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Objective

Objective. The student will be able to: solve systems of equations using elimination with multiplication. Solving Systems of Equations. So far, we have solved systems using graphing , substitution , and elimination .

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Objective

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  1. Objective The student will be able to: solve systems of equations using elimination with multiplication.

  2. Solving Systems of Equations So far, we have solved systems using graphing, substitution, and elimination. This goes one step further and show how to use ELIMINATION with multiplication. What happens when the coefficients are not the same? We multiply the equations to make them the same! You’ll see…

  3. 1) Solve the system using elimination. 2x + 2y = 6 3x – y = 5 Multiply the bottom equation by 2 2x + 2y = 6 (2)(3x – y = 5) 2x + 2y = 6 (+) 6x – 2y = 10 8x = 16 x = 2 2(2) + 2y = 6 4 + 2y = 6 2y = 2 y = 1 (2, 1)

  4. 2) Solve the system using elimination. Multiply the top equation by -4 (-4)(x + 4y = 7) 4x – 3y = 9 -4x – 16y = -28 (+) 4x – 3y = 9 -19y = -19 y = 1 4x – 3y = 9 4x – 3(1) = 9 4x – 3 = 9 4x = 12 x = 3 (3, 1) x + 4y = 7 4x – 3y = 9

  5. Which variable is easier to eliminate? 3x + y = 4 4x + 4y = 6 • y is easier to eliminate. • Multiply the 1st equation by (-4)

  6. 3) Solve the system using elimination. Multiply both equations (3)(3x + 4y = -1) (4)(4x – 3y = 7) 9x + 12y = -3 (+) 16x – 12y = 28 25x = 25 x = 1 3(1) + 4y = -1 3 + 4y = -1 4y = -4 y = -1 (1, -1) 3x + 4y = -1 4x – 3y = 7

  7. 4) What is the best number to multiply the top equation by to eliminate the x’s? 3x + y = 4 6x + 4y = 6 6x + 4y = 6 6x + 4(1) = 6 6x = 2 x = 2/6 x = 1/3 (1/3, 1) • Multiply the 1st equation by (-2) • (-2)(3x + y = 4)  -6x -2y = -8 6x + 4y = 6 2y = -2 y = 1

  8. 5) Solve using elimination. 2x – 3y = 1 x + 2y = -3 2x – 3y = 1 2x – 3(-1) = 1 2x + 3 = 1 2x = -2 x = -1 (-1, -1) • Multiply 2nd equation by (-2) to eliminate the x’s. • (-2)(x + 2y = -3)  -2x – 4y = 6 1st equation  2x – 3y = 1 -7y = 7 y = -1

  9. 5) Solve using elimination. Add down to eliminate x. But look what happens, y is eliminated too. We now have a false statement, thus the system has no solution, it is inconsistent. 2(4x + y = 6 ) -8x – 2y = 13 8x + 2y = 12 -8x – 2y = 13 0 = 25

  10. 6) Elimination Solve the system using elimination. 5x – 2y = –15 3x + 8y = 37 20x – 8y = –60 3x + 8y = 37 23x = –23 x = –1 (4) Since neither variable will drop out if the equations are added together. Multiply one or both of the equations by a constant to make one of the variables have the same number with opposite signs. The best choice is to multiply the top equation by 4 since only one equation would have to be multiplied. Also, the signs on the y-terms are already opposites.

  11. 6) Elimination x = –1 3x + 8y = 37 (second equation) 3(–1) + 8y = 37 –3 + 8y = 37 8y = 40 y = 5 The solution is (–1, 5) To find the second variable, substitute in any equation that contains two variables.

  12. 7) Elimination Solve the system using elimination. 4x + 3y = 8 3x – 5y = –23 20x + 15y = 40 9x – 15y= –69 29x = –29 x = –1 (5) For this system, we must multiply both equations by a different constant in order to make one of the variables “drop out.” (3) It would work to multiply the top equation by –3 and the bottom equation by 4 OR to multiply the top equation by 5 and the bottom equation by 3.

  13. 7) Elimination x = –1 4x + 3y = 8 4(–1) + 3y = 8 –4 + 3y = 8 3y = 12 y = 4 The solution is (–1, 4)

  14. 8) Elimination Method (1) (2) Solve the system. To eliminate x, multiply equation (1) by –2 and equation (2) by 3 and add the resulting equations.

  15. 8) Elimination Method Substitute 2 for yin (1) or (2). The solution is (3, 2)

  16. 9) Solving an Inconsistent System (1) (2) Solve the system Eliminate x by multiplying (1) by 2 and adding the result to (2). Solution set is . Inconsistent System

  17. 10) Solving a System with Dependent Equations (1) (2) Solve the system. Eliminate x by multiplying (1) by 2 and adding the result to (2). Each equation is a solution of the other. Infinite solutions.

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