Whakaoti i te {l}{a=3a+1}{b={(a+1/a)}^2}{c=b}{text{Solvefor}dtext{where}}{d=c}

$$\left. \begin{array} { l } { a = 3 a + 1 }\\ { b = {(a + \frac{1}{a})} ^ {2} }\\ { c = b }\\ { \text{Solve for } d \text{ where} } \\ { d = c } \end{array} \right.$$

Whakaoti mō a, b, c, d

$$d = \frac{25}{4} = 6\frac{1}{4} = 6.25$$

Ngā Upane Otinga

Whakaarohia te whārite tuatahi. Tangohia te 3a mai i ngā taha e rua.

$$a-3a=1$$

Pahekotia te a me -3a, ka -2a.

$$-2a=1$$

Whakawehea ngā taha e rua ki te -2.

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

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

$$b=\left(-\frac{1}{2}+\frac{1}{-\frac{1}{2}}\right)^{2}$$

Whakawehe 1 ki te -\frac{1}{2} mā te whakarea 1 ki te tau huripoki o -\frac{1}{2}.

$$b=\left(-\frac{1}{2}+1\left(-2\right)\right)^{2}$$

Whakareatia te 1 ki te -2, ka -2.

$$b=\left(-\frac{1}{2}-2\right)^{2}$$

Tangohia te 2 i te -\frac{1}{2}, ka -\frac{5}{2}.

$$b=\left(-\frac{5}{2}\right)^{2}$$

Tātaihia te -\frac{5}{2} mā te pū o 2, kia riro ko \frac{25}{4}.

$$b=\frac{25}{4}$$

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

$$c=\frac{25}{4}$$

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

$$d=\frac{25}{4}$$

Kua oti te pūnaha te whakatau.

$$a=-\frac{1}{2} b=\frac{25}{4} c=\frac{25}{4} d=\frac{25}{4}$$