Whakaoti i te 1/2t+1/3=1/6t-1/3

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

Whakaoti mō t

$$t=-2$$

Ngā Upane mō te Whakaoti Whārite Rārangi

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

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

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

$$\frac{1}{3}t+\frac{1}{3}=-\frac{1}{3}$$

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

$$\frac{1}{3}t=-\frac{1}{3}-\frac{1}{3}$$

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

$$\frac{1}{3}t=\frac{-1-1}{3}$$

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

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

Me whakarea ngā taha e rua ki te 3, te tau utu o \frac{1}{3}.

$$t=-\frac{2}{3}\times 3$$

Me whakakore te 3 me te 3.

$$t=-2$$