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Solving Systems of Linear Equations By Elimination

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What is Elimination? To eliminate means to get rid of or remove. You solve equations by eliminating one of the variables (x or y) using addition or subtraction.

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Example 1 Solve the following system of linear equations by elimination. 2x – 3y = 15 5x + 3y = 27 (1) (2) Add equation (1) to equation (2) 7x + 0y = 42 7x = 42 x = 6 By eliminating y, we can now solve for x

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Example 1 Substitute x= 6 into equation (1) to solve for y 2x – 3y = 15 2(6) – 3y = 15 12 – 3y = 15 – 3y = 15 – 12 – 3y = 3 y = -1 Check your solution x = 6 and y = -1 in equation (2) 5x + 3y = 27 5(6) + 3(-1) = 27 30 – 3 = 27 27 = 27 LS = RS Therefore, the solution set = (6,-1)

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5x – 6y = -32 3x + 6y = 48 You Try

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One More 7x + 2y = −19 −x + 2y = 21

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If you have noticed in the last few examples that to eliminate a variable, it’s coefficients must have a sum or difference of zero. Sometimes you may need to multiply one or both of the equations by a nonzero number first so that you can then add or subtract the equations to eliminate one of the variables. 2x + 5y = 177x + 2y = 102x + 5y = -22 6x – 5y = -19-7x + y = -1610x + 3y = 22 We can add these two equations together to eliminate the y variable. We can add these two equations together to eliminate the x variable. What are we going to do with these equations, can’t eliminate a variable the way they are written?

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Multiplying One Equation Solve by Elimination 2x + 5y = -22 10x + 3y = 22 2x + 5y = -22 5(2x + 5y = -22) 10x + 25y = -110 10x + 3y = 22 10x + 3y = 22 - (10x + 3y = 22) 0 + 22y = -132 y = -6

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Step 2 y = -6 Solve for the eliminated variable using either of the original equations. 2x + 5y = -22 Choose the first equation. 2x + 5(-6) = -22 Substitute -6 for y. 2x – 30 = -22 Solve for x. 2x = 8 x = 4 The solution is (4, -6).

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Solve by elimination. -2x + 5y = -32 7x – 5y = 17 2x – 3y = 61 2x + y = -7 3x – 10y = -25 4x + 40y = 20

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5x + 4y = -28 3x + 10y = -13

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Multiplying Both Equations To eliminate a variable, you may need to multiply both equations in a system by a nonzero number. Multiply each equation by values such that when you write equivalent equations, you can then add or subtract to eliminate a variable. 4x + 2y = 14 7x + 3y = -8 In these two equations you cannot use graphing or substitution very easily. However ever if we multiply the first equation by 3 and the second by 2, we can eliminate the y variable. Find the least common multiple LCM of the coefficients of one variable, since working with smaller numbers tends to reduce the likelihood of errors. 4 x 7 = 28 2 x 3 = 6

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4x + 2y = 143(4x + 2y = 14) 12x + 6y = 42 7x – 3y = - 8 2(7x – 3y = -8) 14x – 6y = -16 26x + 0 = 26 26x = 26 x = 1 Solve for the eliminated variable y using either of the original equations. 4x + 2y = 14 4(1) + 2y = 14 4 + 2y = 14 2y = 10 y = 5The solution is (1, 5).

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Closure Solve by your method of choice: 1) 2x + 5y = 172) 7x + 2y = 10 6x + 5y = -9 -7x + y = -16 3) 2x – 3y = 614) 24x + 2y = 52 2x + y = -7 6x – 3y = -36 5) y = 2x6) 9x + 5y = 34 y = x – 1 8x – 2y = -2

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