Solving Systems of Equations by the Substitution and Addition Methods 7.2 Solving Systems of Equations by the Substitution and Addition Methods

Procedure for Solving a System of Equations Using the Substitution Method Solve one of the equations for one of the variables. If possible, solve for a variable with a coefficient of 1. Substitute the expression found in step 1 into the other equation. Solve the equation found in step 2 for the variable. Substitute the value found in step 3 into the equation, rewritten in step 1, and solve for the remaining variable.

Example: Substitution Method Solve the following system of equations by substitution. 5x  y = 5 4x  y = 3 Solve the first equation for y. y = 5x  5 Then we substitute 5x  5 for y in the other equation to give an equation in one variable. 4x  y = 3 4x  (5x  5) = 3 Distibute the negative 4x – 5x + 5 = 3

Example: Substitution Method continued Now solve the equation for x. 4x – 5x + 5 = 3 –x + 5 = 3 –x = –2 x = 2 Substitute x = 2 in the equation solved for y and determine the y value. y = 5x  5 y = 5(2) – 5 y = 10 – 5 y = 5 Thus, the solution is the ordered pair (2, 5).

Example: Substitution Method Solve the following system by Substitution 3x – 4y = 28 x = -2y + 6 3(-2y + 6) – 4y = 28 -6y + 18 – 4y = 28 Since bottom equation is solved for x, substitute (-2y + 6) into the first equation. -10y + 18 = 28 -18 -18 -10y = 10 y = -1

Example: Substitution Method continued 3x – 4y = 28 x = -2y + 6 Substitute y = -1 into either original equation Solution: (8, -1) x = -2(-1) +6 x = 2 + 6 x = 8

Practice problem: Solve by substitution y = 2x – 1 3x – y = -1

Addition Method If neither of the equations in a system of linear equations has a variable with the coefficient of 1, it is generally easier to solve the system by using the addition (or elimination) method. To use this method, it is necessary to obtain two equations whose sum will be a single equation containing only one variable.

Procedure for Solving a System of Equations by the Addition Method 1. If necessary, rewrite the equations so that the variables appear on one side of the equal sign and the constant appears on the other side of the equal sign. 2. If necessary, multiply one or both equations by a constant(s) so that when you add the equations, the result will be an equation containing only one variable. 3. Add the equations to obtain a single equation in one variable.

Procedure for Solving a System of Equations by the Addition Method continued 4. Solve the equation in step 3 for the variable. 5. Substitute the value found in step 4 into either of the original equations and solve for the other variable.

Example: Multiplying Both Equations Solve the system using the elimination method. 6x + 2y = 4 10x + 7y =  8 Solution: In this system, we cannot eliminate a variable by multiplying only one equation and then adding. To eliminate the variable x, we can multiply the first equation by 5 and the second equation by 3. Then we will be able to eliminate the x variable.

Continued, 10x + 7y =  8 6x + 2y = 4 30x + 10y = 20 30x  21y = 24 We can now find x by substituting 4 for y in either of the original equations. Substituting: 6x + 2y = 4 6x + 2(4) = 4 6x  8 = 4 6x = 12 x = 2 The solution is (2, 4).

Practice problem: Solve by Addition Method *hint: since y values are opposites already, you don’t need to multiply x + y = 12 x – y = 2

Practice problem: Solve by Addition Method x + 3y = 7 2x – y = 7

Homework: p. 401 # 3 – 36 (x3) Show all work