Whakaoti i te 5/6(3-x)-1/2(x-4)geq1/2

$$\frac { 5 } { 6 } ( 3 - x ) - \frac { 1 } { 2 } ( x - 4 ) \geq \frac { 1 } { 2 }$$

Whakaoti mō x

$$x\leq 3$$

Ngā Upane Otinga

Whakamahia te āhuatanga tohatoha hei whakarea te \frac{5}{6} ki te 3-x.

$$\frac{5}{6}\times 3+\frac{5}{6}\left(-1\right)x-\frac{1}{2}\left(x-4\right)\geq \frac{1}{2}$$

Tuhia te \frac{5}{6}\times 3 hei hautanga kotahi.

$$\frac{5\times 3}{6}+\frac{5}{6}\left(-1\right)x-\frac{1}{2}\left(x-4\right)\geq \frac{1}{2}$$

Whakareatia te 5 ki te 3, ka 15.

$$\frac{15}{6}+\frac{5}{6}\left(-1\right)x-\frac{1}{2}\left(x-4\right)\geq \frac{1}{2}$$

Whakahekea te hautanga \frac{15}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 3.

$$\frac{5}{2}+\frac{5}{6}\left(-1\right)x-\frac{1}{2}\left(x-4\right)\geq \frac{1}{2}$$

Whakareatia te \frac{5}{6} ki te -1, ka -\frac{5}{6}.

$$\frac{5}{2}-\frac{5}{6}x-\frac{1}{2}\left(x-4\right)\geq \frac{1}{2}$$

Whakamahia te āhuatanga tohatoha hei whakarea te -\frac{1}{2} ki te x-4.

$$\frac{5}{2}-\frac{5}{6}x-\frac{1}{2}x-\frac{1}{2}\left(-4\right)\geq \frac{1}{2}$$

Tuhia te -\frac{1}{2}\left(-4\right) hei hautanga kotahi.

$$\frac{5}{2}-\frac{5}{6}x-\frac{1}{2}x+\frac{-\left(-4\right)}{2}\geq \frac{1}{2}$$

Whakareatia te -1 ki te -4, ka 4.

$$\frac{5}{2}-\frac{5}{6}x-\frac{1}{2}x+\frac{4}{2}\geq \frac{1}{2}$$

Whakawehea te 4 ki te 2, kia riro ko 2.

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

Pahekotia te -\frac{5}{6}x me -\frac{1}{2}x, ka -\frac{4}{3}x.

$$\frac{5}{2}-\frac{4}{3}x+2\geq \frac{1}{2}$$

Me tahuri te 2 ki te hautau \frac{4}{2}.

$$\frac{5}{2}-\frac{4}{3}x+\frac{4}{2}\geq \frac{1}{2}$$

Tā te mea he rite te tauraro o \frac{5}{2} me \frac{4}{2}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\frac{5+4}{2}-\frac{4}{3}x\geq \frac{1}{2}$$

Tāpirihia te 5 ki te 4, ka 9.

$$\frac{9}{2}-\frac{4}{3}x\geq \frac{1}{2}$$

Tangohia te \frac{9}{2} mai i ngā taha e rua.

$$-\frac{4}{3}x\geq \frac{1}{2}-\frac{9}{2}$$

Tā te mea he rite te tauraro o \frac{1}{2} me \frac{9}{2}, me tango rāua mā te tango i ō raua taurunga.

$$-\frac{4}{3}x\geq \frac{1-9}{2}$$

Tangohia te 9 i te 1, ka -8.

$$-\frac{4}{3}x\geq \frac{-8}{2}$$

Whakawehea te -8 ki te 2, kia riro ko -4.

$$-\frac{4}{3}x\geq -4$$

Me whakarea ngā taha e rua ki te -\frac{3}{4}, te tau utu o -\frac{4}{3}. I te mea he tōraro a -\frac{4}{3}, ka huri te ahunga koreōrite.

$$x\leq -4\left(-\frac{3}{4}\right)$$

Whakareatia -4 ki te -\frac{3}{4}.

$$x\leq 3$$