Solusi - Pertidaksamaan linear dengan satu variabel tidak diketahui

$$\left(\frac{5x}{3}\right)+\frac{3}{4}·x>=\left(\frac{9}{4}+\frac{-3}{4}x\right)+\frac{3}{4}x$$

Kelompokkan koefisien-koefisien:

$$\left(\frac{5}{3}+\frac{3}{4}\right)x>=\left(\frac{9}{4}+\frac{-3}{4}·x\right)+\frac{3}{4}x$$

Tentukan penyebut terkecil:

$$\left(\frac{\left(5·4\right)}{\left(3·4\right)}+\frac{\left(3·3\right)}{\left(4·3\right)}\right)x>=\left(\frac{9}{4}+\frac{-3}{4}·x\right)+\frac{3}{4}x$$

Kalikan penyebut:

$$\left(\frac{\left(5·4\right)}{12}+\frac{\left(3·3\right)}{12}\right)x>=\left(\frac{9}{4}+\frac{-3}{4}·x\right)+\frac{3}{4}x$$

Kalikan pembilang:

$$\left(\frac{20}{12}+\frac{9}{12}\right)x>=\left(\frac{9}{4}+\frac{-3}{4}·x\right)+\frac{3}{4}x$$

Gabungkan pecahan:

$$\frac{\left(20+9\right)}{12}·x>=\left(\frac{9}{4}+\frac{-3}{4}·x\right)+\frac{3}{4}x$$

Gabungkan pembilang:

$$\frac{29}{12}·x>=\left(\frac{9}{4}+\frac{-3}{4}·x\right)+\frac{3}{4}x$$

Kelompokkan suku sejenis:

$$\frac{29}{12}·x>=\left(\frac{-3}{4}·x+\frac{3}{4}x\right)+\frac{9}{4}$$

Gabungkan pecahan:

$$\frac{29}{12}·x>=\frac{\left(-3+3\right)}{4}x+\frac{9}{4}$$

Gabungkan pembilang:

$$\frac{29}{12}·x>=\frac{0}{4}x+\frac{9}{4}$$

Pengurangan pembilang nol:

$$\frac{29}{12}x>=0x+\frac{9}{4}$$

Sederhanakan hitungan:

$$\frac{29}{12}x>=\frac{9}{4}$$

$$\left(\frac{29}{12}x\right)·\frac{12}{29}>=\left(\frac{9}{4}\right)·\frac{12}{29}$$

Kelompokkan suku sejenis:

$$\left(\frac{29}{12}·\frac{12}{29}\right)x>=\left(\frac{9}{4}\right)·\frac{12}{29}$$

Kalikan koefisien:

$$\frac{\left(29·12\right)}{\left(12·29\right)}x>=\left(\frac{9}{4}\right)·\frac{12}{29}$$

Sederhanakan pecahan:

$$x>=\left(\frac{9}{4}\right)·\frac{12}{29}$$

Kalikan pecahan:

$$x>=\frac{\left(9·12\right)}{\left(4·29\right)}$$

Sederhanakan hitungan:

$$x>=\frac{27}{29}$$

Penyelesaian:
$$x\ge \frac{27}{29}$$

Notasi interval:
$$\left[\frac{27}{29},\infty \right]$$