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Solving a System of Equations in Two Variables By Elimination Chapter 8.3

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Steps to solve a system of equations using the elimination method 1.The coefficients of one variable must be opposite. 2.You may have to multiply one or both equations by an integer so that step 1 occurs. 3.Add the equations so that a variable is eliminated. 4.Solve for the remaining variable. 5.Substitute the value into one of the original equations to solve for the other variable. 6.Check the solution.

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3x 5x step 1 coefficients of one variable must be opposite. 1. Solve by addition. + y = 7 – 2y = 8

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2( ) 6x + 2y = 14 5x – 2y = 8 step 2 make the y opposites, multiply first equation by 2. 1. Solve by addition. 3x 5x + y = 7 – 2y = 8

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2( ) 6x + 2y = 14 5x – 2y = 8 11x = 22 step 3 add to eliminate the y. 1. Solve by addition. 3x 5x + y = 7 – 2y = 8

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2( ) 6x + 2y = 14 5x – 2y = 8 11x = 22 11 11 x = 2 step 4 solve for x. 1. Solve by addition. 3x 5x + y = 7 – 2y = 8

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2( ) 6x + 2y = 14 5x – 2y = 8 11x = 22 11 x = 2 3(2) -6 -6 (2, 1) 6 + y = 7 y = 1 step 5 substitute into equation 1 and solve for y. 1. Solve by addition. + y= 7 3x 5x + y = 7 – 2y = 8

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4x 3x step 1 coefficients of one variable must be opposite. 2. Solve by addition. + 5y = 17 + 7y = 12

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-4( ) 3( ) 12x + 15y = 51 -12x – 28y = -48 step 2 make the x opposites, multiply first equation by 3, second equation by -4. 2. Solve by addition. 4x 3x + 5y = 17 + 7y = 12

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-4( ) -13y = 3 step 3 add to eliminate the x. 2. Solve by addition. 3( ) 12x + 15y = 51 -12x – 28y = -48 4x 3x + 5y = 17 + 7y = 12

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-4( ) -13 -13 step 4 solve for y. 2. Solve by addition. -13y = 3 3( ) 12x + 15y = 51 -12x – 28y = -48 y = 13 -3 4x 3x + 5y = 17 + 7y = 12

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-4( ) 52x 13( ) step 5 substitute into equation 1 and solve for x. 2. Solve by addition. -13 -13y = 3 3( ) 12x + 15y = 51 -12x – 28y = -48 y = 13 -3 4x + ( ) = 17 13 -15 +15 +15 52x = 236 52 52 x = 13 59 (, ) 13 59 13 -3 – 15= 221 + 5( ) 13 -3 = 17 4x 4x 3x + 5y = 17 + 7y = 12

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( ) 12 xx -2x Before beginning with the steps remove the fractions in the first equation by multiplying 12 to each term. 3. Solve by addition. 8x – 9y= 36 step 1 coefficients of one variable must be opposite. – y = 3 + y= 6

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4( ) 12( ) 3. Solve by addition. 8x – 9y= 36 step 2 make x opposites, multiply second equation by 4. -8x + 4y= 24 xx -2x – y = 3 + y= 6

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4( ) 3. Solve by addition. 12( ) 8x – 9y= 36 -8x + 4y= 24 -5y = 60 step 3 add to eliminate the x. xx -2x – y = 3 + y= 6

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4( ) 12( ) 3. Solve by addition. 8x – 9y= 36 -8x + 4y= 24 -5y = 60 -5 -5 step 4 solve for y. y = - 12 xx -2x – y = 3 + y= 6

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4( ) 12( ) 3. Solve by addition. 8x – 9y= 36 -8x + 4y= 24 -5y = 60 -5 step 5 substitute into equation 2 and solve for x. y = - 12 -2x +12 +12 -2x = 18 -2 -2 + ( - 12)= 6 x = - 9 ( - 9, - 12) xx -2x – y = 3 + y= 6

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( ) ( ) 10 10 0.2x 0.5x Before beginning with the steps remove the decimals by multiplying 10 to each term in each equation. 4. Solve by addition. 2x + 3y= - 1 step 1 coefficients of one variable must be opposite. 5x – y= - 11 + 0.3y= - 0.1 – 0.1y= - 1.1

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3( ) 4. Solve by addition. 2x + 3y= - 1 step 2 make y opposites, multiply second equation by 3. 5x – y= - 11 2x + 3y = - 1 15x – 3y= - 33

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3( ) 4. Solve by addition. 2x + 3y= - 1 5x – y= - 11 2x + 3y = - 1 15x – 3y= - 33 17x = - 34 step 3 add to eliminate the y.

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3( ) 4. Solve by addition. 2x + 3y= - 1 5x – y= - 11 2x + 3y = - 1 15x – 3y= - 33 17x = - 34 17 17 step 4 solve for x. x = - 2

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3( ) 4. Solve by addition. 2x + 3y= - 1 5x – y= - 11 2x + 3y = - 1 15x – 3y= - 33 17x = - 34 17 x = - 2 step 5 substitute into equation 1 and solve for y. 2( - 2) +4 +4 3y = 3 3 3 + 3y= - 1 y = 1 ( - 2, 1) - 4 + 3y = - 1

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Solving a System of Equations in Two Variables By Elimination Chapter 8.3

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