3.2 Day 2 Solving Systems Algebraically 2010
October 18, 2010
3.2 Solving Systems Algebraically Day 2
Objective
solve systems of
equatıons by
elımınatıon.
Nov 26­4:24 PM
1
3.2 Day 2 Solving Systems Algebraically 2010
October 18, 2010
Check Skills You'll Need
Find the additive inverse of each term
1) 4
3) 5x
2) ­x
4) 8y
Oct 1­11:05 AM
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3.2 Day 2 Solving Systems Algebraically 2010
October 18, 2010
So what is elimination?
Elimination is getting rid of one of the
variables by adding both sets of equations.
Let's Try One!
Oct 1­11:03 AM
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3.2 Day 2 Solving Systems Algebraically 2010
October 18, 2010
Solve this system of equations by elimination.
x + y = ­3
x ­ y = ­1
Oct 1­11:04 AM
4
3.2 Day 2 Solving Systems Algebraically 2010
October 18, 2010
Solve this system of equations by elimination.
x + y = ­3
x ­ y = ­1
Step 1: Make sure all the variables are on the left side and are all in the same order.
x + y = ­3
x ­ y = ­1
Step 2: Choose one set of variables and make them have "opposite" signs.
Our y values already have opposite signs!
Step 3: Draw a line under the equations and add them straight down.
x + y = ­3
x ­ y = ­1
Steps 4, 5 & 6
2x = ­4
Oct 1­11:04 AM
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3.2 Day 2 Solving Systems Algebraically 2010
October 18, 2010
Solve this system of equations by elimination.
x + y = ­3
x ­ y = ­1
Step 4: Solve for x.
2x = ­4
x = ­2
(Divide both sides by 2)
Step 5: Substitute your x value into either original equation to solve for x.
x + y = ­3
­2 + y = ­3
y = ­1
(Add 2 to both sides)
Step 6: State your results.
x = ­2
y = ­1
Oct 1­11:04 AM
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3.2 Day 2 Solving Systems Algebraically 2010
October 18, 2010
Solve this system of equations using elimination.
3y = 2x + 2
x ­ 2 = 3y
Step 1: Make sure all the variables are on the left side and are all in the same order.
2x ­ 3y = ­2
x ­ 3y = 2
Step 2: Choose one set of variables and make them have "opposite" signs.
Multiply the bottom equation by ­1.
(­1) (x ­ 3y = 2)
­x +3y = ­2
Step 3: Draw a line under the equations and subtract them straight down.
2x ­ 3y = ­2
­x + 3y = ­2
x = ­4
Step 4: Solve for x.
We already did! x = ­4
Step 5: Substitute your x value into either original equation to solve for x.
x ­ 2 = 3y
­4 ­2 = 3y
­6 = 3y
­2 = y
(divide each side by 3)
Step 6: State your results.
x = ­4
y = ­2
Oct 1­11:04 AM
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3.2 Day 2 Solving Systems Algebraically 2010
October 18, 2010
Sometimes you might need to multiply equations by a number. Let's try one of those!
4x ­ 3y = 11
­5x +2y = ­12
x = 2
y = ­1
Oct 1­11:04 AM
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3.2 Day 2 Solving Systems Algebraically 2010
October 18, 2010
Try one more! ­2c + 9d = 35
6c +7d = 65
Oct 1­11:04 AM
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3.2 Day 2 Solving Systems Algebraically 2010
October 18, 2010
homework:
pg. 128
#'s: 18-29, 43, 54-59, 67, 68
Nov 26­4:24 PM
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