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TUGAS 5

LINEAR EQUATION SYSTEM TWO VARIABLES (SPLDV)
- study porpuse
after studying this chapter, student can hoped :
1. the form explain linear equation system two variables (SPLDV)
2. finish SPLDV
- this chapter in subject matter
1. linear equation with two variables (SPLDV)
2. linear equation system two variables (SPLDV)
Before studying this chapter, to work problem this following!
1. to determine solution the set equation following!
a. 2x = 5
b. 2x – 5 =4 – x
c. 1/(4-2x) = 2
d. (x + 2)/5 = 1
e. ½ = 2/(x – 3)
2. To make graph of y = 2x + 1 !
A. Linear equation with two variable (PLDV)
Before discuss linear equation with two variable, please more fast we to came back remembers form following equation.
1. 2x + 5 = 4
2. 3y + 2 = y + 1
3. x^2 + 5 = 2
you try watch form equation above.
Furthermore, do you still remember finish the way linear equation with one variable (PLSV)? To attention example following!
Example :
To determine solution of equation (2/3)x + 4 = 6!
Solution :
(2/3) + 4 = 6
(2/3) x + 4 – 4 = 6 -4 (twice space between joint minus with 4)
(2/3) x = 2
2/3 (2/3) x = 3/2 x 2 (space between joints match with 3/2)
x = 3
so, the solution of equation’s is (3)
1. discuss linear equation with two variable (PLDV)
can you make equation ? if you answer with true so can written the equation as following.
2 x + 3 y = 6000
At the equation above measles beside variable’s hold on an important one highest also consist of two variable. So the equation it is the equation with two variables.
Example :
1. 2 x + y = 4 is it the linear equation with two variable, because has two variable the highest rank 1 is x and y.
2. x^2 + y = 2 the linear no equation with two variable, because the there rank variable than one is x^2
exercise 4.1
1. mention which of following equation the form linear equation with two variable !
a. 2 x – 3 = x + 2
b. (x/y) = 4
c. 2 x – y = 0
d. x + y^2 = 0
e. (y/x) + 1 = 4
f. 2 x = y + 3
2. To obvious true of false pronouncement following !
a. 2 x + 1 is the linear equation with two variable, because it has the rank variable 1 highest
b. 2 x + y^2 = 5 is the linear equation with two variable, because it has the two variables
c. (x/y) + 1 = 3 the linear no equation with two variable, because x divide y
3. Explain one variable to another one (PLDV)
In PLDV, a certain variable can obvious become the besides form variable. For more, to attention example and explanation this following!
Let, PLDV of 3 x + y = 6

This equation has two variable x and y, the exhausted at same math, is math left. These form can change another to form as the form obvious in variable other variable or the form variable one with besides variable exhausted at different the math. It so happen method’s as following :
X = 2 – (1/3) y
X = -(1/3) y + 2
3 x + y = 6
Y = 6 – 3 x (to move x and coefficient to right math)
Y = -3 x + 6
So, form 3 x + y = 6 can obvious in form
Y = -3 x + 6 or x = - (1/3) y +2
So, a certain PLDV can changed to form the variable one obvious to besides variable
Exercise 4.2
1. To fill points at following obvious until x become true !
a. 3 x + 6 = 9
3 x = 9 - …
3 x = - 6 y + …
X =…+ …
b. 2 x + 4 y – 20 = 0
2 x + 4 y =…
2 x = … - 4 y
2 x =…. + 20
X = … + …
2. To obvious to for form the explain variable y to x following!
a. 3 y = 6 x + 9
b. 2 x – 3 y = 6
c. 4 x + 2 y = 8
d. – 3 x + 4 y = 12
e. 3 x – 2 y + 6 = 0
f. – 3 x – 4 x – 12 = 0
B. Linear equation two variable (SPLDV)
1. Discuss mean linear equation with two variables
To attention example following!
Purchase 2 a books and 5 pencils with total price Rp 4.500,00. Form mentioned the problem if price a book is x and a pencil is y so obtained two the linear equation two variables; it is 3 x + 4 y = 5.000,00 and 2 x + 5 y = 4.500,00.

If there two linear equation with be found two the variable in a problem and each the variable that represent a thing so second this the equation mentioned as the linear equation two variable system. The answer it while form this system is value fastened form each variable fulfill the second equation. Remember, if at equation system this isn’t pair which fulfill that the second equation, so is called the equation system has the answer or the answer association is the empty set.
The kind of the linear equation system, among others :
a. Linear equation system with two variable
Example : 2 x + y = 4
3 x + y = 5
b. Linear equation system with three variable
Example : x + y + z = 6
2 x + y + 3 z = 13
X + 2 y + z = 8
Further more, the discuss this only limited at the linear equation system with two variable.
To attention the linear equation system two variable, it is
2 x + y = 4
3 x + y = 5

From mentioned SPLDF, the first equation consist of the variable x with 2 coefficient, the variable y with 1 coefficient, and 4 constant, the second equation consist of the variable y with 3 coefficients, the variable y with 1 coefficient and 5 constant.
Exercise 4.3
From following equation, which was form the linear equation with two variable ?
1. x + 3 v = 4
2. x^2 + 2 v = 3
3. v
4. x + v = 3
2. the root and the not two variables linear equation system
to attention this the linear equation !
x + y = 3
x – y = 1

if we watch closely, the equation system this consist of two variables.
For x = 1, y = 1, so x + y = 3
1 + 1 = 3 (sentence is false)
x – y = 1
1 – 1 = 1 (sentence is false)
For x = 2, y = 1, so x + y = 3
2 + 1 = 3 (sentence is true)
x – y = 1
2 – 1 = 1 (sentence is true)
For x = 2, y = 1, so x + y = 3
1 + 2 = 3 (sentence is true)
x – y = 1
1 – 2 = 1 (sentence is false)

From mentioned explanation, it appears if x substituted y substituted 1 so obtained at SPLDV the equation became the sentences. The substituted fastened is call the solution SPLDV or the root SPLDV. The values besides 2 and wouldn’t cause at SPLDV the equation become sentences which true. These value from the solution SPLDV>

So, the solution SPLDV or the roots SPLDV is the substituted variable’s until the second equation at SPLDV become be sentences with true. Further more, process of writing SPLDV with use the word “and” as SPLDV : x + y = 3 and x – 1 =1
Can substituted with use sign and
SPLDV : x + y = 3
x – y = 1
remember :
the word “and” at the linear equation system two variable has mean is bound or the relation another the equation.
So, the solution of equation system this must fulfill which is the second the equation.

Exercise 4.4
1.to obvious each this pronouncement true of false !
If SPLDV is x – y = 2 and x+ y = 5 so, the solution of SPLDV is (1,3)
2. try with understand the value x and y, determine solution SPLDV of
a. 2 x + 3 y = 7
3 x – 2 y = 4
b. x + 5 y = 8
2 x + y = 7
3. price 3 apples and 2 manggo is Rp 8.000,00. price 2 apples and 3 manggo is Rp. 7.000,00. how price are 1 apple and mango ? to working base on your way !

3. determining the solution SPLDV
there some the way for finish an the linear equation system with two variables, it is as the following!
a. finishing equation with the method substitute
finish equation with other variable. For more clearly, to attention example following!
Example :
1.to finish the equation system following with the method substitute!
2 x + y = 5 and substituted to y = 3 x
Solution :
Y = 3 x, equation 2 x + y = 5
Result’s can to obtained as following
2 x + y = 5
2 x + 3 y = 5
5 x = 5
X = 1
Because x = 1 so value 1 can substituted at y = 3 x. result’s :
Y = 3 x
Y = 3 (1)
Y = 3
So, root solution’s = (1,3)
2.to determine solution set the equation follow with the method substitute !
X – y - 1 = 0 and 3 x + 2 y = 8
Solution :
For this problem we takes form :
x – y -1 = 0
x – 1 = y or y = x – 1
further more, to substitute y = x -1 at 3 x + 2 y = 8, be obtained :
3 x + 2 y = 8
3 x + 2 (x – 1 ) = 8
3 x + 2 x – 2 = 8
5 x – 2 = 8
5 x = 10
X = 2
If x = 2 substituted at y = x – 1, so be obtained :
Y = x – 1
y = 2 – 1
y = 1
so, root solution’s (2,1)

exercise 4.5
to determine solution the equation system with the method substitute following !
1. x = 2 y, 2 x – y = 3
2. y = x – 1, x + 2 y = 7
3. 3 x – y = 1, 2 x – y = 0
4. x – 2 y = 2, x + 2 y – 6 = 0
5. y = 2 x + 1, 4 x – y = 1
6. x + 2 y = 4, 2 x – 3 y -1 = 0
7. 3 x + 2 y = 12, 2 x – 3 y – 1 = 0
8 . (1/3) x + (1/2) y = 3, (2/3) x – (1/4) y = 1
b. finish the equation with the method substitute
eliminate means’s the cause. Finish the equation system with the method substitute eliminate is the cause one fault obtain value variable.

Further more, for understand solution the equation system with the method eliminate to attention example following !
Example :
1. with the method eliminate, to finish the equation system following !
x + y = 5 and x – y = 1
solution :
the second equation arranged vertical, letter to loss one fault variable’s with count up or subtract. Until be obtained :
with add
x + y = 5
for x = 3 so, x + y = 5
3 + y =5
Y = 2
So, root solution’s (3,2)

With subtract
x + y = 5
for y = 2, so x + 5
x + 2 = 5
x = 3
so root solution’s (3,2)

2. to finish equation with the method eliminate following !
2 x – 3 y = 3 time 1 2 x – 3 y = 3
3 x + y = 10 time 3 9 x + 3 y = 30
So x = 3
For x = 3 so be obtained :
2 x – 3 y = 3
2 (3) – 3 y = 3
6 – 3 y = 3
- 3 y = - 3
Y = 1
So, root solution’s (3.1)

Exercise 4.6
To determine solution the equation system with the method eliminate following !
1. x + y = 4 and 2 x – y = 5
2. 3 x + 2 y = 13 and 2 x + y = 8
3. x + 2 y = - 1 and 2 x – y = 3
4. 3 x + 4 y + 2 = 0 and x + 3 y – 1 = 0
5. 5 x – 2 y = 4 and 3 x = 4 y – 6
6. 4 x – 3 y – 11 = 0 and 3 x – 2 y – 8 = 0
7. 2 x + 91/3) y = 5 and (1/2) x – y = - 2
8. (x – 2 )/3 + y = 2 and (y + 1)/2 + x = 6
c. finish equation system with the method graph


Finish equation system with the method graph is drawing graph the second equation at one draw
At coordinate plane, and the point coordinate graph the second equation this form solution’s.
For than explain, to attention example following !
Solution
The way with as has explain in the lesson drawing graph it is with take some example the fulfill equation x + y = 3 and x – y = 1.
Remember !
Determine solution with the method graph rather difficult if point piece not round number. From picture 4.1 measles the second graph equation mentioned have the looks of in point (2.1). so, root solution (2.1)

The draft from this third method can be found when you study the linear equation system more of two variable, matriks, and the quadrate equation.

Exercise 4.7
To determine solution of the equation system with the method graph following.
1. x = 2, y = 3
2. y = x, y = 2
3. y = x + 1, y = 2 x
4. x + y = 3, x + 2 y = 4
5. x + y = 0, 2 x + y = 1
6. x + 2 y = 8, 2 x – y = 3
7. 3 x + 2 y = 8, 2 x – y = 3
8. 2 x – y = 5, x – 2 y = 1
To try !
To make graph of SPLDV
2 x + y = 6
x – y = 1
according to graph, to determine root solution of SPLDV

4.finish story the problem interrelation with system.
Example :
Price two a pencil and a book is Rp. 1.000,00. if price a book Rp. 4000,00 more expensive of price a pencil, how price is a pencil a book ?
Solution
a. understand problem
be cognizant of price two pencil and one book is Rp 1.000,00. price book Rp 4.000,00 more expensive of price a pencil
question : price one pencil and price one book.
b. Arranging planning
The draft used is solution linear equation system two variable using the method substitute.
c.perform planning
let, price a pencil is x and a book is y, be obtained until linear equation system with two variable as following. 2 x + y = 1000 and y = x + 400. for the finish this equation system, to use the method substitute. y = x + 400 substitute at 2 x + y =1000 be obtained :
2 x + y = 1000
2 x (x + 400) =1000
2 x + x + 400 =1000
3 x + 400 = 1000
3 x = 600
x = 200
for x = 200 so, y = 200 + 400
y = 600
so, price a pencil Rp 200,00 and a book Rp 600,00
d. examination
for x = 200 and y = 600, so
2 x + y = 1.000,00
2 (200) + 600 = 1000 (proven)

y = x + 400
600 = 200 + 400
600 = 600 (proven)

Exercise 4.8
1. if add two number is and different is 1, to determine
2. two a corner interspace straight. If different the second corner is 20 degree, to determine each corner’s
3. a certain point (2,3) and (4,5) exhaust at y = m x + c with two variable m and c, to determine m and c.
4. the cost wash two dress and three wastern-style shirt Rp 6.000,00. if the cost wash a dress and wastern – style shirt Rp 2.500,00, to determine each the cost to wash
5. there two corner interspace right. If one fault corner two times of other’s, to determine each the corner following

5.finish SP nonlinear two variable from to used SPLDV
In more discuss the mathematic, ordernary could nonlinear equation system. This non linear equation system more found in counting matriks and vector. For finish nonlinear equation system, change more ago to for from linear equation system. For more explain to attention this example following.
Example :
To determine solution of 6/x + 2/y = 3
3/x + 8/y = 5
Solution :
6/x + 2/y = 3 time xy 6 y + 2 x = 3 xy…..(1)
3/x + 8/y = 5 time xy 3 y + 8 x = 5 xy….. (2)
So obtained new SPLDV
- 14 x = - 7 xy
Substitute y = z to for (1) , be obtained :
6 y + 2 x = 3 xy
6(2) + 2 x = 3 x (2)
12 + 2 x = 6 x
12 = 6 x – 2 x
12 = 4 x
x = 12/4
x = 3
so root solution is (3,2)

exercise 4.9
to determine solution of SPLDV following
1. 2/x + 3/y = 2
6/x – 3/y = 2
2. 9/2x + 4/y = 5
3/x + 8/y = 6
3. 3/x – 2/y = 2
4/x + 2/y = 5
4. 5/x + 3/2x = 6
3/x + 6/y = 7

SUMMARY

1. linear equation with two variable is equation with two variable the rank one highest.
2. solution SPLDV or root SPLDV is change variable until equation at SPLDV following become the sentence true.
3. the way determine root solution SPLDV, it is :
a. substitute
b. eliminate
c. graph
Reflection
From discuss SPLDV at these chapter, which are you more understand of the third method the root determine SPLDV or solution or solution SPLDV ?
Which aren’t you understand ?
What is problem ?
Can you do inlaid work this the method