Solving Linear Systems The EliminationLinear Combinations Method Introduction

Solving Linear Systems The Elimination/Linear Combinations Method

Introduction • John rented a car and • Define the variables: drove 125 miles on a • Let “d” represent the 2 day trip and was cost per day and “m” charged $95. 75. He represent the cost per drove 350 miles on a mile. different 4 day trip and • Set up the system was charged $226. 50 for the same kind of rental car. • Find the daily fee and cost per mile

Using Substitution to Investigate the Combinations Method Compare the values in the original system to the values in the 3 rd step

In Order to Eliminate… In order to eliminate a value ( get a solution of zero) when we “combine” it with another Opposites value the two values must be _____ The opposite of 4 is -4 -2 is 2 -3/4 is 3/4 2 x is -2 x -4 y is 4 y

The Elimination Method 1. Looking for an opposite variable 2. Combine the equations (Add them together) 3. Solve for the remaining variable 4. Substitute the value back into the easiest equation to solve

Solve the following systems using the elimination method 1. 2. X We eliminate the ____ 3. We eliminate the ____ Y X We eliminate the ____ 6 12 ____Y = ____ 7 14 ____X = ____ 5 15 ____Y = ____ 2 X = ____ 3 Y = ____ Substitute the Y value back into the equation where the math is_____ easiest 3 x + 2(___) 2 =7 4 =7 3 x + ____ 3 3 x = ___ 1 X = ___

The Elimination Method 1. 4. Look for variables that are opposite… and if there aren’t opposites then Multiply one of the equations by a value to get an opposite variable Combine the equations 5. Solve for the remaining variable 6. Substitute the value back into the easiest equation to solve 2. 3. Multiply by (-1)

The Elimination Method 1. Multiply by (2) 3. Look for variables that are opposite Multiply one of the equations by a value to get an opposite variable Combine the equations 4. Solve for the remaining variable 5. Substitute the value back into the easiest equation to solve 2.

Solve the following systems using the elimination method 4. 5. 6. Multiply the 2 nd equation ________ Multiply the 1 st equation ________ Multiply the 2 nd equation ________ -1 by ____to p eliminate_____ 2 by ____to x eliminate_____ -2 by ____to y eliminate_____ 5 = ____ 25 ____r 1 0 ____Y = ____ 5 r = ____ 0 Y = ____ p + 4(___) 5 = 23 20 = 23 p + ____ 3 p = ___ 8 -16 ____X = ____ -2

The Elimination Method 1. Multiply by 2 Multiply by -3 3. Look for variables that are opposite Multiply both of the equations by a value to get an opposite variable Combine the equations 4. Solve for the remaining variable 5. Substitute the value back into the easiest equation to solve 2.

Solve the following systems using the elimination method 7. 8. 9. Multiply the 3 1 st equation by___ _____ Multiply the -7 1 st equation by___ __________ 2 nd equation And the______ 2 by____ 29 -29 = ____X 4 Y = ____ -1 X = ____ 20 = 26 2 x + ____ 6 2 x = ___ 3 x = ___ 2 nd equation And the______ 3 by____ 7 Y = ____ 28 ___ 2 x + 5(___) 4 = 26 Multiply the -4 1 st equation by___ _____ -2 4 ____Y = ____ -2

Elimination Gotcha Solve the system using elimination 1. 4. Make sure both equations are written in same form. (Usually with both x and y variables on the left side) Then solve as you would any other system. Multiply both of the equations by a value to get an opposite variable Combine the equations 5. Solve for the remaining variable 6. Substitute the value back into the easiest equation to solve 2. 3.

Car Rental Problem The car was rented for $26 per day and $0. 35 per mile

Choosing a method Example Method Why Substitution The value of y is known and can be easily substituted into the other equation Linear Combinations 5 y and -5 y are opposites and are easily eliminated. Linear Combinations B can be easily eliminated by multiplying the first equation by -2