2-VARIABLE
LINEAR EQUATION SYSTEM (SPLDV)
Definition of SPLDV
SPLDV is a….
system of equations or forms of
relationships similar to in the form of algebra which has two variables and one rank
and if depicted in a graph it will form a straight line. And because of this, this equation is called a linear equation.
Characteristics of SPLDV
Ø Use the
equal sign relation (=)
Ø Has two
variables
Ø Both of
these variables have first degree (pangkat satu)
Matters relating to SPLDV
A. Tribe
Tribe is a part of an
algebraic form consisting of variables, coefficients and constants. And each tribe is separated by addition or
subtraction punctuation
Example:
6x - y + 4, then the tribes of
the equation are 6x, -y and 4
B. Coefficient
Coefficient is a number that
states the number of a number of similar variables. The coefficient is also
called the number that is in front of the variable, because the writing of a
coef fi cient equation is in front of the variable.
Example:
Mika has 2 pineapple and 5
oranges. If written in the form of equations are:
Answer:
Pineapple = x and Orange = y
The connection is 2x + 5y
Where 2 and 5 are coefficients. And 2 is
the coefficient x and 5 is the coefficient y
C. Variable
Variables, namely variables or
substitutes for a number that is usually denoted by letters such as x and y.
Example:
Mika has 2 pineapple and 5
oranges.
If written in the form of an
equation is
Pineapple = x
Orange = y
The connection is 2x + 5y
D. Constant
Constants are numbers that are
not followed by variables, the value is constant or constant for whatever the
value of the change is
Example:
2x + 5y + 7, from the equation
the constant is 7, because 7 values are fixed and are not affected by any
variable.
Method of Completion of
Two Variable Linear Equation Systems
To complete the spldv
calculation method (a linear two-variable system of equations) it can be solved
with the following 4 methods:
1. Substitution Method
2. Elimination Method
3. Combined Method
(Substitution and Elimination)
4. Graph Method
Following are the steps to
complete spldv using the Substitution method
1. Change one of the equations
to the form x = cy + d or y = ax + b
- a, b, c, and d are the values in the equation
- The trick is you have to look for two equations,find one of the easiest equations.
- a, b, c, and d are the values in the equation
- The trick is you have to look for two equations,find one of the easiest equations.
2. After getting the equation,
substitute the x or y value
3. Complete the equation so that you get the value x or y
4. Get the value of the variable that is not yet known with the results of the previous step.
3. Complete the equation so that you get the value x or y
4. Get the value of the variable that is not yet known with the results of the previous step.
Look for the following resolution sets
for each SPLDV.
5x + 5y = 25
3x + 6y = 24
Answer
5x + 5y = 25 ………. Pers. (1)
3x + 6y = 24 ………. Pers. (2)
From equation (1) we get the y equation as follows.
⇔ 5x + 5y = 25
⇔ 5y = 25 - 5x
⇔ y = 5-x
Then we substitute the y equation to equation (2) as follows.
⇔ 3x + 6 (5 - x) = 24
⇔ 3x + 30 - 6x = 24
⇔ 30 - 3x = 24
⇔ 3x = 30-24
⇔ 3x = 6
⇔ x = 2
Finally, to determine the y value, we substitute the value of x to equation (1) or equation (2) as follows.
⇔ 5 (2) + 5y = 25
⇔ 10 + 5y = 25
⇔ 5y = 25-10
⇔ 5y = 15
⇔ y = 3
So, the set of resolutions from the SPLDV is {(2, 3)}.
5x + 5y = 25
3x + 6y = 24
Answer
5x + 5y = 25 ………. Pers. (1)
3x + 6y = 24 ………. Pers. (2)
From equation (1) we get the y equation as follows.
⇔ 5x + 5y = 25
⇔ 5y = 25 - 5x
⇔ y = 5-x
Then we substitute the y equation to equation (2) as follows.
⇔ 3x + 6 (5 - x) = 24
⇔ 3x + 30 - 6x = 24
⇔ 30 - 3x = 24
⇔ 3x = 30-24
⇔ 3x = 6
⇔ x = 2
Finally, to determine the y value, we substitute the value of x to equation (1) or equation (2) as follows.
⇔ 5 (2) + 5y = 25
⇔ 10 + 5y = 25
⇔ 5y = 25-10
⇔ 5y = 15
⇔ y = 3
So, the set of resolutions from the SPLDV is {(2, 3)}.
steps to complete spldv with the elimination method:
- Elimination method is a method or way to
solve a system of linear
equations in two variables by
eliminating or eliminating one variable (variable) by equating the coefficient of the
equation.
- The way to eliminate one of the variables is
by observing the sign, if the
sign is the same as [(+) with
(+) or (-) with (-)], then to eliminate it by subtracting it. And vice versa
if the mark is different then use
the addition system.
Determine the set of
resolutions of the equation x + 3y = 15 and
3x + 6y = 30
Known :
Equation 1 = x + 3y = 15
Equation 2 = 3x + 6y = 30
The first step is to determine which variables will be eliminated first.
Known :
Equation 1 = x + 3y = 15
Equation 2 = 3x + 6y = 30
The first step is to determine which variables will be eliminated first.
This time we will remove x
first, and so we find the y value. The
methods are:
3x + 6y = 30 : 3
x + 2y = 10. . . . (1)
x + 3y = 15. . . (2)
The Second Step From equations (1) and (2), let's eliminate, so the result:
x + 3y = 15
x + 2y = 10 _
y = 5
3x + 6y = 30 : 3
x + 2y = 10. . . . (1)
x + 3y = 15. . . (2)
The Second Step From equations (1) and (2), let's eliminate, so the result:
x + 3y = 15
x + 2y = 10 _
y = 5
Third Step Next, to find out the value of x, then the way is
as follows:
x + 3y = 15 | x2 | <=> 2x + 6y = 30. . . (3)
3x + 6y = 30 | x1 | <=> 3x + 6y = 30. . ... (4)
The elimination between equations (3) with (4), the result being:
3x + 6y = 30
2x + 6y = 30 _
x = 0
Then, the solution is HP = {0. 5}
x + 3y = 15 | x2 | <=> 2x + 6y = 30. . . (3)
3x + 6y = 30 | x1 | <=> 3x + 6y = 30. . ... (4)
The elimination between equations (3) with (4), the result being:
3x + 6y = 30
2x + 6y = 30 _
x = 0
Then, the solution is HP = {0. 5}
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