Jawaban PTS Math Ganjil Kelas X
Ruben Valentino Hasibuan/24/ X IPA 1
1.|2x-7| = 3
2x - 7 = 3 , 2x - 7 = -3
x = 5 , x = 2
2. | 2x - 1 | = | x + 4 |
(2x-1)² = (x+4)²
(2x-1)² - (x+4)² = 0
(2x-1 + x+4)(2x-1 - (x+4)) = 0
(3x + 3)(x - 5) = 0
3x + 3 = 0 atau x - 5 = 0
3x = -3 x = 5
x = -3 : 3
x = -1
3. | x + 7 / 2x - 1 | = 2
x + 7 / 2x -1 = 2 , x + 7 / 2x - 1 = -2
x = 3 , x = -1
4. 2 | -2x -2 | -3 = 13
2 |-2x -2 | = 13 + 3
| -2x -2 | = 16/2
| -2x -2 | = 8
-2x -2 = 8 , -2x -2 = -8
x = -5 , x = 3
5. | 4x + 2 | ≥ 6
4x + 2 ≥ 6 , 4x + 2 ≤ -6
x ≥ 1 , x ≤ -2
6. |3x+2|²+|3x+2|-2=0
p² + p – 2 = 0
(p+2)(p-1) = 0
p + 2 = 0
p = -2 (nilai mutlak tidak negatif )
atau
p-1 = 0
p = 1
|3x + 2| = 1
=> 3x + 2 = 1
3x = 1 - 2
3x = -1
x = -1/3
=> -(3x + 2) = 1
3x + 2 = -1
3x = -1-2
3x = -3
x = -1
7. | 2x -2 | ^2 > 6 | 2x -1 | + 7
(2x-1)^2 > 6 |2x -1| +7
(2x-1)^2 -6 |2x-1| > 7
(2x-1)^2 -6(2x-1) > 7 , (2x-1)^2 -6(-(2x-1)) > 7
x > 4 , x < -3
8.| x -1 |^2 + 2 | x -1 | -3 = 0
( | x-1 | -1 )( | x -1 | +3) = 0
| x-1 | -1 = 0 , | x-1| +3 = 0
x = 2 , x = 0
9.x + 3 < 0
x < 3
x + 3 < 2
x < 2 -3
x < -1
x + 3 = 0
x > -3
(3x -1) / (x+3) > 2
3x - 1 > 2(x +3)
3x -1 > 2x + 6
3x - 2x > 6 + 1
x > 7
{ x > 7 , -3 < x < -1, x < 3 }
10.|x + 2| < |x + 3|
(x + 2)² < (x + 3)²
(x + 2)² - (x + 3)² < 0
(x + 2 + x + 3)(x + 2 - x - 3) < 0
(2x + 5) (-1) < 0
2x + 5 > 0
2x > -5
x > -5/2
11.| 3x +2 | ≤ 5
= -5 ≤ 3x + 2 ≤ 5
= -5 -2 ≤ 3x ≤ 5 -2
= -7 ≤ 3x ≤ 3
= -7/3 ≤ x ≤ 3/3
= -7/3 ≤ x ≤ 1 ( Jadi x > -7/3 atau x ≤ 1)
12.misal = jarak minimarket baru ke mini market b = x
Jarak minimarket B = 50 km posisi minimarket c
| x - 70 | > 30
a) x - 70 < -30
x < 50
b) x - 70 > 30
x > 100
jadi agar jarak minimarket baru ke minimarket b > 30 km maka minimarket baru harus didirikan sebelum km ke 50 atau di daerah setelah kilometer ke 100
13.|x – 5| = 1
x – 5 = ± 1
x – 5 = 1 atau x – 5 = –1
x = 5 + 1 x = 5 – 1
x = 6
x =4
14.
15.x-10y=23 X1
6x-10y=38
x=-15/-5
x=3
3-10y=23
-10y=23-3
-10y=20
y=20/-10
y=-2
16.
17.
18.
19.
20.x + y = 110 (1)
x = z – 22 (2)
y = 5/6 z
(3)
Substitusikan persamaan (2) dan (3) ke persamaan (1),
diperoleh:
x + y = 110
↔ (z – 22) + 5/6 z = 110
↔ z – 22 + 5/6 z = 110
↔ 11/6 z = 110 + 22
↔ z = 6/11 × 132 = 72
Substitusikan nilai z = 72 ke persamaan (2) dan (3),
diperoleh:
x = z – 22 = 72 – 22 = 50
y = 5/6
z = 5/6 (72) = 60
21.x + y + z = -6 x + y + z = -6
x + y - 2z = 3 x - 2y + z = 9
______________ - _______________-
3z = -9 3y = -15
z = -3 y = -5
x + y + z = -6
x - 5 -3 = -6
x - 0 = -6
x= 2
HP = {2, -5, -3 }
22. 3x + y + 2z = 17200 | x 2 | 6x + 2y + 4z = 34400
2x + 2y + 3z = 19700 | x 1 | 2x + 2y + 3z = 19700
__________________ -
4x + z = 14 700
2x + 2y + 3z =19700 4x + z = 14700
x + 2y + 2z = 14000 x + z = 5700
__________________ - ________________ -
x + 2 = 5700 3x = 9000 x = 3000
x + z = 5700 3x + y + 2z = 17200
3000 + z = 5700 9000 + y + 5400 = 17200
z = 2700 y + 14400 = 17200
y = 2800
HP= {Buku = 300 , spidol = 2800 , tinta = 2700 }
23. 6z = 3
z = 3/6
z= 1/2
3y - 2 = -5
3t + d= 131.000
3d + 2u = 360.000
3d + 2u = 360.000
2u = 360.000 - 3d
u = 180.000 - 3/2 d 3t + d = 131.000
3t = 131.000 - d
t = 131.000/3 - d/3
5t + 2d + u = 305.000
5 ( 131.000/3 - d/3) + 2d + 180.000 - 3/2 d = 305.000
655.000/3 - 5/3 d + 2d + 180.000 - 3/2 d = 305.000 (dikali 6)
1.310.000 - 10 d + 12d + 1.080.000 - 9d = 1.830.000
- 7 d = 1.830.000 - 1.310.000 - 1.080.000
- 7 d = - 560.000
d= 80.000
akan diperoleh harga telur = t = 17.000/kg, daging = d =
80.000/kg dan udang = u = 60.000/kg
2a+b = c … (2)
a+b-c = -1 … (3)
a+b+c = 11
a+b-c = -1
menjadi
2c = 12
c = 6
2a+b = 6
2a = 6 - b
a = 6 - b/2 …(4)
a+b-c = -1
(6 - b/2) + b - 6 = -1
6 - b + 2b - 12/2 = -1
-6 + b = -2 (dpt -2 karena 2 di ruas kiri pembagian pindah ke
ruas kanan menjadi perkalian)
b = -2+6
b = 4
a = 6 - b/2
a = 6 - 4/2
a = 2/2
a = 1
|x – 5| = 1
x – 5 = ± 1
x – 5 = 1 atau x – 5 = –1
x = 5 + 1 x = 5 – 1
x = 6
x =4
Maaf bu ada sebagian nomor yang belum dikerjakan karena saya kurang mengerti dan belum menemukan jawabannya bu.
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