1. Chapter 8 Section 3 Solving System of Equations by the Addition Method

2. Learning Objective Solve a system of equations by the Addition Method Key Vocabulary:  Addition (or elimination) method

3. Solve Systems of Equations Addition Method 1.If necessary rewrite each equation so that the terms containing variables appear on the left side of the equal sign and the constants appear on the right side of the equal sign 2.If necessary multiply one or both equations by a constant(s) so that when the remaining equations are added the resulting sum will contain only one variable 3.Add the equations resulting in one equation 4.Solve for the variable 5.Substitute the value found into one of the original equations, and solve that equation for the other variable 6.Check

4. Example: Equation 1 x + y = 7 Equation 2 4x – y = 3 Solve Systems of Equations Addition Method Step 1: Variables are already on one side and constants are on the other side Step 2 adding will already result in one variable Step 3 Add the equations resulting in one equation 4x – y = 3 x + y = 7 5x = 10 Step 4 Solve for the variable 5x = 10 x = 10/5 x = 2 x + y = 7 2 + y = 7 y = 7 – 2 y = 5 Step 5 Substitute the value found into one of the original equations, and solve that equation for the other variable Solution is (2, 5) Consistent What do you think the ordered pair (2,5) represents on the graph?

5. Example: Equation 1 x + 2y = 2 Equation 2 x + 3y = 6 -1(x +2y = 2) -x – 2y = -2 Solve Systems of Equations Addition Method Step 1: Variables are already on one side and constants are on the other side Step 2 multiply one or both equations by a constant(s) Step 3 Add the equations resulting in one equation x + 3y = 6 -x – 2y = -2 y = 4 Step 4 Already solved for the variable x + 2y = 2 x + 2(4) = 2 x + 8 = 2 x = 2 – 8 x = -6 Step 5 Substitute the value found into one of the original equations, and solve that equation for the other variable Solution is (-6, 4) Consistent Both sides of the equation has to be multiplied by the given number.

6. Example: Equation 1 3x + y = 8 Equation 2 x + 5y = -2 -3(x + 5y = -2) -3x – 15y = 6 Solve Systems of Equations Addition Method Step 1: Variables are already on one side and constants are on the other side Step 2 multiply one or both equations by a constant(s) Step 3 Add the equations resulting in one equation 3x + y = 8 -3x – 15y = 6 -14y = 14 Step 4 Solve for the variable -14y = 14 y = -1 x + 5y = -2 x + 5(-1) = -2 x - 5 = -2 x = -2 + 5 x = 3 Step 5 Substitute the value found into one of the original equations, and solve that equation for the other variable Solution is (3, -1) Consistent Both sides of the equation has to be multiplied by the given number. I could have multiplied the first equation by -5 and got the same results.

7. Example: Equation 1 2x + 9y = 5 Equation 2 8x - 3y = -6 8(2x + 9y = 5) 16x + 72y = 40 -2(8x – 3y = -6) -16x + 6y = 12 Solve Systems of Equations Addition Method Step 1: Variables are already on one side and constants are on the other side Step 2 multiply 1 st equation by 8 and the 2 nd equation by -2 Step 3 Add the equations resulting in one equation 16x + 72y = 40 -16x + 6y = 12 78y = 52 Step 4 Solve for the variable 78y = 52 y = 52/78 (26) y = ⅔ 8x - 3y = -6 8x – 3(2/3) = -6 8x - 2 = -6 8x = -6 + 2 8x = -4 x = -4/8 x = - ½ Step 5 Substitute the value found into one of the original equations, and solve that equation for the other variable Solution is (- ½, ⅔) Consistent

8. Example: Equation 1 3x + 5y = -6 Equation 2 -2x + 7y = 4 2(3x + 5y = -6) 6x + 10y = -12 3(-2x + 7y = 4) -6x + 21y = 12 Solve Systems of Equations Addition Method Step 1: Variables are already on one side and constants are on the other side Step 2 multiply 1 st equation by 2 and the 2 nd equation by 3 Step 3 Add the equations resulting in one equation 6x + 10y = -12 -6x + 21y = 12 31y = 0 Step 4 Solve for the variable 31y = 0 y = 0/31 y = 0 3x + 5y = -6 3x – 5(0) = -6 3x = -6 3x = -6/3 x = -2 Step 5 Substitute the value found into one of the original equations, and solve that equation for the other variable Solution is (- 2, 0) Consistent

9. Example: Equation 1 3x - 2y = 1 Equation 2 -6x + 4y = 5 2(3x - 2y = 1) 6x - 4y = 2 Solve Systems of Equations Addition Method Step 1: Variables are already on one side and constants are on the other side Step 2 multiply 1 st equation by 2 and the 2 nd equation by 3 Step 3 Add the equations resulting in one equation 6x + 4y = 5 -6x – 4y = 2 0 = 7 False, No solution Inconsistent Parallel lines Slope intercept form would show us that they have the Same slope and different y-intercepts -2y = -3x + 1 y = 3/2 x – ½ 4y = 6x + 5 y = 3/2 x + 5/4

10. Example: Equation 1 y = 1/3 x + 2 -1/3 x + y = 2 Equation 2 3y – x = 6 -x + 3y = 6 -3(-1/3x + y = 2) x – 3y = -6 Solve Systems of Equations Addition Method Step 1: Variables on one side and constants are on the other side Step 2 multiply 1 st equation by 2 and the 2 nd equation by 3 Step 3 Add the equations resulting in one equation x – 3y = -6 -x + 3y = 6 0 = 0 True, same line Dependent Infinite number of solution The answer would have been clear if we had put the 2 nd equations in slope intercept form. We could then see that they have the same slope and same y-intercept 3y = x + 6 y = 1/3 x + 2

11. Example: Equation 1 4x + 5y = 3 Equation 2 2x - 3y = 1 -2(2x - 3y = 1) -4x + 6y = -2 Solve Systems of Equations Addition Method Step 1: Variables are already on one side and constants are on the other side Step 2 multiply 1 st equation by 2 and the 2 nd equation by 3 Step 3 Add the equations resulting in one equation Step 4 Solve for the variable 11y =1 y = 1/11 2x - 3y = 1 2x - 3(1/11) = 1 2x – 3/11 = 1 2x = 3/11 + 1 2x = 14/11 x = 14/11 (½) x = 14/22 x = 7/11 Step 5 Substitute the value found into one of the original equations, and solve that equation for the other variable Solution is (7/11, 1/11) Consistent 4x + 5y = 3 -4x + 6y = -2 11y = 1

12. Remember The objective is to obtain two equations who sum will be an equation containing only one variable Contemplate which variable will be easiest to eliminate and what multiplication will be needed to make the elimination possible Neatness and organization will help when solving by substitution and by addition method There is always more than one way to solve a problem The solution should be and ordered pair. Check by substituting the solution back into the original equations.

13. HOMEWORK 8.3 Page 511 - 512 #7, 9, 11, 13, 17, 19, 23, 33