Whakaoti i te 1/3+1/2xleq5/6x-5/3

$$\frac { 1 } { 3 } + \frac { 1 } { 2 } x \leq \frac { 5 } { 6 } x - \frac { 5 } { 3 }$$

Whakaoti mō x

$$x\geq 6$$

Ngā Upane Otinga

Tangohia te \frac{5}{6}x mai i ngā taha e rua.

$$\frac{1}{3}+\frac{1}{2}x-\frac{5}{6}x\leq -\frac{5}{3}$$

Pahekotia te \frac{1}{2}x me -\frac{5}{6}x, ka -\frac{1}{3}x.

$$\frac{1}{3}-\frac{1}{3}x\leq -\frac{5}{3}$$

Tangohia te \frac{1}{3} mai i ngā taha e rua.

$$-\frac{1}{3}x\leq -\frac{5}{3}-\frac{1}{3}$$

Tā te mea he rite te tauraro o -\frac{5}{3} me \frac{1}{3}, me tango rāua mā te tango i ō raua taurunga.

$$-\frac{1}{3}x\leq \frac{-5-1}{3}$$

Tangohia te 1 i te -5, ka -6.

$$-\frac{1}{3}x\leq \frac{-6}{3}$$

Whakawehea te -6 ki te 3, kia riro ko -2.

$$-\frac{1}{3}x\leq -2$$

Me whakarea ngā taha e rua ki te -3, te tau utu o -\frac{1}{3}. I te mea he tōraro a -\frac{1}{3}, ka huri te ahunga koreōrite.

$$x\geq -2\left(-3\right)$$

Whakareatia te -2 ki te -3, ka 6.

$$x\geq 6$$