Whakaoti i te {l}{f{(x)}={(x+3)}}{g=f{(3)}}{h=g}{i=h}{text{Solvefor}jtext{where}}{j=i}

$$\left. \begin{array} { l } { f {(x)} = {(x + 3)} }\\ { g = f {(3)} }\\ { h = g }\\ { i = h }\\ { \text{Solve for } j \text{ where} } \\ { j = i } \end{array} \right.$$

Whakaoti mō f, x, g, h, j

$$j=i$$

Ngā Upane Otinga

Whakaarohia te whārite tuawhā. Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$h=i$$

Whakaarohia te whārite tuatoru. Me kōkuhu ngā uara tāupe mōhiotia ki te whārite.

$$i=g$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$g=i$$

Whakaarohia te whārite tuarua. Me kōkuhu ngā uara tāupe mōhiotia ki te whārite.

$$i=f\times 3$$

Whakawehea ngā taha e rua ki te 3.

$$\frac{i}{3}=f$$

Whakawehea te i ki te 3, kia riro ko \frac{1}{3}i.

$$\frac{1}{3}i=f$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$f=\frac{1}{3}i$$

Whakaarohia te whārite tuatahi. Me kōkuhu ngā uara tāupe mōhiotia ki te whārite.

$$\frac{1}{3}ix=x+3$$

Tangohia te x mai i ngā taha e rua.

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

Pahekotia te \frac{1}{3}ix me -x, ka \left(-1+\frac{1}{3}i\right)x.

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

Whakawehea ngā taha e rua ki te -1+\frac{1}{3}i.

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

Me whakarea te taurunga me te tauraro o \frac{3}{-1+\frac{1}{3}i} ki te haumi hiato o te tauraro, -1-\frac{1}{3}i.

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

Mahia ngā whakarea i roto o \frac{3\left(-1-\frac{1}{3}i\right)}{\left(-1+\frac{1}{3}i\right)\left(-1-\frac{1}{3}i\right)}.

$$x=\frac{-3-i}{\frac{10}{9}}$$

Whakawehea te -3-i ki te \frac{10}{9}, kia riro ko -\frac{27}{10}-\frac{9}{10}i.

$$x=-\frac{27}{10}-\frac{9}{10}i$$

Kua oti te pūnaha te whakatau.

$$f=\frac{1}{3}i x=-\frac{27}{10}-\frac{9}{10}i g=i h=i j=i$$