represent dollars made and the value of y on the dotted red line represent dollars spent. Loss

  8. Systems of Linear Equations A system of linear equationsconsists of two or more linear equations. The solution of a system of linear equations in two variables is any ordered pair that solves both of the linear equations.

  9. Solution of a System Determine whether the given point is a solution of the following system. point: (– 3, 1) system: x – y = – 4 2x+ 10y = 4 • Plug the values into the equations. First equation: – 3 – 1 = – 4 true Second equation: 2(– 3) + 10(1) = – 6 + 10 = 4 true • Since the point (– 3, 1) produces a true statement in both equations, it is a solution.

  10. Solution of a System Determine whether the given point is a solution of the following system point: (4, 2) system: 2x – 5y = – 2 3x+ 4y = 4 Plug the values into the equations First equation: 2(4) – 5(2) = 8 – 10 = – 2 true Second equation: 3(4) + 4(2) = 12 + 8 = 20  4 false Since the point (4, 2) produces a true statement in only one equation, it is NOT a solution.

  11. A system of equations is a set of two or more equations that have variables in common. Remember, when solving systems of equations, you are looking for a solution that makes each equation true.

  12. A system of linear equations is a set of two or more equations with the same variable. The solution of a system in x and y is any ordered pair (x, y) that satisfies each of the equations in the system.The solution of a system of equations is the intersection of the graphs of the equations.

  13. Two equations, such as those listed below, are called a system of linear equations. A solution to a system of linear equations is an ordered pair that satisfies all equations in the system. For example, (3, 4) satisfies the system x + y = 7 (3 + 4 is, indeed, 7.) x – y = -1 (3 – 4 is indeed, -1.) Thus, (3, 4) satisfies both equations and is a solution of the system. The solution can be described by saying that x = 3 and y = 4. The solution can also be described using set notation. The solution set to the system is {(3, 4)} - that is, the set consisting of the ordered pair (3, 4). Systems of Linear Equations and Their Solutions

  14. Solution Because 4 is the x-coordinate and -1 is the y-coordinate of (4, -1), we replace x by 4 and y by -1. x + 2y = 2 x – 2y = 6 4 + 2(-1) = 2 4 – 2(-1) = 6 4 + (-2) = 2 4 – (-2) = 6 2 = 2 true 4 + 2 = 6 6 = 6 true The pair (4, -1) satisfies both equations: It makes each equation true. Thus, the pair is a solution of the system. The solution set to the system is {(4, -1)}. ? ? ? ? ? Example: Determining Whether an Ordered Pair Is a Solution of a System Determine whether (4, -1) is a solution of the system x + 2y = 2 x – 2y = 6.

  15. y y y x x x Exactly one solution No Solution (parallel lines) Infinitely many solutions (lines coincide) The Number of Solutions to a System of Two Linear Equations The number of solutions to a system of two linear equations in two variables is given by one of the following. Number of SolutionsWhat This Means Graphically Exactly one ordered-pair solution The two lines intersect at one point. No solution The two lines are parallel. Infinitely many solutions The two lines are identical.

  16. 3 Methods for Solving Systems • Substitution Method (used mostly when one of the equations has a variable with a coefficient of 1 or -1 or a variable is already isolated) • Linear Combination Method • Graphing Method (used only when the graph is easily completed and normally with whole number points)

  17. Solving Systems of Equations Algebraically Substitution Method

  18. The Substitution Method • Step 1 Solve one equation for x (or y). • Step 2 Substitute the expression from Step 1 into the other equation. • Step 3 Solve for y (or x). • Step 4 Take the value of y (or x) found in Step 3 and substitute it into one of the original equations. Then solve for the other variable. • Step 5 The ordered pair of values from Steps 3 and 4 is the solution. If the system has no solution, a contradictory statement will result in either Step 3 or 4.

  19. The substitution method is used to eliminate one of the variables by replacement when solving a system of equations.Think of it as “grabbing” what one variable equals from one equation and “plugging” it into the other equation.

  20. Solve this system of equations using the substitution method. Step 1 3y – 2x = 11 Y + 2x = 9 Solve one of the equations for either “x” or “y”. In this example it is easier to solve the second equation for “y”, since it only involves one step. Y = 9 – 2x

  21. Step 2Replace the “y” value in the first equation by what “y” now equals (y = 9 – 2x).Grab the “y”value and plug it into the other equation.3y – 2x = 113(9 – 2x) – 2x = 11

  22. Step 3Solve this new equation for “x”.3(9 – 2x) – 2x = 1127 – 6x – 2x = 1127 – 6x – 2x = 1127 – 8x = 11-8x = -16x = 2

  23. Step 4Now that we know the “x” value(x = 2), we place it into either of the ORIGINALequations in order to solve for “y”.Pick the easierone to work with!Y + 2x = 9y + 2(2) = 9y + 4 = 9y = 5

  24. Step 5Check:substitute x = 2 and y = 5 into BOTH ORIGINAL equations. If these answers are correct BOTH equations will be true!3y – 2x = 113(5) – 2(2) = 1115 – 4 = 11 11 = 11 True?Y + 2x = 95 + 2(2) = 95 + 4 = 99 = 9 True?

  25. 3x-y=13 2x+2y= -10 Solve the 1st eqn for y. 3x-y=13 -y= -3x+13 y=3x-13 Now substitute 3x-13 in for the y in the 2nd equation. 2x+2(3x-13)= -10 Now, solve for x. 2x+6x-26= -10 8x=16 x=2 Now substitute the 2 in for x in for the equation from step 1. y=3(2)-13 y=6-13 y=-7 Solution: (2,-7) Plug in to check soln. Ex: Solve using substitution method

  26. Solving Systems of Equations Elimination method

  27. Elimination Method • Multiply one or both equations by a real number so that when the equations are added together one variable will cancel out. • Add the 2 equations together. Solve for the remaining variable. • Substitute the value form step 2 into one of the original equations and solve for the other variable. • Write the solution as an ordered pair (x,y).

  28. 2x-6y=19 -3x+2y=10 Multiply the entire 2nd eqn. by 3 so that the y’s will cancel. 2x-6y=19 -9x+6y=30 Now add the 2 equations. -7x=49 and solve for the variable. x=-7 Substitute the -7 in for x in one of the original equations. 2(-7)-6y=19 -14-6y=19 -6y=33 y= -11/2 Now write as an ordered pair. (-7, -11/2) Plug into both equations to check. Ex: Solve using lin. combo. method.

  29. 9x-3y=15 -3x+y= -5 Which method? Substitution! Solve 2nd eqn for y. y=3x-5 9x-3(3x-5)=15 9x-9x+15=15 15=15 OK, so? What does this mean? Both equations are for the same line! ¸ many solutions Means any point on the line is a solution. Ex: Solve using either method.

  30. 6x-4y=14 -3x+2y=7 Which method? Linear combo! Multiply 2ndeqn by 2. 6x-4y=14 -6x+4y=14 Add together. 0=28 Huh? What does this mean? It means the 2 lines are parallel. No solution Since the lines do not intersect, they have no points in common. Ex: Solve using either method.

  31. You can use the Addition and SubtractionProperties of Equality to solve a system by the elimination method. You can add or subtract equations to eliminate (getting rid of) a variable. Step 1 5x – 6y = -32 3x + 6y = 48 Eliminate ybecause the sum of the coefficients of y is zero 5x – 6y = -32 3x + 6y = 48 8x + 0 = 16 x = 2 If you add the two equations together, the +6y and -6y cancel each other out because of the Property of Additive Inverse Addition Property of Equality Solve for x

  32. Step 2Solve for the eliminated variable y using either of the original equations.3x + 6y = 483(2) + 6y = 486 + 6y = 486y = 42y = 7 Choose the 2nd equation Remember x = 2 Substitute 2 for x Simplify. Then solve for y.

  33. Since x = 2 and y = 7, the solution is (2, 7) Check 5x – 6y = -32 3x + 6y = 48 5(2) – 6(7) = -32 3(2) + 6(7) = 48 10 – 42 = -32 6 + 42 = 48 -32 = -32 48 = 48 Remember, the order pair (2, 7) must make both equations true. True True

  34. Multiplying One Equation Be careful when you subtract. All the signs in the equationthat isbeing subtracted change. -10x – 3y = -22 Solve by Elimination 2x + 5y = -22 10x + 3y = 22 Step 1 2x + 5y = -22 5(2x + 5y = -22) 10x + 25y = -110 10x + 3y = 22 10x + 3y = 22-(10x + 3y = 22) 0 + 22y = -132 y = -6 To prepare for eliminating x, multiply the first equation by 5. Subtract the equations to eliminate x. Start with the given system. Ask: Is one coefficient a factor of the other coefficient for the same variable? NEXT

  35. Step 2Solve for the eliminated variable using either of the original equations. 2x + 5y = -22 Choose the first equation.2x +5(-6) = -22 Substitute -6 for y. 2x – 30 = -22Solve for x. 2x = 8 x = 4The solution is (4, -6).

  36. Solve by elimination.-2x + 5y = -327x – 5y = 173x – 10y = -254x + 40y = 202x – 3y = 612x + y = -7 Ask: Is one coefficient a factor of the other coefficient for the same variable?

  37. Multiplying Both Equations To eliminate a variable, you may need to multiply both equations in a system by a nonzero number. Multiply each equation by values such that when you write equivalent equations, you can then add or subtract to eliminate a variable. 4x + 2y = 14 7x + 3y = -8 In these two equations you cannot use graphing or substitution very easily. However ever if we multiply the first equation by 3 and the second by 2, we can eliminate the y variable. Find the least common multiple LCM of the coefficients of one variable, since working with smaller numbers tends to reduce the likelihood of errors. 4 x 7 = 28 2 x 3 = 6 NEXT

  38. Add the equations to eliminate y. 4X + 2Y = 14 3(4X + 2Y = 14) 12X + 6Y = 427X – 3Y = - 8 2(7X – 3Y = -8) 14X – 6Y = -16 26X + 0 = 26 26X = 26X = 1Solve for the eliminated variable y using either of the original equations.4x + 2y = 144(1) + 2y = 14 4 + 2y = 14 2y = 10y = 5 The solution is (1, 5). Start with the given system. To prepare to eliminate y, multiply the first equation by 3 and the second equation by 2.

  39. Answer these questions. Can I eliminate a variable by adding or subtracting the given equations? Do It. YES Or NO Can I multiply one of the equations by a number, and then add or subtract the equations? YES or NO Do It. Multiply both equations by different numbers. then add or subtract the equations

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