Ch 2 – Systems of Linear Equations & Inequalities 2.1 – Solving Systems of Equations in Two Variables

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

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.

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.

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.

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!

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!

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.

Draw a graphical picture of the four vocabulary words.

There are 3 ways to solve linear systems: 1. Graphically 2. Substitution 3. Elimination

Graphically When solving graphically, you look for the intersection of the two lines (if there is any)

Substitution The goal of substitution is to put one equation into the other so there is only one variable to solve for.

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.

Solve the system graphically x = 5 4x + 5y = 20

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.

Solve the linear system by substitution. x + 3y = 0 2x + 6y = 5

Solve by elimination. 3x – 5y = -8 x + 2y = 1

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

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.

Solve the following linear systems. 1. x = 2y – 8 2x – y = 7

2. 3x + 2y = 10 5x - 7y = -4

3. 6x – 4y = 14 -3x + 2y = 7

4. 9x – 3y = 15 -3x + y = -5

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?