1. Elimination using Multiplication Honors Math – Grade 8

2. 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.

3. 1 3x + 4y=-2 3(2) + 4y=-2 6 + 4y=-2 4y=-8 y=-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 “–y” and the second equation has “+4y,” multiply the first equation by 4. The solution is (2, -2) + The y variable is eliminated because -4 + 4 = 0 Solve the equation x = 2 2. Since the coefficients of y are 4 and -4 (opposites), add the equations together. 1. Write the equations in column form and multiply. 4() Distribute on both sides of = 3. Now substitute x = 2 in either equation and solve.

4. 2 6x - 2y=10 6x – 2(4)=10 6x – 8=10 6x=18 x=3 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. The solution is (3, 4) - The x variable is eliminated because 6 - 6 = 0 Solve the equation y = 4 2. Since the coefficients of x are 6 and 6 (the same), subtract the equations. 1. Write the equations in column form and multiply. 2( ) Distribute on both sides of = 3. Now substitute y = 4 in either equation and solve.

5. 3 3x + 4y=-25 3x + 4(-4)=-25 3x – 16=-25 3x=-9 x=-3 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. The solution is (-3, -4) - The x variable is eliminated because 6 - 6 = 0 Solve the equation y = -4 2. Since the coefficients of x are 6 and 6 (the same), subtract the equations. 1. Write the equations in column form and multiply. 3( ) 3. Now substitute y = -4 in either equation and solve. Method 1 Eliminate x To eliminate x, think what is the LCM of 2 and 3? 2( ) Distribute on both sides of = 6

6. 3 3x + 4y=-25 3(-3) + 4y=-25 -9 + 4y=-25 4y=-16 y=-4 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. The solution is (-3, -4) + The y variable is eliminated because 12 + -12 = 0 Solve the equation x = -3 2. Since the coefficients of y are 12 and -12 (opposites), add the equations. 1. Write the equations in column form and multiply. 4( ) 3. Now substitute x = -3 in either equation and solve. Method 2 Eliminate y To eliminate y, think what is the LCM of 4 and 3? 3( ) Distribute on both sides of = 12

7. 4 6a + 2b=2 6a + 2(4)=2 6a + 8=2 6a=-6 a= 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. The solution is (-1, 4) - The a variable is eliminated because 12 - 12 = 0 Solve the equation b = 4 2. Since the coefficients of a are 12 and 12 (the same), subtract the equations. 1. Write the equations in column form and multiply. 3( ) 3. Now substitute b = 4 in either equation and solve. Method 1 Eliminate a To eliminate a, think what is the LCM of 6 and 4? 2( ) Distribute on both sides of = 12

8. 4 6a + 2b=2 6(-1) + 2b=2 -6 + 2b=2 2b=8 b=4 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. The solution is (-1, 4) - The b variable is eliminated because 6 – 6 = 0 Solve the equation a = -1 2. Since the coefficients of b are 6 and 6 (the same), subtract the equations. 1. Write the equations in column form and multiply. 2( ) 3. Now substitute a = -1 in either equation and solve. Method 2 Eliminate b To eliminate b, think what is the LCM of 2 and 3? 3( ) Distribute on both sides of = 6