Untitled13228.notebook
December 12, 2016
Warm up
Solve each system using substitution:
y = 4x ‐3
y + 5x = 4
y = ‐2x + 2
Y = 2x + 15
y = 7x – 20
3x ‐17 = 2y
Dec 11­10:39 AM
I can manipulate an equation as much as I want as long as I do the same to both sides
5=5
3x = 15
Dec 11­10:42 AM
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Untitled13228.notebook
December 12, 2016
We will eliminate y because the sum of the coefficients is zero
I can ELIMINATE a variable by adding my second equation to the first!
This strategy is called the elimination method.
5x – 6y = ‐32
3x + 6y = 48
Dec 11­10:50 AM
2y + x = 7
‐2y + 3x = ‐3
Dec 11­10:51 AM
2
Untitled13228.notebook
December 12, 2016
2x + 5y = ‐22
10x + 3y = 22
Dec 11­10:53 AM
‐2x + 15y = ‐32
7x – 5y = 17
Dec 11­10:53 AM
3
Untitled13228.notebook
December 12, 2016
3x – 10y = ‐25
4x + 40y = 20
Dec 11­10:55 AM
challenge
4x – 3y = 11
3x – 5y = ‐11
Dec 11­10:55 AM
4
Untitled13228.notebook
challenge
December 12, 2016
6x + 3y = 27
‐4x +7y = 27
Dec 11­10:56 AM
Steps we take for elimination 1 If needed, manipulate an equation such that one variable can be cancelled out
2
add or subtract your two equations
3
solve for your variable
4
once you have a value for one variable, substitute in that value into an equation and find the second variable.
Dec 11­10:56 AM
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Untitled13228.notebook
December 12, 2016
homework:
page 356 #'s 1­6, 9­14
Dec 11­11:05 AM
Dec 6­5:22 PM
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