Transcript 2 - Images
SYSTEMS OF EQUATIONS Method #2- Elimination Previously, we learned how to solve systems of equations by using substitution. Why did we learn that method first? Because it’s the most difficult! Let’s learn the easiest…elimination. Example 1: 2x 3 y 12 x 3y 6 _____________ 3x = 18 x=6 This system of equations could be solved by eliminating the y variable through addition. 2(6) + 3y = 12 12 + 3y = 12 3y = 0 (6,0) y=0 Then, you would plug the x in just like you did with substitution. Example 2: 3x + 2y = 16 + 3x – y = 10 _____________ 3y = 6 y=2 3x + 2(2) = 16 3x + 4 = 16 3x = 12 x=4 (4, 2) You can solve this one by multiplying one of the equations by a negative one. Again, to finish this problem, you would plug the y back into one of the equations. Why did we put the 4 first in the parentheses? Always write your answers in alphabetical order! Example 3: Here’s one for you to practice on: - 3x –+ 2y = -14 5x – 2y = 22 _____________ 2x =8 x=4 The solution is (4, -1). 3x – 2y = 14 3(4) – 2y = 14 12 – 2y = 14 -2y = 2 y = -1 Example 4: Here’s another: 3x + 5y = 10 -3x – 5y = 2 _____________ 0 = 12 There are no solutions. Solve the following systems of equations by using elimination. 1) 2x + 4y = 18 x – 4y = -15 2) 3x – y = -3 3x + 4y = -18 3) 6x + 3y = -3 2x + 3y = -5 Answers: 1) (1, 4) 2) (-2, -3) 3) (1/2, -2) Example 5: Since this system of equations can’t be solved by elimination with addition, how can we solve it? -3(x 4 y 17) 3x 2 y 9 If the top equation was multiplied by -3, then the first term would be -3x. The bottom equation could then be added to the top equation eliminating the variable x. -3x – 12y = -51 3x – 2y = 9 -14y = -42 y=3 x + 4(3) = 17 x + 12 = 17 x=5 (5, 3) x 4 y 17 3x 2 y 9 We solved this system of equations by multiplying the top equation by 3. Could we have used a different factor for the multiplication? We could have multiplied the bottom equation by 2 to get 6x 4 y 18 The system of equations would then become x 4 y 17 Elimination by addition would then be used to solve this system of equations. 6x 4 y 18 4x 3 y 12 -4( x 2 y 14) 4x – 3y = 12 (+) -4x – 8y = - 56 11y 44 Multiply the bottom equation by -4 to get a new system of equations. Add the bottom equation to the top equation. Solve for y. Solve for x by substituting the value for y into one of the equations. 11 y 44 11 11 y4 x 2 y 14 x 2( 4 ) 14 x 8 14 x 6 (6, 4) Example 7: Since this system of equations can’t be solved by multiplying one row by a number, how can we solve it? -3( 2x + 3y = 6 ) 2( 3x – 4y = -25) If the top equation was multiplied by -3, then the first term would be -6x. Then multiply the bottom equation by 2, and the first term becomes a 6x. -6x – 9y = -18 6x – 8y = -50 -17y = -68 y=4 2x + 3(4) = 6 2x + 12 = 6 2x = -6 x = -3 (-3, 4) Example 8: Since this system of equations can’t be solved by multiplying one row by a number, how can we solve it? -3( 7x + 2y = 4 ) 2( 5x + 3y = 6 ) -21x – 6y = -12 10x + 6y = 12 -11x = 0 x=0 7(0) + 2y = 4 0 + 2y = 4 2y = 4 y=2 (0, 2) Solve the following systems of equations by using elimination. 1) 2x + y = 6 x – 4y = -15 2) 3x – y = -3 2x + 5y = -19 4) 3x + 5y = 7 6x + y = 5 6) 3x – 2y = -10 5x + 2y = 10 Answers: 3) x + 2y = -1 2x + 3y = 2 5) 2x + 3y = 1 Page 269/ 7-15 odd 3x + 4y = 2 7) 3x + 4y = 0 8) 5x – 3y = -27 5x – 4y = -8 1) (1, 4) 2) (-2, -3) 5) (2, -1) 6) (0, 5) Page 276/ 7-15 odd 3) (7, -4) 7) (-1, ¾) 2x + 7y = 22 4) (2/3, 1) 8) (-3, 4)