1. U SING A LGEBRAIC M ETHODS TO S OLVE S YSTEMS In this lesson you will study two algebraic methods for solving linear systems. The first method is called substitution. THE SUBSTITUTION METHOD 1 2 3 Substitute value from Step 2 into revised equation from Step 1. Solve. Solve one of the equations for one of its variables. Substitute expression from Step 1 into other equation and solve for other variable.

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

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

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

5. U SING A LGEBRAIC M ETHODS TO S OLVE S YSTEMS 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. 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.

6. THE LINEAR COMBINATION METHOD U SING A LGEBRAIC M ETHODS TO S OLVE S YSTEMS 1 2 3 Substitute value obtained in Step 2 into either original equation and solve for other variable. Multiply one or both equations by a constant to obtain coefficients that d iffer only in sign for one of the variables. Add revised eq uations from Step 1. Combine like terms to eliminate one of the variables. Solve for remaining variable.

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

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

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

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

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

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

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