Transcript Solving Systems Using Elimination

Solving Systems Using
Elimination
Solve each system using substitution.
ALGEBRA 1 LESSON 9-5
(For help, go to Lesson 7-2.)
1.
y = 4x – 3
y = 2x + 13
2. y + 5x = 4
y = 7x – 20
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3. y = –2x + 2
3x – 17 = 2y
Solving Systems Using
Elimination
ALGEBRA 1 LESSON 9-5
Solutions
1. y = 4x – 3
y = 2x + 13
Substitute 4x – 3 for y in the second equation.
y = 2x + 13
4x – 3 = 2x + 13
4x – 2x – 3 = 2x – 2x + 13
2x – 3 = 13
2x = 16
x=8
y = 4x – 3 = 4(8) – 3 = 32 – 3 = 29
Since x = 8 and y = 29, the solution is (8, 29).
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Solving Systems Using
Elimination
ALGEBRA 1 LESSON 9-5
Solutions (continued)
2. y + 5x = 4
y = 7x – 20
Substitute 7x – 20 for y in the first equation.
y + 5x = 4
7x – 20 + 5x = 4
12x – 20 = 4
12x = 24
x=2
y = 7x – 20 = 7(2) – 20 = 14 – 20 = –6
Since x = 2 and y = –6, the solution is (2, –6).
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Solving Systems Using
Elimination
ALGEBRA 1 LESSON 9-5
Solutions (continued)
3. y = –2x + 2
3x – 17 = 2y
Substitute –2x + 2 for y in the second equation.
3x – 17 = 2y
3x – 17 = 2(–2x + 2)
3x – 17 = –4x + 4
7x – 17 = 4
7x = 21
x = 3
y = –2x + 2 = –2(3) + 2 = –6 + 2  –4
Since x = 3 and y = –4, the solution is (3, –4).
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Solving Systems Using
Elimination
Solve by elimination.
3x + 6y = –6
ALGEBRA 1 LESSON 9-5
–5x – 2y = –14
Step 1: Eliminate one variable.
Start with the given
system.
3x + 6y = –6
–5x – 2y = –14
To prepare to eliminate
y, multiply the second
equation by 3.
3x + 6y = –6
3(–5x – 2y = –14)
Step 2: Solve for x.
–12x = 48
x= 4
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Add the equations to
eliminate y.
3x + 6y = –6
–15x – 6y = –42
–12x – 0 = –48
Solving Systems Using
Elimination
ALGEBRA 1 LESSON 9-5
(continued)
Step 3: Solve for the eliminated variable using either of the original
equations.
3x + 6y = –6
Choose the first equation.
3(4) + 6y = –6
Substitute 4 for x.
12 + 6y = –6
Solve for y.
6y = –18
y = –3
The solution is (4, –3).
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Solving Systems Using
Elimination
Suppose the band sells cans of popcorn for $5 per can and
ALGEBRA 1 LESSON 9-5
cans of mixed nuts for $8 per can. The band sells a total of 240 cans
and receives a total of $1614. Find the number of cans of popcorn
and the number of cans of mixed nuts sold.
Define: Let
p
= number of cans of popcorn sold.
Let
n
= number of cans of nuts sold.
Relate: total number of cans
total amount of sales
Write:
5 p
p
+ n
= 240
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+ 8 n = 1614
Solving Systems Using
Elimination
ALGEBRA 1 LESSON 9-5
(continued)
Step 1: Eliminate one variable.
Start with the given
system.
p + n = 240
5p + 8n = 1614
To prepare to eliminate
p, multiply the first
equation by 5.
5(p + n = 240)
5p + 8n = 1614
Step 2: Solve for n.
–3n = –414
n = 138
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Subtract the equations
to eliminate p.
5p + 5n = 1200
5p + 8n = 1614
0 – 3n = –414
Solving Systems Using
Elimination
ALGEBRA 1 LESSON 9-5
(continued)
Step 3: Solve for the eliminated variable using either of the original
equations.
p + n = 240
Choose the first equation.
p + 138 = 240
Substitute 138 for n.
p = 102
Solve for p.
The band sold 102 cans of popcorn and 138 cans of mixed nuts.
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Solving Systems Using
Elimination
Solve by elimination.
3x + 5y = 10
ALGEBRA 1 LESSON 9-5
5x + 7y = 10
Step 1: Eliminate one variable.
Start with the given
system.
3x + 5y = 10
5x + 7y = 10
To prepare to eliminate
x, multiply one equation
by 5 and the other
equation by 3.
5(3x + 5y = 10)
3(5x + 7y = 10)
Step 2: Solve for y.
4y = 20
y = 5
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Subtract the equations
to eliminate x.
15x + 25y = 50
15x + 21y = 30
0 + 4y = 20
Solving Systems Using
Elimination
ALGEBRA 1 LESSON 9-5
(continued)
Step 3: Solve for the eliminated variable x using either of the original
equations.
3x + 5y = 10
Use the first equation.
3x + 5(5) = 10
Substitute 5 for y.
3x + 25 = 10
3x = –15
x = –5
The solution is (–5, 5).
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Solving Systems Using
Elimination
ALGEBRA 1 LESSON 9-5
Solve using elimination.
1. –6x + 5y = 4
2. 7p + 5q = 2
8p – 9q = 17
3x + 4y = 11
(1, 1)
(1, 2)
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