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Solving Linear Systems

Solving Linear Systems. The Elimination/Linear Combinations Method. John rented a car and drove 125 miles on a 2 day trip and was charged $95.75. He drove 350 miles on a different 4 day trip and was charged $226.50 for the same kind of rental car. Find the daily fee and cost per mile.

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Solving Linear Systems

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  1. Solving Linear Systems The Elimination/Linear Combinations Method

  2. John rented a car and drove 125 miles on a 2 day trip and was charged $95.75. He drove 350 miles on a different 4 day trip and was charged $226.50 for the same kind of rental car. Find the daily fee and cost per mile Define the variables: Let “d” represent the cost per day and “m” represent the cost per mile. Set up the system Introduction

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

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

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

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

  7. 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 Solve for the remaining variable Substitute the value back into the easiest equation to solve The Elimination Method Multiply by (-1)

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

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

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

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

  12. Elimination Gotcha Solve the system using elimination • 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 • Solve for the remaining variable • Substitute the value back into the easiest equation to solve

  13. Car Rental Problem The car was rented for $26 per day and $0.35 per mile

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

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