Simultaneous equations in two variables. Finding solution to systems of linear equations by graphing. Solving systems of linear equations by elimination and substitution method.
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2. Linear Equation in Two Variables
1. π΄π₯ + π΅π¦ + πΆ = 0
2. ππ₯ + ππ¦ = π
3. π¦ = ππ₯ + π
3. Joe and Alan have 12 one-baht coins together.
Let π₯ be the number of one-baht coins Joe has
Let π¦ be the number of one-baht coins Allan has
π₯ + π¦ = 12
1,11 2,10 3,9 4,8 5,7 6,6
7,5 8,4 9,3 10,2 11,1
How many one-baht coins does Joe have? Alan?
4. Joe and Alan have 12 one-baht coins together.
How many one-baht coins does Joe have? Alan?
Joe has two more coins than Alan.
π₯ = π¦ + 2 or π₯ β π¦ = 2
3,1 4,2 5,3 6,4 7,5 8,6
9,7 10,8 β―
π₯ + π¦ = 12
1,11 2,10 3,9 4,8 5,7 6,6
7,5 8,4 9,3 10,2 11,1
simultaneously
6. Systems of Linear Equations in two variables
A system of linear equations in two variables has two variables and two
or more equations.
Let π, π, π, π, π and π be any real numbers and that π and π are not
simultaneously zero; π and π are also not simultaneously zero.
We call ππ₯ + ππ¦ = π and ππ₯ + ππ¦ = π a system of linear equations in
two variables with two equations.
The solution to the system of equations is the values of π₯ and π¦ that
make both equations true simultaneously and can be written in the
form of an ordered pair π₯, π¦
7. Solutions to systems of linear equations in two variables
GRAPHING
1. Let π₯ and π¦ be any real numbers. Plot the graph of the following
systems of linear equations in two variables on the same coordinate
plane and find the solution to the system of equations.
π₯ + π¦ = 7
π₯ β 2π¦ = 1
π₯ + π¦ = 7
π₯
π¦
β1
8
0
7
1
6
π₯ β 2π¦ = 1
π₯
π¦
β3
β2
0
β
1
2
3
1
9. Solutions to systems of linear equations in two variables
GRAPHING
2. Let π₯ and π¦ be any real numbers. Plot the graph of the following
systems of linear equations in two variables on the same coordinate
plane and find the solution to the system of equations.
2π₯ β π¦ = 5
3π₯ + π¦ = 10
2π₯ β π¦ = 5
π₯
π¦
β1
β7
0
β5
1
β3
3π₯ + π¦ = 10
π₯
π¦
β1
13
0
10
1
7
11. Solutions to systems of linear equations in two variables
GRAPHING
3. Let π₯ and π¦ be any real numbers. Plot the graph of the following
systems of linear equations in two variables on the same coordinate
plane and find the solution to the system of equations.
π₯ + 2π¦ = 8
2π₯ + 4π¦ = 20
π₯ + 2π¦ = 8
π₯
π¦
β2
5
0
4
2
3
2π₯ + 4π¦ = 20
π₯
π¦
β2
6
0
5
2
4
13. Solutions to systems of linear equations in two variables
GRAPHING
4. Let π₯ and π¦ be any real numbers. Plot the graph of the following
systems of linear equations in two variables on the same coordinate
plane and find the solution to the system of equations.
π₯ β 2π¦ = 4
3π₯ β 6π¦ = 12
π₯ β 2π¦ = 4
π₯
π¦
β2
β3
0
β2
2
β1
3π₯ β 6π¦ = 12
π₯
π¦
β2
β3
0
β2
2
β1
15. Find the solution to the following systems of linear equations in two
variables by plotting the graphs on the coordinate plane.
π₯ + π¦ = 8
π₯ β π¦ = β2
π₯ + π¦ = 8
π₯
π¦
0
8
8
0
π₯ β π¦ = β2
π₯
π¦
0
2
β2
0
17. Find the solution to the following systems of linear equations in two
variables by plotting the graphs on the coordinate plane.
2π₯ β π¦ = β1
π₯ β 3π¦ = 2
2π₯ β π¦ = β1
π₯
π¦
0
1
β
1
2
0
π₯ β 3π¦ = 2
π₯
π¦
0
β
2
3
2
0
19. Find the solution to the following systems of linear equations in two
variables by plotting the graphs on the coordinate plane.
π₯ β π¦ = β3
2π₯ β 2π¦ = β6
π₯ β π¦ = β3
π₯
π¦
0
3
β3
0
2π₯ β 2π¦ = β6
π₯
π¦
0
3
β3
0
21. Find the solution to the following systems of linear equations in two
variables by plotting the graphs on the coordinate plane.
π₯ β π¦ = 1
π₯ β π¦ = 4
π₯ β π¦ = 1
π₯
π¦
0
β1
1
0
π₯ β π¦ = 4
π₯
π¦
0
β4
4
0
23. Determine whether the following
graphs of systems of linear equations
have no solution, one solution, or
infinitely many solutions.
π₯
π¦
0
One solution
24. Determine whether the following
graphs of systems of linear equations
have no solution, one solution, or
infinitely many solutions.
π₯
π¦
0
Infinitely many solutions
25. Determine whether the following
graphs of systems of linear equations
have no solution, one solution, or
infinitely many solutions.
π₯
π¦
0
No solution
26. Determine whether the following
graphs of systems of linear equations
have no solution, one solution, or
infinitely many solutions.
π₯
π¦
0
One solution
27. Determine whether the following
graphs of systems of linear equations
have no solution, one solution, or
infinitely many solutions.
π₯
π¦
0
No solution
28. Determine whether the following
graphs of systems of linear equations
have no solution, one solution, or
infinitely many solutions.
π₯
π¦
0
One solution
30. The rental rate for a parking space in the parking lot owned by Sehun is 35 baht per day plus a 30 baht fee,
while the rate for a space in the parking lot owned by Suho is 30 baht per day plus a 45 baht fee. Plot a graph
of a parking space rental charges versus parking periods of the two lots.
π¦ = 30 + 35π₯
π¦ = 45 + 30π₯
Let π¦ be the rental charge
Let π₯ be the number of days
100
120
140
160
180
200
220
20
40
60
80
1 2 3 4 5 6
Number of days
Rental Charge baht
0
π¦ = 30 + 35π₯
Sehun
π¦ = 45 + 30π₯
Suho
1. If you need to rent a parking space for less than 3
days, which parking lot should you choose?
2. If you need to rent a parking space longer than 3
days, which parking lot should you choose?
3. For how many days will the rent for the parking
space of the two lots be the same?
Answer: Sehunβs
Answer: Suhoβs
Answer: 3 days
31. Bambam knows that the temperature in degree Fahrenheit Β°πΉ and degree Celsius Β°πΆ have a linear
relationship but doesnβt know the equation. He knows that the melting point of water is 0Β°πΆ or 32Β°πΉ and
that the boiling point of water is 100Β°πΆ or 212Β°πΉ. Help Bambam find the temperature of 140Β°πΉ in degree
Celsius.
100
120
140
160
180
200
220
20
40
60
80
20 40 60 80 100
Β°πΆ
Β°πΉ
1200
0,32
100,212
Answer: The temperature of 140Β°πΉ is
60 degrees in Celsuis
32. Jung Kook needs to reinstall wiring in his garage. He asks two electricians for service rates. The first
electrician charges 200 baht for the trip and 200 baht per working hour. The second electrician charges 300
baht for the trip and 150 baht per working hour. Find the number of hours this electrical wiring installation
should be done that make the two electricians charge the same.
800
900
1000
1 2 3 4 5
Number of hours
Service charge baht
6
500
400
600
700
1100
1200
1300
1400
0
π¦ = 200 + 200π₯
π¦ = 300 + 150π₯
Answer: Two hours
34. Solving linear equations in Two Variables by
ELIMINATION
ππ₯ + ππ¦ = π
ππ₯ + ππ¦ = π
STEPS:
1. Use the multiplicative property of equality to make the coefficients in front of the variable
in the two equations opposite numbers.
2. Use the additive property of equality to eliminate the intended variable with opposite
coefficients from the equations. The resulting equation is a linear equation with only one
variable.
3. Find the solution to the linear equation with one variable in (2).
4. Find the value of the eliminated variable by substituting the value of variable in (2) for the
equation or .1 2
1
2
35. Find the solution to the following system of equations.
2π₯ β 3π¦ = 7
3π₯ + 2π¦ = 4
1
2
We will eliminate the variable π¦ by making its coefficients in the two equations opposite
numbers, using the multiplicative property of equality. Step 1
1 Γ 2, 4π₯ β 6π¦ = 14
2 Γ 3, 9π₯ + 6π¦ = 12
3
4
Then, we use the additive property of equality. Step 2
3 + 4 , 4π₯ β 6π¦ + 9π₯ + 6π¦ = 14 + 12
4π₯ β 6π¦ + 9π₯ + 6π¦ = 14 + 12
Next, find the solution. Step 3
13π₯ = 26
13π₯ = 2
Lastly, substitute π₯ with 2 in 1 Step 4
2π₯ β 3π¦ = 7
36. Find the solution to the following system of equations.
2π₯ β 3π¦ = 7
3π₯ + 2π¦ = 4
1
2
We will eliminate the variable π¦ by making its coefficients in the two equations opposite
numbers, using the multiplicative property of equality. Step 1
1 Γ 2, 4π₯ β 6π¦ = 14
2 Γ 3, 9π₯ + 6π¦ = 12
3
4
Then, we use the additive property of equality. Step 2
3 + 4 , 4π₯ β 6π¦ + 9π₯ + 6π¦ = 14 + 12
4π₯ β 6π¦ + 9π₯ + 6π¦ = 14 + 12
Next, find the solution.
13π₯ = 26
13π₯ = 2
Lastly, substitute π₯ with 2 in 1 Step 4
2 2 β 3π¦ = 7
4 β 3π¦ = 7
4 β 3π¦ = 3
4 β 3π¦ = β1
Therefore, the solution to the system of equations is 2, β1 .
Step 3
37. +
Γ
Γ
Find the solution to the following system of equations.
3
2
π₯ β
3
4
π¦ =
15
4
4
3
π₯ β
5
3
π¦ = 3
1
2
To make all the coefficients of π₯ and π¦ whole numbers, we multiply by the least common
multiple of 2 and 4, which is 4, and by 3.
1
2
1 Γ 4, 6π₯ β 3π¦ = 15 3
2 Γ 3, 4π₯ β 5π¦ = 9 4
We will eliminate the variable π¦. Step 1
3 5, 30π₯ β 15π¦ = 75 5
4 β3, β12π₯ + 15π¦ = β27 6
Then, we use the additive property of equality. Step 2
5 6 , 30π₯ β 15π¦ + β12π₯ + 15π¦ = 75 + β27
Next, find the solution. Step 3
30π₯ β 12π₯ = 48
18π₯ = 48
18π₯ =
48
18
=
8
3
Lastly, substitute π₯ with
8
3
in 3 Step 4
6π₯ β 3π¦ = 15
38. +
Γ
Γ
Find the solution to the following system of equations.
3
2
π₯ β
3
4
π¦ =
15
4
4
3
π₯ β
5
3
π¦ = 3
1
2
To make all the coefficients of π₯ and π¦ whole numbers, we multiply by the least common
multiple of 2 and 4, which is 4, and by 3.
1
2
1 Γ 4, 6π₯ β 3π¦ = 15 3
2 Γ 3, 4π₯ β 5π¦ = 9 4
We will eliminate the variable π¦. Step 1
3 5, 30π₯ β 15π¦ = 75 5
4 β3, β12π₯ + 15π¦ = β27 6
Then, we use the additive property of equality. Step 2
5 6 , 30π₯ β 15π¦ + β12π₯ + 15π¦ = 75 + β27
Next, find the solution. Step 3
30π₯ β 12π₯ = 48
18π₯ = 48
18π₯ =
48
18
=
8
3
Lastly, substitute π₯ with
8
3
in 3 Step 4
6
8
3
β 3π¦ = 15
16π₯ β 3π¦ = 15
16 β 3π¦ = β1
16 β 3π¦ =
1
3
Therefore, the solution to the system of equations is
8
3
,
1
3
.
40. Solving linear equations in Two Variables by
Substitution
ππ₯ + ππ¦ = π
ππ₯ + ππ¦ = π
STEPS:
1. Choose one of the equations in order to write one variable in terms of the other variable:
π¦ in terms of π₯ or π₯ in terms of π¦.
2. Substitute the variable from (1) in the other equation.
3. Solve the equation in (2) to get the value of one variable.
4. Use the value of the variable obtained from (3) to substitute back into one of the original
two equations and solve for the second variable. The values of the two variables are the
solution to the system of linear equations in two variables.
1
2
41. 3π₯ β π¦ = 3π₯ β 17
Find the solution to the following system of equations.
3π₯ β π¦ = 17
π₯ + 2π¦ = 8
1
2
From 1 3π₯ β π¦ = 17, we can write π¦ in terms of π₯ as
3 Step 1
Substitute π¦ from 3 in 2 Step 2
π₯ + 2π¦ = 8
π₯ + 2 3π₯ β 17 = 8, which is an equation in one variable
Solve for π₯. Step 3
π₯ + 2 3π₯ β 17 = 8
π₯ + 6π₯ β 34 = 8
7π₯ = 42
7π₯ = 6
Substitute π₯ with 6 in 3 Step 4
3π₯ β π¦ = 3π₯ β 17
3π₯ β π¦ = 3 6 β 17
3π₯ β π¦ = 18 β 17
3π₯ β π¦ = 1
Therefore, the solution to the system of linear equations
is 6,1
42. π₯ = 3 β π¦
Find the solution to the following system of equations.
3π₯ = 9 β 2π¦
π₯ + π¦ = 3
1
2
From 2 π₯ + π¦ = 3, we can write π₯ in terms of π¦ as
3 Step 1
Substitute π₯ from 3 in 1 Step 2
3π₯ = 9 β 2π¦
3 3 β π¦ = 9 β 2π¦, which is an equation in one variable
Solve for π¦. Step 3
3 3 β π¦ = 9 β 2π¦
9 β 3π¦ = 9 β 2π¦
0 = π¦
π¦ = 0
Substitute π¦ with 0 in 3 Step 4
π₯ = 3 β π¦
π₯ = 3 β 0
π₯ = 3
Therefore, the solution to the system of linear equations
is 3,0