1. 1 Chapter 3, Section 2 Solving Systems of Equations Algebraically

2. 2 Three Algebraic Methods Substitution: Solve one equation for one variable, then SUBSTITUTE that value into THE OTHER equation. Works best with systems when a coefficient of a variable is 1 or -1. Elimination: Add or subtract one equation to or from the other to ELIMINATE one of the variables. Works only when the absolute values of the coefficients of either variable are equal. Elimination by Multiplication: The absolute values of the coefficients of either variable are MADE equal BY MULTIPLICATION, then the equations are added or subtracted to ELIMINATE one of the variables. With ALL THREE of these methods, once a value for one variable is found, it can be substituted into ANY two-variable equation to find the value for the other variable.

3. 3 Special Cases When using these methods, you may run into two special cases:  You may get an expression that is ALWAYS true, e.g. 4 = 4, and that indicates a consistent, dependent set of solutions.  You may get an expression that is NEVER true, e.g. 9 = 0, and that indicates an inconsistent set of solutions.

4. 4 Example 1 Solve the system of equations using substitution: x + 4y = 26 x - 5y = -10 We see that a coefficient of one of the variables is 1, so substitution is a wise choice. First EquationSecond Equation x + 4y = 26 x - 5y = -10 substitute → (26 - 4y) - 5y = -10 26 - 4y - 5y = -10 26 - 9y = -10 36 = 9y 4 = y x = 26 - 16 x = 10 Solution: (10,4) -- you should verify that it solves both equations! x = 26 - 4y x = 26 - 4(4)←substitute

5. 5 Alg 1 Review: How Elimination Works…. You’ve added and subtracted numbers before, but you can also add and subtract equations. Watch: 3 + 5 = 87 + 4 = 11 2 + 11 = 132 + 5 = 7 5 + 16 = 215 - 1 = 4 After you get over you initial shock, you may say, “So?” (+)(-)

6. 6 Alg 1 Review: How Elimination Works (continued) So, let’s add these two equations! -3x + 4y = 12 3x - 6y = 18 -2y = 30  y = -15 AMAZING!! By adding these two equations, we got rid of one of the variables!! Now, let’s substitute that value for y into EITHER equation. (I’ll show that it doesn’t matter which you choose by substituting it into BOTH.) -3x + 4(-15) = 123x - 6(-15) = 18 -3x - 60 = 123x + 90 = 18 -3x = 723x = -72 x = -24 x = -24 So, (-24,-15) is the solution to this system of equations. (+)

7. 7 Example 2 Solve the system of equations using elimination: x + 2y = 10 x - y = 4 Since we see the absolute values of the x coefficients is the same, elimination will be the preferred method. (Substitution would work, too, though not as effortlessly.) Since the coefficients are both positive we will subtract the second equation from the first: x + 2y = 10 (-) x - y = 4Careful--subtracting integers!! 3y = 6 y = 2 Now that we have a value for y, we’ll substitute it into the 2 nd equation: x - y = 4 x - 2 = 4 x = 6 The solution to this system is (6,2). Verify that it solves both equations.

8. 8 Example 3 Use elimination by multiplication to solve the system of equations: 2x + 3y = 12 5x - 2y = 11 Because of the coefficients, substitution is not the easy choice it once was, and straight elimination won’t work either. Let’s make the absolute values of the y coefficients the same -- the LCM of 2 and 3: 2(2x + 3y = 12) 3(5x - 2y = 11) 19x = 57 x = 3 We can now substitute x = 3 into either of the original equation and find that y = 2. Verify that (3,2) solves both equations. 4x + 6y = 24 15x - 6y = 33 (+)

9. 9 Example 4 Solve the system of equations by elimination: -3x + 5y = 12 6x - 10y = -21 Let’s try making the absolute values of the coefficients of y the same: 2(-3x + 5y = 12) 6x - 10y = -21 0 = 3 Remember what the special cases were: When you came up with an equation that cannot be true, that means the solutions to this system are inconsistent. There is no solution to this system of equations. -6x + 10y = 24 6x - 10y = -21 (+)