Solve Systems of Equations by the Addition/Elimination Method

Transcription

1 Solve Systems of Equations by the Addition/Elimination Method When solving systems, we have found that graphing is very limited. We then considered the substitution method which has its limitations as well. The substitution method works well when one equation is an x= or y= but can become rather involved when both equations are in standard form, especially if there is no lone variable. When both equations are in standard form, the addition/elimination method is very useful. When using the addition/elimination method, the name explains what we do. Basically, we add the two equations, eliminating one of the variables. In order for one of variable to eliminate, they must be exact opposites. Example 1: Solve the system: x + 3y = -5 x 3y = 7 x + 3y = -5 x 3y = 7 Add the two equations, eliminating the y s x = Divide both sides by x = y = -5 Substitute 1 into either equation y = Divide both sides by 3 y = - The solution to this system is (1, -) To check the answer: 1 + 3(-) = = -5-5 = -5 True

2 1 3(-) = = 7 7 = 7 True Since (1, -) satisfies both equations, it is the solution to the system Example : Solve the system: 3x 4y = 8 5x + 4y = -4 3x 4y = 8 5x + 4y = -4 Add the two equations, eliminating the y s 8x = Divide both sides by 8 x = - 5(-) + 4y = -4 Substitute - into either equation y = Add 10 to both sides 4y = Divide both sides by 4 y = - 7 The solution to this system is (-, - 7 ) To check the answer: 3(-) 4(- 7 ) = = 8 8 = 8 True 5(-) + 4(- 7 ) = = -4-4 = -4 True Since (-, - 7 ) satisfies both equations, it is the solution to the system.

3 In examples 1 and, one variable term was an exact opposite of the other; 3y and - 3y in example 1, and 4y and -4y in example. Generally, though, we won t have exact opposites. We will have to manipulate one equation, or both equations, by multiplying by some value(s) to get exact opposites. Example 3: Solve the system: -6x + 5y = x + 3y = 3(x + 3y) = (3) Multiply the second equation by 3 6x + 9y = 6 to get +6y to get exact opposites -6x + 5y = 6x + 9y = 6 Add the new equation to first equation 14y = Divide both sides by 14 y = -6x + 5() = Substitute into the first equation -6x + 10 = Subtract 10 from both sides -6x = Divide both sides by -6 x = - The solution to this system is (-, ) To check the answer: -6(-) + 5() = = = True (-) + 3() = = = True Since (-, ) satisfies both equations, it is the solution to the system

4 Example 4: Solve the system: 3x + 6y = -9 x + 9y = -6 (3x + 6y) = -9() Multiply the first equation by to get +6x 6x + 1y = -18-3(x + 9y) = -6(-3) Multiply the second equation by -3 to get -6x -6x 7y = 78 6x + 1y = -18-6x 7y = 78 Add the resulting equations -15y = Divide both sides by -15 y = -4 3x + 6(-4) = -9 Substitute -4 into either original equation 3x 4 = Add 4 to both sides 3x = Divide both sides by 3 x = 5 The solution to this system is (5, -4) To check the answer: 3(5) + 6(-4) = = -9-9 = -9 True (5) + 9(-4) = = -6-6 = -6 True Since (5, -4) satisfies both equations, it is the solution to the system

5 Example 5: Solve the system: 3(x + y) = 8x 7 y = 11 5x First we need to get both equations in standard form: 3(x + y) = 8x 7 3x + 6y = 8x 7 Distribute 3-8x -8x Subtract 8x from both sides -5x + 6y = -7 y = 11 5x + 5x + 5x Add 5x to both sides 5x + y = 11-5x + 6y = -7 5x + y = 11 Add the two new equations 8y = Divide both sides by 8 y = 1 5x + ( 1 ) = 11 Substitute 1 into either equation 5x + 1 = Subtract 1 from both sides 5x = Divide both sides by 5 x = The solution to this system is (, 1 ) To check the answer: 3( + ( 1 )) = 8() 7 3( + 1) = (3) = 9 9 = 9 True ( 1 ) = 11 5() 1 = = 1 True

6 Since (, 1 ) satisfies both equations, it is the solution to the system Example 6: Solve the system: 3 x y = 1 4 x 3 4 y = 6 3( x + 1 y) = (3) Multiply by the LCD to clear the fractions 3 3 x + y = 6 4( 1 x 3 y) = 6(4) Multiply by the LCD to clear the fractions 4 4 x 3y = 4 x + y = 6 x 3y = 4 Now use the resulting equations to solve the system 3(x + y) = 6(3) Multiply the first equation by 3 to get +6x 6x + 3y = 18 6x + 3y = 18 x 3y = 4 Add the two equations 7x = Divide both sides by 7 x = 6 (6) + 1 y = Substitute 6 into either equation y = Subtract 4 from both sides 1 3 y = - (3) 1 y = -(3) 3 Multiply both sides by 3 y = -6 The solution to this system is (6, -6)

7 To check the answer: 3 (6) (-6) = 4 = = True 1 4 (6) 3 4 (-6) = = 6 1 = 6 6 = 6 True Since (6, -6) satisfies both equations, it is the solution to the system Just as with substitution, if the variables all disappear from the problem, a true statement will indicate infinite solutions and a false statement will indicate no solution. Example 5: Solve the system: x 5y = 3-6x + 15y = -9 3(x 5y) = 3(3) Multiply the first equation by 3 6x 15y = 9 6x 15y = 9-6x + 15y = -9 Add the new equation to the second equation 0 = 0 This is a true statement, so there are infinite solutions, meaning the two equations are equivalent and represent the same line.

8 Example 6: Solve the system: 1x 18y = 4 6x 9y = 15 -(6x 9y) = 15(-) Multiply the second equation by - -1x + 18y = 30 1x 18y = 4-1x + 18y = 30 Add the new equation to the first equation 0 = 54 This is a false statement, so there is no solution, meaning the two equations represent two different parallel lines.