THE SUBSTITUTION METHOD USING ALGEBRAIC METHODS TO SOLVE SYSTEMS In this lesson you will study two algebraic methods for solving linear systems. The first method is called substitution. THE SUBSTITUTION METHOD 1 Solve one of the equations for one of its variables. 2 Substitute expression from Step 1 into other equation and solve for other variable. 3 Substitute value from Step 2 into revised equation from Step 1. Solve.

y  5 3 x + 4y  – 4 Equation 1 x + 2y  2 x  – 2y + 2 3x + 4y  – 4 The Substitution Method Solve the linear system using the substitution method. 3 x + 4y  – 4 Equation 1 x + 2y  2 Equation 2 SOLUTION Solve Equation 2 for x. x + 2y  2 Write Equation 2. x  – 2y + 2 Revised Equation 2. Substitute the expression for x into Equation 1 and solve for y. 3x + 4y  – 4 Write Equation 1. 3(– 2y + 2) + 4y  – 4 Substitute – 2y + 2 for x. y  5 Simplify.

3 x + 4y  – 4 Equation 1 3x + 4y  – 4 3(– 2y + 2) + 4y  – 4 y  5 The Substitution Method Solve the linear system using the substitution method. 3 x + 4y  – 4 Equation 1 x + 2y  2 Equation 2 3x + 4y  – 4 3(– 2y + 2) + 4y  – 4 y  5 Substitute – 2y + 2 for x. Write Equation 1. Simplify. Substitute the expression for x into Equation 1 and solve for y. Substitute the value of y into revised Equation 2 and solve for x. x  – 2y + 2 Write revised Equation 2. x  – 2(5) + 2 Substitute 5 for y. x  – 8 Simplify. The solution is (– 8, 5).

3 x + 4y  – 4 Equation 1 3x + 4y  – 4 x + 2y  2 The Substitution Method Solve the linear system using the substitution method. 3 x + 4y  – 4 Equation 1 x + 2y  2 Equation 2 CHECK Check the solution by substituting back into the original equation. 3x + 4y  – 4 Write original equations. x + 2y  2 3 (– 8) + 4 (5)  – 4 ? – 8 + 2 (5)  2 ? Substitute x and y. – 4  – 4 Solution checks. 2  2

USING ALGEBRAIC METHODS TO SOLVE SYSTEMS CHOOSING A METHOD In the first step of the previous example, you could have solved for either x or y in either Equation 1 or Equation 2. It was easiest to solve for x in Equation 2 because the x-coefficient was 1. In general you should solve for a variable whose coefficient is 1 or –1. If neither variable has a coefficient of 1 or –1, you can still use substitution. In such cases, however, the linear combination method may be better. The goal of this method is to add the equations to obtain an equation in one variable.

THE LINEAR COMBINATION METHOD USING ALGEBRAIC METHODS TO SOLVE SYSTEMS THE LINEAR COMBINATION METHOD 1 Multiply one or both equations by a constant to obtain coefficients that d iffer only in sign for one of the variables. 2 Add revised eq uations from Step 1. Combine like terms to eliminate one of the variables. Solve for remaining variable. 3 Substitute value obtained in Step 2 into either original equation and solve for other variable.

Solve the linear system using the linear combination method. The Linear Combination Method: Multiplying One Equation Solve the linear system using the linear combination method. 2 x – 4y  13 Equation 1 4 x – 5y  8 Equation 2 SOLUTION Multiply the first equation by – 2 so that x-coefficients differ only in sign. 2 x – 4y  13 • – 2 – 4x + 8y  – 26 4 x – 5y  8 4 x – 5y  8 3y  –18 Add the revised equations and solve for y. y  – 6

( ) Solve the linear system using the linear combination method. The Linear Combination Method: Multiplying One Equation Solve the linear system using the linear combination method. 2 x – 4y  13 Equation 1 4 x – 5y  8 Equation 2 y  – 6 Add the revised equations and solve for y. Substitute the value of y into one of the original equations. You can check the solution algebraically using the method shown in the previous example. CHECK 2 x – 4y  13 Write Equation 1. 2 x – 4(– 6)  13 Substitute – 6 for y. 2 x + 24  13 Simplify. x  – 11 2 Solve for x. The solution is – , – 6 . ( 11 2 )

Solve the linear system using the linear combination method. The Linear Combination Method: Multiplying Both Equations Solve the linear system using the linear combination method. 7 x – 12 y  – 22 Equation 1 – 5 x + 8 y  14 Equation 2 SOLUTION Multiply the first equation by 2 and the second equation by 3 so that the coefficients of y differ only in sign. 7 x – 12 y  – 22 • 2 14 x – 24y  – 44 – 5 x + 8 y  14 • 3 – 15 x + 24y  42 – x  – 2 Add the revised equations and solve for x. x  2

Check the solution algebraically or graphically. The Linear Combination Method: Multiplying Both Equations Solve the linear system using the linear combination method. 7 x – 12 y  – 22 Equation 1 – 5 x + 8 y  14 Equation 2 x  2 Add the revised equations and solve for x. Substitute the value of x into one of the original equations. Solve for y. – 5 x + 8 y  14 Write Equation 2. – 5 (2) + 8 y  14 Substitute 2 for x. y = 3 Solve for y. The solution is (2, 3). Check the solution algebraically or graphically.

Solve the linear system Linear Systems with Many or No Solutions x – 2 y  3 2 x – 4 y  7 Solve the linear system SOLUTION Since the coefficient of x in the first equation is 1, use substitution. x – 2 y  3 Solve the first equation for x. x  2 y + 3

Solve the linear system Linear Systems with Many or No Solutions x – 2 y  3 2 x – 4 y  7 Solve the linear system x  2 y + 3 Solve the first equation for x. Substitute the expression for x into the second equation. 2 x – 4 y  7 Write second equation. 2(2 y + 3) – 4 y  7 Substitute 2 y + 3 for x. 6  7 Simplify. Because the statement 6 = 7 is never true, there is no solution.

Solve the linear system Linear Systems with Many or No Solutions 6 x – 10 y  12 – 15 x + 25 y  – 30 Solve the linear system SOLUTION Since no coefficient is 1 or –1, use the linear combination method. 6 x – 10 y  12 • 5 30 x – 50 y  60 – 15 x + 25 y  – 30 • 2 – 30 x + 50 y  – 60 Add the revised equations. 0  0 Because the equation 0 = 0 is always true, there are infinitely many solutions.