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Elimination using Multiplication Honors Math – Grade 8 Elimination using Multiplication Some systems of equations cannot be solved simply by adding or subtracting the equations. In such cases, one or both equations must first be multiplied by a number before the system can be solved by elimination. 2x y 6 3x 4 y 2 2x y 6 3x 4 y 2 1 Adding or Subtracting the equations will not eliminate a variable. Therefore, we must multiply one or both equations to change the coefficients to enable elimination. Since the first equation has “–y” and the second equation has “+4y,” multiply the first equation by 4. 1. Write the equations in column form and multiply. 4(2 x y6) 3 x 4 y 2 Distribute on both sides of = 8 x 4 y 24 +3x 4 y 2 The y variable is eliminated because -4 + 4 = 0 11x 22 Solve the equation 2. Since the coefficients of y are 4 and -4 (opposites), add the equations together. x=2 3. Now substitute x = 2 in either equation and solve. The solution is (2, -2) 3x + 4y = -2 3(2) + 4y = -2 6 + 4y = -2 4y = -8 y = -2 6 x 2 y 10 3x 7 y 19 2 Adding or Subtracting the equations will not eliminate a variable. Therefore, we must multiply one or both equations to change the coefficients to enable elimination. Since the first equation has “6x” and the second equation has “3x,” multiply the second equation by 2. 1. Write the equations in column form and multiply. 6 x 2 y 10 2(3 x 7 y 19) Distribute on both sides of = 6 x 2 y 10 - 6 x 14 y 38 2. Since the coefficients of x are 6 and 6 (the same), subtract the equations. 12 y 48 y=4 3. Now substitute y = 4 in either equation and solve. The solution is (3, 4) The x variable is eliminated because 6 - 6 = 0 Solve the equation 6x - 2y = 10 6x – 2(4) = 10 6x – 8 = 10 6x = 18 x = 3 3 Method 1 Eliminate x 3x 4 y 25 2 x 3 y 6 Adding or Subtracting the equations will not eliminate a variable. Therefore, we must multiply one or both equations to change the coefficients to enable elimination. To eliminate x, think what is the LCM of 2 and 3? 1. Write the equations in column form and multiply. 2(3 x 4 y 25) 3(2 x 3 y 6 ) 2. Since the coefficients of x are 6 and 6 (the same), subtract the equations. - 6 x 8 y 50 6 x 9 y 18 Distribute on both sides of = 17 y 68 y = -4 3. Now substitute y = -4 in either equation and solve. The solution is (-3, -4) 6 The x variable is eliminated because 6 - 6 = 0 Solve the equation 3x + 4y = -25 3x + 4(-4) = -25 3x – 16 = -25 3x = -9 x = -3 3 Method 2 Eliminate y 3x 4 y 25 2 x 3 y 6 Adding or Subtracting the equations will not eliminate a variable. Therefore, we must multiply one or both equations to change the coefficients to enable elimination. To eliminate y, think what is the LCM of 4 and 3? 1. Write the equations in column form and multiply. 9 x 12 y 75 +8 x 12 y 24 3(3 x 4 y 25) 4(2 x 3 y 6 ) 2. Since the coefficients of y are 12 and -12 (opposites), add the equations. Distribute on both sides of = 17 x 51 x = -3 3. Now substitute x = -3 in either equation and solve. The solution is (-3, -4) 12 The y variable is eliminated because 12 + -12 = 0 Solve the equation 3x + 4y = -25 3(-3) + 4y = -25 -9 + 4y = -25 4y = -16 y = -4 4 Method 1 Eliminate a 6a 2b 2 4a 3b 8 Adding or Subtracting the equations will not eliminate a variable. Therefore, we must multiply one or both equations to change the coefficients to enable elimination. To eliminate a, think what is the LCM of 6 and 4? 1. Write the equations in column form and multiply. 2( 6a 2b 2) 3( 4a 3b 8 ) 2. Since the coefficients of a are 12 and 12 (the same), subtract the equations. Distribute on both sides of = 12a 4b 4 12a 9b 24 5b 20 b=4 3. Now substitute b = 4 in either equation and solve. The solution is (-1, 4) 12 The a variable is eliminated because 12 - 12 = 0 Solve the equation 6a + 2b = 2 6a + 2(4) = 2 6a + 8 = 2 6a = -6 a = -1 4 Method 2 Eliminate b 6a 2b 2 4a 3b 8 Adding or Subtracting the equations will not eliminate a variable. Therefore, we must multiply one or both equations to change the coefficients to enable elimination. To eliminate b, think what is the LCM of 2 and 3? 1. Write the equations in column form and multiply. 3( 6a 2b 2) 2( 4a 3b 8 ) 2. Since the coefficients of b are 6 and 6 (the same), subtract the equations. Distribute on both sides of = 18a 6b 6 8a 6b 16 10a 10 a = -1 3. Now substitute a = -1 in either equation and solve. The solution is (-1, 4) 6 The b variable is eliminated because 6–6 =0 Solve the equation 6a + 2b = 2 6(-1) + 2b = 2 -6 + 2b = 2 2b = 8 b = 4