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Elimination Method Day 1

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Presentation on theme: "Elimination Method Day 1"— Presentation transcript:

1 Elimination Method Day 1
Today’s Objective: I can solve a system using elimination.

2 x + y = 7 x – y = 3 (5, 2) x – y = 3 + y + y x = 3 + y x + y = 7
1. Solve for 1 variable x + y = 7 x – y = 3 2. Substitute the variable you just solved for into the other equation AND solve. Substitution review 3. Solve for the other variable (substitute) 4. Write the ordered pair Step 2 Step 1 Step 3 x + y = 7 ( ) + y = 7 3 + y + y = 7 3 + 2y = 7 2y = 4 y = 2 x= 3 + y x = 3 + ( ) x = 3 + 2 x = 5 x – y = 3 + y y x = 3 + y 3 + y 2 Step 4 (5, 2)

3 x + y = 7 x – y = 3 2x = 10 x = 5 (5, 2) x+ y = 7 ( ) + y = 7 -5 -5
Solve each system using ELIMINATION Identify one variable in each equation that has the same number but with opposite signs. Step 2 2x = 10 2. Add the 2 equations 3. Solve for one variable x = 5 Solve for the other variable. (substitute into one of the original equations). Step 3 x+ y = 7 ( ) + y = 7 y = 2 Step 4 5. Write the ordered pair. 5 (5, 2) Step 5

4 2x – y = -5 -2x – 5y = 11 -6y = 6 (-3, -1) y = -1 -2x – 5y = 11
Identify one variable in each equation that has the same number but with opposite signs. 2. Add the 2 equations 3. Solve for one variable Step 2 -6y = 6 Solve for the other variable. (substitute into one of the original equations). y = -1 Step 3 5. Write the ordered pair. -2x – 5y = 11 -2x – 5( ) = 11 -2x + 5 = 11 -2x = 6 x = -3 Step 4 -1 (-3, -1) Step 5

5 10x – 35y = 15 -10x + 8 y = 12 Identify one variable in each equation that has the same number but with opposite signs. 2. Add the 2 equations 3. Solve for one variable Step 2 -27y = 27 Solve for the other variable. (substitute into one of the original equations). y = -1 Step 3 5. Write the ordered pair. -10x + 8y = 12 -10x + 8( ) = 12 -10x – 8 = 12 -10 x = 20 x = -2 Step 4 -1 (-2, -1) Step 5

6 I can solve a system using elimination.
Assignment Book pg 382: 7, 8,11, 54, 55 Select answers 7. (4,5) (22,7) 11. ( 3,15)

7 2x – 3y = 5 x – 2y = 4 -2( ) 2x – 3y = 5 -2x + 4y = -8 y = -3
Step 4 (substitute) x – 2y = 4 x – 2( ) = 4 x + 6 = 4 x = -2 -2( ) -3 Rewrite both equations 2x – 3y = 5 -2x + 4y = -8 y = -3 Step 2 (add the equations) Step 5 (ordered pair) ( -2, -3) Step 3 Solve for 1 variable y = -3

8 4( ) 2x + 3y = 10 3x – 4y = -2 3( ) 8x + 12y = 40 9x – 12y = -6
4( ) 2x + 3y = 10 3x – 4y = -2 Step 4 2x + 3y = 10 2( ) + 3y = 10 4 + 3y = 10 3y = 6 y = 2 3( ) 2 8x + 12y = 40 9x – 12y = -6 17x = 34 Step 2 Step 5 (2, 2) Step 3 x = 2

9 2( ) 3x + 2y = 17 2x + 5y = 26 -3( ) 6x + 4y = 34 -6x – 15y = -78
2( ) 3x + 2y = 17 2x + 5y = 26 Step 4 3x + 2y = 17 3x + 2( ) = 17 3x + 8 = 17 3x = 9 x = 3 -3( ) 4 6x + 4y = 34 -6x – 15y = -78 -11y = -44 Step 2 Step 5 (3, 4) Step 3 y = 4

10 Assignment day 2 Book pg 382: 9,10,12,15-20,22-24 (note: , look at previous notes for 1 solution, infinitely many solutions, or no solutions) 9. (1,5) (3,1) (2,-1) 10.(6,3) (1,2) (4,7) 17. (5,3) 12. (4,3) (3,4)


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