SOAL DAN PEMBAHASAN BUKU SISWA MATEMATIKA KLS 8 TH. 2019 HAL. 145
garis lurus ?
a. x + 3y = 0
b. x2 + 2y = 5
c. 3y + 3x = 32
d. y/3 + 3x = 12
f. y2 + x2 = 12
Pembahasan:
Titik potong dengan sumbu-X maka y = 0
x + 3y = 0
x + 3(0) = 0
x + 0 = 0
x = 0
Jadi, titik potong sumbu-X adalah (0, 0)
Titik potong dengan sumbu-Y maka x = 0
x + 3y = 0
0 + 3y = 0
3y = 0 – 0
3y = 0
y = 0
Jadi, titik potong sumbu-Y adalah (0, 0)
Jika x = -3 maka x + 3y = 0
-3 + 3y = 0
3y = 3
y = 1 sehingga koordinatnya (-3 , 1)
Jika titik (0, 0) dan (-3, 1) dihubungkan maka terbentuklah garis lurus dari persamaan
x + 3y = 0 seperti gambar berikut ini
Jadi persamaan x + 3y = 0 termasuk persamaan garis lurus
b. x2 + 2y = 5
|
x
|
-2
|
-1
|
0
|
1
|
2
|
|
y
|
0,5
|
2
|
2,5
|
2
|
0,5
|
|
(x , y)
|
(-2 , 0,5)
|
(-1 , 2)
|
(0 , 2,5)
|
(1 , 2)
|
(2 , 0,5)
|
Jika x = -2 maka x2 + 2y = 5
(-2)2 + 2y = 5
4 + 2y = 5
2y = 5 – 4
2y = 1
y = ½ = 0,5
Jika x = -1 maka x2 + 2y = 5
(-1)2 + 2y = 5
1 + 2y = 5
2y = 5 – 1
2y = 4
y = 4/2 = 2
Jika x = 0 maka x2 + 2y = 5
02 + 2y = 5
2y = 5
y = 5/2 = 2,5
Jika x = 1 maka x2 + 2y = 5
12 + 2y = 5
1 + 2y = 5
2y = 5 – 1
2y = 4
y = 4/2 = 2
Jika x = 2 maka x2 + 2y = 5
22 + 2y = 5
4+ 2y = 5
2y = 5 – 4
2y = 1
y = 1/2
Jadi persamaan x2 + 2y = 5 bukan persamaan garis lurus
c. 3y + 3x = 32
|
x
|
-2
|
-1
|
0
|
1
|
2
|
|
y
|
5
|
4
|
3
|
2
|
1
|
|
(x , y)
|
(-2 , 5)
|
(-1 , 4)
|
(0 , 3)
|
(1 , 2)
|
(2 , 0,5)
|
Jika x = -2 maka 3y + 3x = 32
3y + 3(-2) = 9
3y – 6 = 9
3y = 9 + 6
3y = 15
y = 15/3 = 5
Jika x = -1 maka 3y + 3x = 32
3y + 3(-1) = 9
3y – 3 = 9
3y = 9 + 3
3y = 12
y = 12/3 = 4
Jika x = 0 maka 3y + 3x = 32
3y + 3(0) = 9
3y + 0 = 9
3y = 9
y = 9/3 = 3
Jika x = 1 maka 3y + 3x = 32
3y + 3(1) = 9
3y + 3 = 9
3y = 9 – 3
3y = 6
y = 6/3 = 2
Jika x = 2 maka 3y + 3x = 32
3y + 3(2) = 9
3y + 6 = 9
3y = 9 – 6
3y = 3
y = 3/3 = 1
Jadi persamaan 3y + 3x = 32 termasuk persamaan garis lurus
d. y/3 + 3x = 12
|
x
|
-2
|
-1
|
0
|
1
|
2
|
|
y
|
54
|
45
|
36
|
27
|
18
|
|
(x , y)
|
(-2 , 54)
|
(-1 , 45)
|
(0 , 36)
|
(1 , 27)
|
(2 , 18)
|
y/3 + 3(-2) = 12
y/3 – 6 = 12
y/3 = 12 + 6
y/3 = 18
y = 18 x 3 = 54
Jika x = -1 maka y/3 + 3x = 12
y/3 + 3(-1) = 12
y/3 – 3 = 12
y/3 = 12 + 3
y/3 = 15
y = 15 x 3 = 45
Jika x = 0 maka y/3 + 3x = 12
y/3 + 3(0) = 12
y/3 + 0 = 12
y/3 = 12
y = 12 x 3 = 36
Jika x = 1 maka y/3 + 3x = 12
y/3 + 3(1) = 12
y/3 + 3 = 12
y/3 = 12 – 3
y/3 = 9
y = 9 x 3 = 27
Jika x = 2 maka y/3 + 3x = 12
y/3 + 3(2) = 12
y/3 + 6 = 12
y/3 = 12 – 6
y/3 = 6
y = 6 x 3 = 18
Jadi persamaan y/3 + 3x = 12 termasuk persamaan garis lurus
f. y2 + x2 = 12
|
x
|
-2
|
-1
|
0
|
1
|
2
|
|
y
|
2,8
|
3,3
|
3,5
|
3,3
|
2,8
|
|
(x , y)
|
(-2 , 2,8)
|
(-1 , 3,3)
|
(0 , 3,5)
|
(1 , 3.3)
|
(2 , 2,8)
|
Jika x = -2 maka y2 + x2 = 12
y2 + (-2)2 = 12
y2 + 4 = 12
y2 = 12 – 4
y2 = 8
y = √8
y = 2,8
Jika x = -1 maka y2 + x2 = 12
y2 + (-1)2 = 12
y2 + 1 = 12
y2 = 12 – 1
y2 = 11
y = √11
y = 3,3
Jika x = 0 maka y2 + x2 = 12
y2 + 02 = 12
y2 + 0 = 12
y2 = 12
y = √12
y = 3,5
Jika x = 1 maka y2 + x2 = 12
y2 + 12 = 12
y2 + 1 = 12
y2 = 12 – 1
y = √11
y = 3,3
Jika x = 2 maka y2 + x2 = 12
y2 + (2)2 = 12
y2 + 4 = 12
y2 = 12 – 4
y2 = 8
y = √8
y = 2,8
Jika x = 3 maka y2 + x2 = 12
y2 + (3)2 = 12
y2 + 9 = 12
y2 = 12 – 9
y2 = 3
y = √3
y = 1,7
Jadi persamaan y2 + x2 = 12 bukan persamaan garis lurus
2. Diketahui persamaan garis 2y = 3x – 6 lengkapilah tabel berikut
Pembahasan:
2y = 3x – 6
|
x
|
-4
|
-2
|
0
|
2
|
4
|
6
|
|
y
|
-9
|
-6
|
-3
|
0
|
3
|
6
|
|
(x , y)
|
(-4 , -9)
|
(-2 , -6)
|
(0 , -3)
|
(2 , 0)
|
(4 , 3)
|
(6 , 6)
|
2y = 3(-4) – 6
2y = -12 – 6
2y = -18
y = -18/2
y = -9
Jika x = -2 maka 2y = 3x – 6
2y = 3(-2) – 6
2y = -6 – 6
2y = -12
y = -12/2
y = -6
Jika x = 0 maka 2y = 3x – 6
2y = 3(0) – 6
2y = 0 – 6
2y = -6
y = -6/2
y = -3
Jika x = 2 maka 2y = 3x – 6
2y = 3(2) – 6
2y = 6 – 6
2y = 0
y = 0
Jika x = 4 maka 2y = 3x – 6
2y = 3(4) – 6
2y = 12 – 6
2y = 6
y = 6/2
y = 3
Jika x = 6 maka 2y = 3x – 6
2y = 3(6) – 6
2y = 18 – 6
2y = 12
y = 12/2
y = 6
3. Gambarlah garis yang memiliki persamaan berikut.
a. 2x = 6y
b. 3x – 4 = 4y
d. y + 3x – 4 = 0
Pembahasan:
a. 2x = 6y
|
x
|
-2
|
-1
|
0
|
1
|
2
|
|
y
|
-0,7
|
-0,3
|
0
|
0,3
|
0,7
|
|
(x , y)
|
(-2 , -0,7)
|
(-1 , -0,3)
|
(0 , 0)
|
(1 , 0,3)
|
(2 , 0,7)
|
Jika x = -2 maka 2x = 6y
2(-2) = 6y
-4 = 6y
y = -4/6 = -0,7
Jika x = -1 maka 2x = 6y
2(-1) = 6y
-2 = 6y
y = -2/6 = -0,3
Jika x = 0 maka 2x = 6y
2(0) = 6y
0 = 6y
y = 0
Jika x = 1 maka 2x = 6y
2(1) = 6y
2 = 6y
y = 2/6 = 0,3
Jika x = 2 maka 2x = 6y
2(2) = 6y
4 = 6y
y = 4/6 = 0,7
b. 3x – 4 = 4y
|
x
|
0
|
1,3
|
|
y
|
-1
|
0
|
|
(x , y)
|
(0 , -1)
|
(1,3 , 0)
|
Jika x = 0 maka 3x – 4 = 4y
3(0) – 4 = 4y
0 – 4 = 4y
– 4 = 4y
y = -4/4
y = -1
Jika y = 0 maka 3x – 4 = 4y
3x – 4 = 0
3x = 0 + 4
3x = 4
x = 4/3 = 1,3
c. 4x + 2y = 6
|
x
|
0
|
1,5
|
|
y
|
3
|
0
|
|
(x , y)
|
(0 , 3)
|
(1,5 , 0)
|
Jika x = 0 maka 4x + 2y = 6
4(0) + 2y = 6
0 + 2y = 6
2y = 6
y = 6/2 = 3
Jika y = 0 maka 4x + 2y = 6
4x + 2(0) = 6
4x + 0 = 6
4x = 6
x = 6/4 = 1,5
