Download 7.3 - Solving Systems by Elimination

Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
Solve by elimination.
2x + 3y = 11
–2x + 9y = 1
Step 1: Eliminate x because the sum of the coefficients is 0.
2x + 3y = 11
–2x + 9y =1
0 + 12y = 12
Addition Property of Equality
y=1
Solve for y.
Step 2: Solve for the eliminated variable x using either original equation.
2x + 3y = 11
2x + 3(1) = 11
2x + 3 = 11
2x = 8
x=4
Choose the first equation.
Substitute 1 for y.
Solve for x.
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Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
(continued)
Since x = 4 and y = 1, the solution is (4, 1).
Check: See if (4, 1) makes true the equation not used in Step 2.
–2(4) + 9(1) 1
Substitute 4 for x and 1 for y into the
second equation.
–8 + 9
1
1 = 1
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You try:
Use elimination to solve:
6x – 3y = 3 and -6x + 5y = 3
Eliminate the x: 6x – 3y = 3
-6x + 5y = 3
2y = 6
Now solve for y divide by 2 and y = 3
Find x by using either original equation
6x – 3 (3) = 3
6x – 9 = 3
+9 +9
6x = 12
x =2
and y = 3
( 2,3)
Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
Solve by elimination.
3x + 6y = –6
–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 7-3
(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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You try:
Use multiplication to eliminate
-2x + 15y = -32 and 7x - 5y =17
Which one do you multiply? You could get rid of the y if you multiply the 2nd
equation by 3
3( 7x – 5y = 17) = 21x – 15y = 51
Now solve for x
-2x + 15y = -32
21x – 15y = 51
19x = 19
x = 1 Now plug in 1 for x and solve for y ** Use one of the original equations**
7(1) – 5y = 17
7- 5y = 17
-7
-7
-5y = 10 so y = -2
Answer (1, -2)
Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
pages 356–359 Exercises
11. (–2, – 5 )
2
1. (1, 3)
12. (1, 14)
2. (2, –2)
13. ( 1 , 1)
2
3. (5, –17)
14. (–2, 3)
4. (–3, 4)
17. (–1, –3)
5. (–9, 1 )
18. (2.5, 1)
6. (– 1 , 10)
19. (2, –2)
9. (–5, 1)
20. (10, 8)
21. (–1, 3/2)
22. (1, 5)
2
2
10. (11, –3)
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Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
Solve using elimination.
1. 3x – 4y = 7
2x + 4y = 8
(3, 0.5)
3. –6x + 5y = 4 (1, 2)
2. 5m + 3n = 22 (2, 4)
5m + 6n = 34
4. 7p + 5q = 2
8p – 9q = 17
3x + 4y = 11
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(1, 1)