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

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

Whakaoti mō a, b, c, d

$$d=-3$$

Ngā Upane Otinga

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

$$c=\left(2-1\right)\left(2-1\right)+\left(2-2\left(-1\right)\right)\left(2+3\left(-1\right)\right)$$

Whakareatia te 2-1 ki te 2-1, ka \left(2-1\right)^{2}.

$$c=\left(2-1\right)^{2}+\left(2-2\left(-1\right)\right)\left(2+3\left(-1\right)\right)$$

Tangohia te 1 i te 2, ka 1.

$$c=1^{2}+\left(2-2\left(-1\right)\right)\left(2+3\left(-1\right)\right)$$

Tātaihia te 1 mā te pū o 2, kia riro ko 1.

$$c=1+\left(2-2\left(-1\right)\right)\left(2+3\left(-1\right)\right)$$

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

$$c=1+\left(2+2\right)\left(2+3\left(-1\right)\right)$$

Tāpirihia te 2 ki te 2, ka 4.

$$c=1+4\left(2+3\left(-1\right)\right)$$

Whakareatia te 3 ki te -1, ka -3.

$$c=1+4\left(2-3\right)$$

Tangohia te 3 i te 2, ka -1.

$$c=1+4\left(-1\right)$$

Whakareatia te 4 ki te -1, ka -4.

$$c=1-4$$

Tangohia te 4 i te 1, ka -3.

$$c=-3$$

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

$$d=-3$$

Kua oti te pūnaha te whakatau.

$$a=2 b=-1 c=-3 d=-3$$