Whakaoti i te 1/3(x+1)-x=2/9

$$\frac { 1 } { 3 } ( x + 1 ) - x = \frac { 2 } { 9 }$$

Whakaoti mō x

$$x=\frac{1}{6}\approx 0.166666667$$

Ngā Upane mō te Whakaoti Whārite Rārangi

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

$$\frac{1}{3}x+\frac{1}{3}-x=\frac{2}{9}$$

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

$$-\frac{2}{3}x+\frac{1}{3}=\frac{2}{9}$$

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

$$-\frac{2}{3}x=\frac{2}{9}-\frac{1}{3}$$

Ko te maha noa iti rawa atu o 9 me 3 ko 9. Me tahuri \frac{2}{9} me \frac{1}{3} ki te hautau me te tautūnga 9.

$$-\frac{2}{3}x=\frac{2}{9}-\frac{3}{9}$$

Tā te mea he rite te tauraro o \frac{2}{9} me \frac{3}{9}, me tango rāua mā te tango i ō raua taurunga.

$$-\frac{2}{3}x=\frac{2-3}{9}$$

Tangohia te 3 i te 2, ka -1.

$$-\frac{2}{3}x=-\frac{1}{9}$$

Me whakarea ngā taha e rua ki te -\frac{3}{2}, te tau utu o -\frac{2}{3}.

$$x=-\frac{1}{9}\left(-\frac{3}{2}\right)$$

Me whakarea te -\frac{1}{9} ki te -\frac{3}{2} mā te whakarea taurunga ki te taurunga me te tauraro ki te tauraro.

$$x=\frac{-\left(-3\right)}{9\times 2}$$

Mahia ngā whakarea i roto i te hautanga \frac{-\left(-3\right)}{9\times 2}.

$$x=\frac{3}{18}$$

Whakahekea te hautanga \frac{3}{18} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 3.

$$x=\frac{1}{6}$$