Whakaoti i te 2/3*(x+1)=1/6*(x-2)

$$\frac { 2 } { 3 } \cdot ( x + 1 ) = \frac { 1 } { 6 } \cdot ( x - 2 )$$

Whakaoti mō x

$$x=-2$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te \frac{2}{3} ki te x+1.

$$\frac{2}{3}x+\frac{2}{3}=\frac{1}{6}\left(x-2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te \frac{1}{6} ki te x-2.

$$\frac{2}{3}x+\frac{2}{3}=\frac{1}{6}x+\frac{1}{6}\left(-2\right)$$

Whakareatia te \frac{1}{6} ki te -2, ka \frac{-2}{6}.

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

Whakahekea te hautanga \frac{-2}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Tangohia te \frac{1}{6}x mai i ngā taha e rua.

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

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

$$\frac{1}{2}x+\frac{2}{3}=-\frac{1}{3}$$

Tangohia te \frac{2}{3} mai i ngā taha e rua.

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

Tā te mea he rite te tauraro o -\frac{1}{3} me \frac{2}{3}, me tango rāua mā te tango i ō raua taurunga.

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

Tangohia te 2 i te -1, ka -3.

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

Whakawehea te -3 ki te 3, kia riro ko -1.

$$\frac{1}{2}x=-1$$

Me whakarea ngā taha e rua ki te 2, te tau utu o \frac{1}{2}.

$$x=-2$$