Download Substitution Elimination and Word Problems

Algebra 2
4.2,
4.3a
Machhapuchhare
SAT Question: When it is 7:00 am in Seattle, it is 10:00 am
in Philadelphia. A plane is scheduled to leave Philadelphia at
11:30 am (Philadelphia time) and to arrive in Seattle at 4:15
pm (Seattle time). How many hours are scheduled for the
trip?
3
A. 4
4
3
B. 5 4
C.
1
6
4
D.
1
7
4
E.
3
7
4
11:30 to 4:15 is 4 ¾ hours.
Add three hours for the time change.
E is the answer.
The substitution method is best to use when one
equation is solved for one of the variables, or when
one equation has a variable with a coefficient of 1.
2 x  1 y  6

3x  4 y  4
 y  2 x  6
3x  4 y  4
3x  4(2 x  6)  4
3x  8x  24  4
5x  24  4
5x  20
x4
y  2(4)  6
y  8  6
y  2
(4, 2)
Example 2
2 y  1 x  1

3 y  2 x  12
 x  2 y  1
3 y  2 x  12
3 y  2( 2 y  1)  12
3 y  4 y  2  12
7 y  2  12
7 y  14
y2
x  2 y  1
x  2(2)  1
x  3
(3, 2)
Example 3
5 x  3 y  6

 1x  y  1

x  y 1
5x  3 y  6
5( y  1)  3 y  6
5y  5  3y  6
8y  5  6
8 y  11
11
y
8
x  y 1
11
x  1
8
3
x
8
 3 11  
 ,  
 8 8  
The linear combination method works best when both
equations are in the form Ax + By = C, and especially when
none of the variables have a coefficient of 1.
We use the properties of addition and multiplication to solve
using linear combinations.
Steps:
1. Write both equations in the form Ax + By = C
2. Clear fractions or decimals.
3. Choose a variable to eliminate.
4. Eliminate the variable by multiplying by an appropriate
number to make the two variables add to zero; then add
the equations together.
5. Check by substituting answer into original equations.
Example 1:
3x  4 y  1

3x  2 y  0
3x  4 y  1
3x  2 y  0
2 y  1
1
y
2
3 x  4 y  1
1
3 x  4    1
2
3x  2  1
3x  1
1
x
3
 1 1  
 ,  
 3 2  
3x  3 y  15
Example 2:
2 x  6 y  22
2) 3 x  3 y  15
2 x  6 y  22
6 x  6 y  30
2 x  6 y  22
4 x  8
x2
3 x  3 y  15
3(2)  3 y  15
6  3 y  15
3y  9
y 3
 2,3
Example 3:
6 x  2 y  16

12 x  5 y  31
2) 6 x  2 y  16
12 x  5 y  31
12 x  4 y  32
12 x  5 y  31
 y  1
y 1
6 x  2 y  16
6 x  2(1)  16
6x  2  16
6x  18
x  3
 3,1
Example 4:
5 x  4 y  11

3x  5 y  23
5) 5 x  4 y  11
4) 3 x  5 y  23
25 x  20 y  55
12 x  20 y  92
37 x  37
x  1
5 x  4 y  11
5(1)  4 y  11
5  4 y  11
4 y  16
y4
 1, 4
Example 5:
0.3x  0.5 y  0.1

0.01x  0.4 y  0.38
10)  0.3x  0.5 y  0.1
100) 0.01x  0.4 y  0.38
3x  5 y  1
3) x  40 y  38
3x  5 y  1
3x  120 y  114
115 y  115
y 1
3x  5 y  1
3x  5(1)  1
3x  5  1
3x  6
x2
 2,1
Example 5:
2
1
 2 x  3 y  1

3 x  1 y  2
 4
3
1 3 2 2
x(6)  y (6)  1(6)
2
3
3 3 1 4
x(12)  y (12)  2(12)
4
3
3x  4 y  6
9 x  4 y  24
12 x  30
5
x
2
5
3   4 y  6
2
15
 4y  6
2
15
(2)  4 y (2)  6(2)
2
15  8 y  12
5
x
2
15  8 y  12
8 y  3
3
y
8
 5 3  
 ,  
 2 8  
Example of simple word problem: Eight small boxes plus
5 large boxes cost $184. A large box costs $3.00 more than
a small box. What is the cost of each size of box?
x = small box
y = large box
8 x  5 y  184
y  3 x
8x  5(3  x)  184
8 x  15  5 x  184
13x  169
x  13
y  3  13
y  16
$13 for small box; $16 for large box
Classwork:
2, 10, 14, 18, 22/166;
2/171
Get ready for a “Small Quiz”
to be written
on your grade sheet.
THE END
Quiz. Copy the problems and write the
answer.
Find the solution by graphing:
3x  2 y  4
2x  y  5
Put your grade paper on the front of
your row, quiz side down.