Question

Tentukan himpunan penyelesaian dari pertidaksamaan berikut ,jika variabel pada himpunan bilangan bulat..

1. 1per3(5y - 1)< 1per2 (2y + 1)

2. 1per2 t - 1 < 1per3(t - 4)

3. 2per3(p - 1) > 1per5 p+2

4. 1per3X + 1 ≥ 1per2(x - 1)

5. 1per3 (x + 2) > 2 + 3x per 2

6. 1per2 (x + 5) - 1per4 (x + 1) > 3


7. t -2 per4 - t - 4 per 6≤ 2per3


tolong di jwb pkai cara...pliss

Answer 1

1. 1per3(5y - 1)< 1per2 (2y + 1)
=>⅓(5y-1) < ½(2y+1)
=> 5/3y-1/3 < y + 1/2
_____________ × 6
=> 10y - 2 < 6y + 3
=> 10y-6y < 3+2
=> 4y < 5
=> y < 5/4
=> {y | y < 5/4}

2. 1per2 t - 1 < 1per3(t - 4)
=> ½t - 1 < ⅓(t-4)
=> ½t - 1 < ⅓t - 4/3
____________ × 6
=> 3t-6 < 2t - 8
=> 3t-2t < -8+6
=> t < -2
=> {t | t < -2}

3. 2per3(p - 1) > 1per5 p+2
=> ⅔(p-1) > 1/5p + 2
=> ⅔p-⅔ > 1/5p + 2
____________ × 15
=> 10p - 10 > 3p + 30
=> 10p - 3p > 30 + 10
=> 7p > 40
=> p > 40/7
=> {p | p > 40/7}

4. 1per3X + 1 ≥ 1per2(x - 1)
=> ⅓x + 1 ≥ ½(x-1)
=> ⅓x + 1 ≥ ½x-½
___________ × 6
=> 2x + 6 ≥ 3x - 3
=> 2x - 3x ≥ -3 - 6
=> -x ≥ -9
=> x ≤ 9
=> {x | x ≤ 9}

5. 1per3 (x + 2) > 2 + 3x per 2
=> ⅓(x+2) > 2 + 3x/2
=> ⅓x + ⅔ > 2 + 3/2x
______________ × 6
=> 2x + 4 > 12 + 9x
=> 2x - 9x > 12 - 4
=> -7x > 8
=> x < 8/7
=> {x | x < 8/7}

6. 1per2 (x + 5) - 1per4 (x + 1) > 3
=> ½(x+5) - ¼(x+1) > 3
=> ½x+5/2 - ¼x - ¼ > 3
_______________ × 4
=> 2x + 10 - x - 1 > 3
=> x + 9 > 3
=> x > 3 - 9
=> x > -6
=> {x | x > -6}

7. t -2 per4 - t - 4 per 6≤ 2per3
=> t-2/4 - (t-4)/6 ≤ ⅔
_____________ × 12
=> 3(t-2) - 2(t-4) ≤ 8
=> 3t-6 - 2t + 8 ≤ 8
=> t + 2 ≤ 8
=> t ≤ 6
=> {t | t ≤ 6}