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Solve the following system using the elimination method.
5x + y = 11 6x + 2y = 14 Notice that the x and y terms do not contain opposites. – 10x – 2y = – 22 If the 1st equation is multiplied by – 2, the y terms will become opposites. 6x + 2y = 14 – 4x = – 8 Add the two equations and solve. x = 2 5x + y = 11 Solve for y. 5(2) + y = 11 → y = 11 → y = 1 Solution: x = and y = 1
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10.05 Solving Systems of Equations by Elimination (2 Multipliers)
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Remember that to solve a system of equations using the elimination method, it must have at least one set of opposite variable terms. Some systems require that both equations be changed in order to create opposites. To solve this type of system, multiply the coefficients of “x” together and find this number. Then multiply the 1st by some factor that results in a positive value of that number. Multiply the 2nd equation by some factor that results in a negative value of that number. Solve as usual.
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Solve the following system using the elimination method.
4x + 5y = 3 5x + 2y = 8 Notice that the x and y terms do not contain opposites. Both equations need to be changed in order to create opposites. 20x y = 15 Multiply the 1st equation by 5. Multiply the 2nd equation by – 4. – 20x – 8y = – 32 Add the two equations and solve. 17y = – 17 → y = – 1 Solve for x. 4x + 5y = 3 4x + 5(– 1) = 3 → 4x – 5 = 3 → 4x = 8 → x = 2 Solution: x = and y = – 1
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Solve the following system using the elimination method.
3x + 5y = 22 4x – 2y = 12 Notice that the x and y terms do not contain opposites. Both equations need to be changed in order to create opposites. 12x y = 88 Multiply the 1st equation by 4. Multiply the 2nd equation by – 3. – 12x + 6y = – 36 Add the two equations and solve. 26y = 52 → y = 2 Solve for x. 3x + 5y = 22 3x + 5(2) = 22 → 3x = 22 → 3x = 12 → x = 4 Solution: x = and y = 2
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Solve the following system using the elimination method.
2x + 4y = 6 3x + 6y = 9 Notice that the x and y terms do not contain opposites. Both equations need to be changed in order to create opposites. 6x y = 18 Multiply the 1st equation by 3. Multiply the 2nd equation by – 2. – 6x – 12y = – 18 Add the two equations and solve. 0 = 0 The variables have cancelled and the statement is true. There are infinite solutions to the system.
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Solve the following system using the elimination method.
– 2x – 2y = 4 3x + 3y = – 7 Notice that the x and y terms do not contain opposites. Both equations need to be changed in order to create opposites. 6x + 6y = – 12 Multiply the 1st equation by – 3. Multiply the 2nd equation by – 2. – 6x – 6y = 14 Add the two equations and solve. 0 = 2 The variables have cancelled and the statement is false. There are no solutions to the system.
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Solve the following system using the elimination method.
Try This: Solve the following system using the elimination method. 2x – 3y = 7 5x + 2y = – 11 Notice that the x and y terms do not contain opposites. Both equations need to be changed in order to create opposites. 10x – 15y = 35 Multiply the 1st equation by 5. Multiply the 2nd equation by – 2. – 10x – 4y = 22 Add the two equations and solve. – 19y = 57 → y = – 3 2x – 3y = 7 Solve for x. 2x – 3(– 3) = 7 → 2x = 7 → 2x = – 2 → x = – 1 Solution: x = – and y = – 3
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