Whakaoti i te {l}{f{(1-x/2+x)}=1-4}{g=f{(3)}}{h=g}{i=h}{text{Solvefor}jtext{where}}{j=i}

$$\left. \begin{array} { l } { f {(\frac{1 - x}{2 + x})} = 1 - 4 }\\ { g = f {(3)} }\\ { h = g }\\ { i = h }\\ { \text{Solve for } j \text{ where} } \\ { j = i } \end{array} \right.$$

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.

$$\frac{1}{3}i\times \frac{1-x}{2+x}=1-4$$

Tē taea kia ōrite te tāupe x ki -2 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te x+2.

$$\frac{1}{3}i\left(1-x\right)=x+2+\left(x+2\right)\left(-4\right)$$

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

$$\frac{1}{3}i-\frac{1}{3}ix=x+2+\left(x+2\right)\left(-4\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te x+2 ki te -4.

$$\frac{1}{3}i-\frac{1}{3}ix=x+2-4x-8$$

Pahekotia te x me -4x, ka -3x.

$$\frac{1}{3}i-\frac{1}{3}ix=-3x+2-8$$

Tangohia te 8 i te 2, ka -6.

$$\frac{1}{3}i-\frac{1}{3}ix=-3x-6$$

Me tāpiri te 3x ki ngā taha e rua.

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

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

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

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

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

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

$$x=\frac{-6-\frac{1}{3}i}{3-\frac{1}{3}i}$$

Me whakarea te taurunga me te tauraro o \frac{-6-\frac{1}{3}i}{3-\frac{1}{3}i} ki te haumi hiato o te tauraro, 3+\frac{1}{3}i.

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

Mahia ngā whakarea i roto o \frac{\left(-6-\frac{1}{3}i\right)\left(3+\frac{1}{3}i\right)}{\left(3-\frac{1}{3}i\right)\left(3+\frac{1}{3}i\right)}.

$$x=\frac{-\frac{161}{9}-3i}{\frac{82}{9}}$$

Whakawehea te -\frac{161}{9}-3i ki te \frac{82}{9}, kia riro ko -\frac{161}{82}-\frac{27}{82}i.

$$x=-\frac{161}{82}-\frac{27}{82}i$$

Kua oti te pūnaha te whakatau.

$$f=\frac{1}{3}i x=-\frac{161}{82}-\frac{27}{82}i g=i h=i j=i$$