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Example 1: -5x – 6y = 8 5x + 2y = 4 -4y = 12 y = -3 (2, -3) Example 5: -5x – 6y = 8 -5x – 6(-3) = 8 -5x + 18 = 8 -5x = -10 x=2 Example 2: 6x – 4y = 14 -3x + 4y = 1 3x = 15 x=5 6x – 4y = 14 6(5) – 4y = 14 30 – 4y = 14 -4y = -16 y=4 3x – 2y = 3 (-x + y = 1 ) x 3x – 2y = 3 3x – 2(6) = 3 3x – 12 = 3 3x = 15 x=5 Example 6: (2x + 5y = 3 )x-2 3x + 10y = -3 2x + 5y = 3 2(9) + 5y = 3 18 + 5y = 3 5y = -15 y = -3 (5,4) Example 3: 4x + 3y = 2 5x + 3y = -2 -1x = 4 x = -4 4(-4) + 3y = 2 -16 + 3y = 2 3y = 18 y=6 (4x + 5y = 35 )x3 (-3x + 2y = -9 )x4 4x + 5y = 35 4x + 5(3) = 35 4x + 15 = 35 4x = 20 x=5 Example 4: 6x = 24 x=4 (4,5) 1. 2. 3. 4. 5. 5x + 6y = 50 5(4) + 6y = 50 20 + 6y = 50 6y = 30 y=5 y=6 (5,6) -4x -10y = -6 3x + 10y = -3 -1x = -9 x=9 (9,-3) Example 7: (-4,6) 5x + 6y = 50 -x + 6y = 26 3 3x – 2y = 3 -3x + 3y = 3 12x + 15y = 105 -12x + 8y = -36 23y = 69 y=3 (5,3) Summary: Write both equations in Standard Form (Ax + By = C) If necessary, multiply one (or both) of the equations by an integer, so one of the variables will eliminate. Add the columns (like terms) Solve the equation for the variable that remains Substitute the variable (from step 4) into the other equation to solve for the other variable. Solving Linear Systems using Elimination © Lisa Davenport 2014 Directions Print pages 1 & 2 (3 & 4 for the answer key). On my printer, I use the option to print double sided and to flip along the short edge. Once photocopied have students fold down the top portion so the rectangular tabs are fully visible. They should lie just above the title “Solving Linear Functions using Graphing”. The final product should look like this: