5.2 Solving Systems of Linear Equations by Substitution

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5.2 Solving Systems of Linear
Equations by Substitution
Solving Linear Systems by Substitution
Example 1:
Solving a System of Linear Equations by Substitution
Solve the system of linear equations by substitution.
y = -2x – 9
6x – 5y = -19
Equation 1
Equation 2
Step 1: Equation 1 is already solved for y.
Step 2: Substitute -2x – 9 in for y in Equation 2
and solve for x.
6x – 5(-2x – 9) = -19
6x + 10x + 45 = -19
16x + 45 = -19
- 45 - 45
16x = -64
x = -4
Step 3: Substitute -4 in for x in Equation 1 and
solve or y.
y = -2(-4) – 9
y=8–9
y = -1
The solution is (-4,-1).
Remember to check your work by substituting
the values back in the original equations!
You try!
Solve the system of linear equations by substitution. Check your
solution!
1) y = 3x + 14
y = -4x
(-2,8)
2) x = 6y – 7
4x + y = -3
(-1,1)
Example 2:
Solving a System of Linear Equations by Substitution
Solve the system of linear equations by substitution.
-x + y = 3
3x + y = -1
Equation 1
Equation 2
Step 1: Solve Equation 1 for y.
y = x +3
Step 2: Substitute x + 3 in for y in Equation 2
and solve for x.
3x + (x + 3) = -1
4x + 3 = -1
-3 -3
4x = -4
x = -1
Step 3: Substitute -1 in for x in Equation 1 and
solve or y.
-(-1) + y = 3
1+y=3
y=2
The solution is (-1,2).
Remember to check your work by substituting
the values back in the original equations!
You try!
3) –x + y = -4
4x – y = 10
(2,-2)
Example 3:
Solving Real-Life Problems
A drama club earns $1040 from a production. A total of 64 adult
tickets and 132 student tickets are sold. An adult ticket cost twice
as much as a student ticket. Write a system of linear equations to
represent the situation. What is the cost of each ticket type?
64 โˆ™ ๐‘Ž๐‘‘๐‘ข๐‘™๐‘ก ๐‘ก๐‘–๐‘๐‘˜๐‘’๐‘ก ๐‘๐‘Ÿ๐‘–๐‘๐‘’ + 132 โˆ™ ๐‘ ๐‘ก๐‘ข๐‘‘๐‘’๐‘›๐‘ก ๐‘ก๐‘–๐‘๐‘˜๐‘’๐‘ก ๐‘๐‘Ÿ๐‘–๐‘๐‘’ = 1040
๐ด๐‘‘๐‘ข๐‘™๐‘ก ๐‘ก๐‘–๐‘๐‘˜๐‘’๐‘ก ๐‘๐‘Ÿ๐‘–๐‘๐‘’ = 2 โˆ™ ๐‘ ๐‘ก๐‘ข๐‘‘๐‘’๐‘›๐‘ก ๐‘ก๐‘–๐‘๐‘˜๐‘’๐‘ก ๐‘๐‘Ÿ๐‘–๐‘๐‘’
Let x be the price (in dollars) of an adult ticket.
Let y be the price (in dollars) of a student ticket.
๐‘†๐‘ฆ๐‘ ๐‘ก๐‘’๐‘š:
64๐‘ฅ + 132๐‘ฆ = 1040
x = 2y
๐‘†๐‘ฆ๐‘ ๐‘ก๐‘’๐‘š:
64๐‘ฅ + 132๐‘ฆ = 1040
x = 2y
Step 1: Equation 2 is already solved for x.
Step 2: Substitute 2y in for x in Equation 1
and solve for y.
64(2y) + 132y = 1040
128y + 132y = 1040
260y = 1040
y=4
Equation 1
Equation 2
Step 3: Substitute 4 in for y in Equation 2 and
solve or x.
x = 2(4)
x=8
The solution is (8,4). This means that an adult
ticket cost $8 and a student ticket cost $4.
Remember to check your work by substituting
the values back in the original equations!
You try!
4) There are a total of 64 students in a drama club and a yearbook
club. The drama club has 10 more students than the yearbook club.
Write a system of linear equations that represents this situation.
How many students are in each club?
# ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘ข๐‘‘๐‘’๐‘›๐‘ก๐‘  ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘‘๐‘Ÿ๐‘Ž๐‘š๐‘Ž ๐‘๐‘™๐‘ข๐‘ + # ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘ข๐‘‘๐‘’๐‘›๐‘ก๐‘  ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘ฆ๐‘’๐‘Ž๐‘Ÿ๐‘๐‘œ๐‘œ๐‘˜ ๐‘๐‘™๐‘ข๐‘ = 64
#๐‘œ๐‘“ ๐‘ ๐‘ก๐‘ข๐‘‘๐‘’๐‘›๐‘ก๐‘  ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘‘๐‘Ÿ๐‘Ž๐‘š๐‘Ž ๐‘๐‘™๐‘ข๐‘ − 10 = # ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘ข๐‘‘๐‘’๐‘›๐‘ก๐‘  ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘ฆ๐‘’๐‘Ž๐‘Ÿ๐‘๐‘œ๐‘œ๐‘˜ ๐‘๐‘™๐‘ข๐‘
Let x be the number of students in the drama club.
Let y be the number of students in the yearbook club.
๐‘†๐‘ฆ๐‘ ๐‘ก๐‘’๐‘š:
๐‘ฅ + ๐‘ฆ = 64
x -10 = y
The solution is (37,27). This means
that there are 37 students in the
drama club and 27 students in the
yearbook club.
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