1. Slide 7- 1 Copyright © 2012 Pearson Education, Inc.

2. 3.2 Solving by Substitution or Elimination ■ The Substitution Method ■ The Elimination Method

3. Slide 3- 3 Copyright © 2012 Pearson Education, Inc. The Substitution Method Graphing can be an imprecise method for solving systems of equations. We are now going to look at ways of finding exact solutions using algebra. One method for solving systems is known as the substitution method. It uses algebra instead of graphing and is thus considered an algebraic method.

4. Slide 3- 4 Copyright © 2012 Pearson Education, Inc. Example Solve the system Solution The second equation says that y and  2x + 5 represent the same value. Thus, in the first equation we can substitute  2x + 5 for y. 3x  2y = 4 3x  2(  2x + 5) = 4 3x + 4x  10 = 4 7x  10 = 4 7x = 14 x = 2 Equation (1) Substituting Using the distributive property Combining like terms Adding 10 to both sides Dividing both sides by 7

5. Slide 3- 5 Copyright © 2012 Pearson Education, Inc. continued We have found the x-value of the solution. To find the y-value, we return to the original pair of equations. Substituting into either equation will give us the y- value. We choose equation (2): y =  2x + 5 y =  2(2) + 5 y =  4 + 5 y = 1 The ordered pair (2, 1) appears to be the solution. We check: Equation (2) Substituting for x Simplifying

6. Slide 3- 6 Copyright © 2012 Pearson Education, Inc. continued check (2, 1) 3x  2y = 4 y =  2x + 5 3(2)  2(1) 4 1  2(2) + 5 6  2 1  4 + 5 4 = 4 True 1 = 1 True Since (2, 1) checks, it is a solution. CAUTION! A solution of a system of equations in two variables is an ordered pair of numbers. Once you have solved for one variable, don’t forget the other. A common mistake is to solve for only one variable.

7. Slide 3- 7 Copyright © 2012 Pearson Education, Inc. Example Solve: x = 3  y (1) 5x + 3y = 5 (2) Solution We substitute x = 3  y for x in the second equation: 5x + 3y = 5 5(3  y) + 3y = 5 15  5y + 3y = 5 15  2y = 5  2y =  10 y = 5 Next, we substitute 5 for y in equation (1) of the original system: Equation (2) Substituting Using the distributive law Solving for y

8. Slide 3- 8 Copyright © 2012 Pearson Education, Inc. continued find x, y = 5 x = 3  y x = 3  5 x =  2 We check the ordered pair (  2, 5). x = 3  y 5x + 3y = 5  2 = 3  5 5(  2)+3(5) = 5  2 =  2  10 + 15 = 5 The pair (  2, 5) is the solution. The graph is shown at the right as another check. (  2, 5) x = 3  y 5x + 3y = 5

9. Slide 3- 9 Copyright © 2012 Pearson Education, Inc. Solving for the Variable First Sometimes neither equation has a variable alone on one side. In that case, we solve one equation for one of the variables and then proceed as before. Example Solve: We can solve either equation for either variable. Since the coefficient of x is one in equation (1), it is easier to solve that equation for x.

10. Slide 3- 10 Copyright © 2012 Pearson Education, Inc. continued We substitute x = 6  y for x in equation (2) of the original pair and solve for y: Remember to use parentheses when you substitute. Equation (2) Substituting Using the distributive law Combining like terms Subtracting 30 from both sides Dividing both sides by  3

11. Slide 3- 11 Copyright © 2012 Pearson Education, Inc. continued To find x we substitute 22/3 for y in equation (1), (2), or (3). Because it is generally easier to use an equation that has already been solved for a specific variable, we decide to use equation (3): If you check the ordered pair you will find it is a solution of the system. The solution is

12. Slide 3- 12 Copyright © 2012 Pearson Education, Inc. Example Solve Solution We solved this system graphically. The lines are parallel and the system has no solution. Let’s see what happens if we try to solve this system by substitution. We obtain a false equation. We state that the system has no solution and thus is inconsistent. Equation (1) Substituting for y Subtracting 3x/4 from both sides

13. Slide 3- 13 Copyright © 2012 Pearson Education, Inc. Example Solve Solution We solved this system earlier. The lines coincide, so the system has an infinite number of solutions. If we use substitution to solve the system, we can use y = 2  x/2 for y in equation (1). This last equation is true for any choice of x. When the solution leads to an equation that is true for all real numbers, we state that the system has an infinite number of solutions. Equation (1) Substituting for y

14. Slide 3- 14 Copyright © 2012 Pearson Education, Inc. Solving by the Elimination Method The elimination method for solving systems of equations makes use of the addition principle. Example Solve the system According to equation (2), x  3y and 7 are the same answer. Thus we can add x  3y to the left side of the equation(1) and 7 to the right side:

15. Slide 3- 15 Copyright © 2012 Pearson Education, Inc. continued Add The resulting equation has just one variable:5x = 15 Dividing both sides by 5, we find that x = 3. Next, we substitute 3 for x in either of the original equations: 4x + 3y = 8 4(3) + 3y = 8 12 + 3y = 8 3y =  4 y =  4/3 Next, we check the ordered pair (3,  4/3).

16. Slide 3- 16 Copyright © 2012 Pearson Education, Inc. continued Check: 4x + 3y = 8 x  3y = 7 4(3) + 3(  4/3) 8 3  3(  4/3) 7 12  4 3 + 4 8 = 8 7 = 7True The graph shown to the right also serves as a check. Since (3,  4/3) checks in both equations, it is the solution.

17. Slide 3- 17 Copyright © 2012 Pearson Education, Inc. Example Solve Solution Adding the two equations as they appear will not eliminate a variable. However, if the 3y were  3y in one equation, we could eliminate y. We multiply both sides of equation (2) by  1 to find an equivalent equation and then add:

18. Slide 3- 18 Copyright © 2012 Pearson Education, Inc. continued Next, we substitute 2 for x in either of the original equations: We can check the ordered pair (2, 3). The graph shown at the right is also a check. (2, 3) checks in both equations, it is a solution

19. Slide 3- 19 Copyright © 2012 Pearson Education, Inc. When deciding which variable to eliminate, we inspect the coefficients in both equations. If one coefficient is a multiple of the coefficient of the same variable in the other equation, that is the easiest variable to eliminate. Example Solve: Solution: No terms are opposites, but if both sides of equation (1) are multiplied by 2, the coefficients of y will be opposites.

20. Slide 3- 20 Copyright © 2012 Pearson Education, Inc. continued We then substitute 2 for x in either equation (1) or (2): Adding Solving for x Equation (1) Substituting and solving for y We leave it to you to confirm that (2,  1) checks and is the solution.

21. Slide 3- 21 Copyright © 2012 Pearson Education, Inc. Sometimes both equations must be multiplied to find the least common multiple of two coefficients. Example Solve: Solution: It is often helpful to write both equations in Standard form before attempting to eliminate a variable: Since neither coefficient of x is a multiple of the other and neither coefficient of y is a multiple of the other, we use the multiplication principle with both equations. We can eliminate the x term by multiplying both sides of equation (3) by 3 and both sides of equation (4) by  2.

22. Slide 3- 22 Copyright © 2012 Pearson Education, Inc. continued We substitute  2 for y in equation (3): The solution (  2,  2) checks. Substituting in equation (3) Multiplying both sides of equation (3) by 3 Multiplying both sides of equation (4) by  2

23. Slide 3- 23 Copyright © 2012 Pearson Education, Inc. No Solution Example Solve: Solution: To eliminate y we multiply equation (2) by  1. Then we add: Note that in eliminating y, we eliminated x as well. The resulting equation 0 = 4, is false for any pair (x, y), so there is no solution.

24. Slide 3- 24 Copyright © 2012 Pearson Education, Inc. Infinite Number of Solutions Example Solve: Solution: To eliminate x, we multiply both sides of equation (1) by  3 and then add the two equations: Again, we have eliminated both variables. The resulting equation, 0 = 0, is always true, indicating that the equations are dependent.

25. Slide 3- 25 Copyright © 2012 Pearson Education, Inc. Decimals or Fractions Example Solve: Solution: The number 5 is the LCD for equation (1). Thus we multiply both sides of equation (1) by 5 to clear fractions:

26. Slide 3- 26 Copyright © 2012 Pearson Education, Inc. continued The resulting system is: Solving the system: Substitute 4 for y in equation (2) to find x. The solution of this system is (5, 4). Multiply equation (2) by  1, then adding

27. Slide 3- 27 Copyright © 2012 Pearson Education, Inc. We now have three distinctly different ways to solve a system. Each method has certain strengths and weaknesses. MethodStrengthsWeaknesses GraphicalSolutions are displayed visually. Works with any system that can be graphed. Inexact when solutions involve numbers that are not integers or are very large and off the graph. SubstitutionAlways yields exact solutions. Easy to use when a variable is alone on one side of an equation. Introduces extensive computations with fractions when solving more complicated systems. Solutions are not graphically displayed. EliminationAlways yields exact solutions. Easy to use when fractions or decimals appear in the system. The preferred method for systems of 3 or more variables (see section 3.4) Solutions are not graphically displayed.