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Ch 2 – Systems of Linear Equations & Inequalities 2.1 – Solving Systems of Equations in Two Variables
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HEARTS You are in charge of teaching the following words: CONSISTENT, INDEPENDENT, DEPENDENT, INCONSISTENT ◦Graphical Example ◦Properties of the slopes ◦Algebraic Example ◦Number of solutions
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DIAMONDS You are in charge of teaching an example of solving a system by graphing. You will find an example, work it out, and explain the solution. It has to be an example with one solution.
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CLUBS You are in charge of teaching an example of solving a system by substitution. You will find an example, work it out, and explain the solution. It has to be an example with one solution.
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SPADES You are in charge of teaching an example of solving a system by elimination. You will find an example, work it out, and explain the solution. It has to be an example with one solution.
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PHASE 1 GET INTO GROUPS BY THE SUIT OF YOUR CARD LOOK THROUGH THE BOOK AND BECOME AN “EXPERT” ON YOUR TOPIC YOU HAVE 10 MINUTES!
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PHASE 2 NOW WE WILL REGROUP BY THE FOLLOWING TO TEACH EACH OTHER OUR TOPIC 2 & 3 4 & 5 6 & 7 8 YOU HAVE 15 MINUTES!
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Vocabulary 1. Consistent: 2. Independent: 3. Dependent: 4. Inconsistent: A system of equations with at least one solution. A system of equations with exactly one solution. A system of equations with infinitely many solutions. aka coincide A system of equations with no solutions.
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Draw a graphical picture of the four vocabulary words.
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There are 3 ways to solve linear systems: 1. Graphically 2. Substitution 3. Elimination
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Graphically When solving graphically, you look for the intersection of the two lines (if there is any)
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Substitution The goal of substitution is to put one equation into the other so there is only one variable to solve for.
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Elimination The goal with elimination is to get one variable to cancel by adding the two equations together or a multiple of the two equations. Then you will only have 1 equation with 1 unknown and you can solve for it.
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Solve the system graphically x = 5 4x + 5y = 20
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Steps for Substitution 1. Solve 1 of the equations for 1 of its variables 2. Substitute the expression from Step 1 into the other equation and solve. 3. Substitute the value from Step 2 into an equation and solve for the remaining variable.
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Solve the linear system by substitution. x + 3y = 0 2x + 6y = 5
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Solve by elimination. 3x – 5y = -8 x + 2y = 1
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Think Opposite! 2 terms are given below. Find what you could multiply one (or two) term(s) by so that the sum of the two terms is zero. ◦3x, x ◦-4y, 2y ◦5x, 3x
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Steps to Solve by Linear Combination 1. Make sure all like terms are lined up. 2. Multiply one or both equations by a constant so 1 variable will cancel when the 2 equations are added. 3. Add the two equations together and solve. 4. Substitute answer into an original equation to find the other variable.
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Solve the following linear systems. 1. x = 2y – 8 2x – y = 7
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2. 3x + 2y = 10 5x - 7y = -4
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3. 6x – 4y = 14 -3x + 2y = 7
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4. 9x – 3y = 15 -3x + y = -5
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Madison is thinking about leasing a car for two years. The dealership says that they will lease her the car she has chosen for $326 per month with only $200 down. However, if she pays $1600 down, the lease payment drops to $226 per month. a) What is the break-even point in the two lease plans that Madison is considering? b) If Madison keeps the lease for 24 months, which lease should she choose?
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