3.2 Day 1 Solving Systems Algebraically 2010
October 13, 2010
3.2 Solving Systems Algebraically Day 1
Objective
solve systems of
equatıons by
substıtutıon.
Nov 26­4:24 PM
1
3.2 Day 1 Solving Systems Algebraically 2010
October 13, 2010
Check Skills You'll Need
Substitute 2y ­ 1 for x in the equation. Solve for y. 1) x +2y = 3
2) y ­ 2x = 8
3) 2y + 3x = ­5
Oct 1­11:05 AM
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3.2 Day 1 Solving Systems Algebraically 2010
October 13, 2010
Solve this system of equations by substitution
x ­ 2y = ­11
x + y = 4
Nov 26­4:28 PM
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3.2 Day 1 Solving Systems Algebraically 2010
Solve this system of equations by substitution
Step 1: Solve the second equation for x.
October 13, 2010
x ­ 2y = ­11
x + y = 4
x + y = 4
­y ­y (subtract y from both sides)
x = 4 ­ y
Step 2: Substitute what you get for x into the 1st equation.
x ­ 2y = ­11
(4 ­ y) ­ 2y = ­11
Step 3: Solve for y.
Steps 4 &5
4 ­ y ­ 2y = ­11
4 ­ 3y = ­11
­4 ­4
(subtract 4 from both sides)
­3y = ­15
(now divide both sides by ­3)
y = 5
Nov 26­4:28 PM
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3.2 Day 1 Solving Systems Algebraically 2010
October 13, 2010
x ­ 2y = ­11
x + y = 4
Step 4: Take your new y value and plug it in either original equation to get x.
(substitute this into the first equation)
y = 5
x ­ 2y = ­11
x ­ 2(5) = ­11
x ­ 10 = ­11
+10 +10 (add 10 to both sides)
x = ­1
Step 5: State your results.
x = ­1 y = 5
Nov 26­5:34 PM
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3.2 Day 1 Solving Systems Algebraically 2010
Solve this system of equations by substitution.
October 13, 2010
a + c = 272
7a +4c = 1694
a = 202
c = 70
Nov 26­5:40 PM
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3.2 Day 1 Solving Systems Algebraically 2010
October 13, 2010
Which equation would you solve for what variable?
3x ­ y = 5
4x + y = 9
3a ­ 5b = 10
a + 7b = 12
r + 2t = 5
2r +t = ­2
Make it as easy on yourself as possible!
Nov 26­6:00 PM
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3.2 Day 1 Solving Systems Algebraically 2010
October 13, 2010
homework:
pg. 128
#'s: 1-13, 77, 79-81
Nov 26­4:24 PM
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