2. Adding two numbers together which have the same absolute value but are opposite in sign results in a value of zero. This same principle can be applied to variables. Below is a zero pair for the variable, “x”. Each box has the same absolute value but is opposite in sign. 1 Each pair, then makes a zero pair.0 = 0 x -x

3. We can apply the idea of a zero pair to a system of equations so that one of the variables can be eliminated. Zero pair The + y and – y make a zero pair, we have ___________ signs and we can eliminate the variable y by ___________ the equations. Look at the following system of equations. What do you observe? Opposite Adding

4. Add the following two equations together by adding like terms together. ….Now solve for “x”. 2x = 8 2 Now pick one of the two equations and substitute the value of “x” into that equation and solve for “y”. (1) x + y = 6 4 + y = 6 Write the solution as an ordered pair. Solution: (4,2) x = 4 y = 2 +

5. Solution : (4,2) To check your work, substitute the values for the variables into each equation and determine if it is true. Since LS = RS for both equations, therefore the solution is correct. RS = 6 LS = RS  RS = 2 LS = RS  LS = 6 LS = 4 + 2 LS = 4 - 2 LS = 2

6.  Using addition worked when the signs of the variables were opposite. Let’s take a look at the following system of equations:  However, in a system of equations in which the signs of the variables are the same, instead of adding, you will SUBTRACT one equation from the other. SUBTRACT The + y and + y make a zero pair, we have ___________ signs and we can eliminate the variable y by ___________ the equations. Same Subtracting

7. Subtract the bottom equation from the top equation. - Since we have the value for “x”. Pick one of the equations and substitute the value 3 in place of “x” and solve for “y”. (1) 3x + y = 6 y = -3 Write the solution as an ordered pair. Solution: (3,-3) (3)(3) + y = 6 9 + y = 6 y = 6 - 9

8. Solution: (3,-3) To check your work, substitute the solution values in place of the variables in each equation. Since LS = RS for both equations, therefore the solution is correct. 

9. Example 3 : Solve the linear system by elimination. 2x + 3y = 2 (1) 5x – y = 22 (2) x 3 15x – 3y = 66 (2) 2x + 3y = 2 (1) The signs are the same so we 17x = 68 x = 4 Sub x = 4 in (1) 2(4) + 3y = 2 8 + 3y = 2 3y = – 6 y = – 2 Solution: (4, – 2) Notice the coefficients don’t match as they are but we can make them match by multiplying 1 (or both) of the equations by numbers so that they will match. We can make the coefficients of either x or y match, choose y here and modify the 2 nd equation by multiplying by ADD + 3

10. Example 4: Solve the linear system by elimination. 3x + 4y = 11 (1) 5x + 7y = 20 (2) – 15x – 20y = – 55 (1) 15x + 21y = 60 (2) y = 5 Sub in (1) 3x + 4(5) = 11 3x + 20 = 11 3x = – 9 x = – 3 Sol: (– 3, 5) Here we must change both equations to make the coefficients match. If we choose to eliminate the x’s, we must multiply (1) by ___ and (2) by ___ -5 3 x -5 x 3 The signs are the opposite so we _____ADD +

11. 1. Arrange the two equations so that the like terms are in vertical columns. 2. If the signs of the variables are opposite, add the two equations together to eliminate one of the variables. 3. If the signs of the variables are the same, then subtract one of the equations from the other equation. 5. Solve for the variable. 6. Substitute the value of this variable into one of the equations and solve for the other variable. 7. Check the solution by substituting the values of the two variables into the other ORIGINAL equation. 8. Write the solution. If there is one solution, write it as an ordered pair. If neither variable has the same coefficient, multiply one or each equation by the number that will give the same coefficient for one of the variables (then ADD or SUBTRACT) 4.