Whakaoti i te {l}{3*x+x^2=(x+1)^2}{y=1}{z=y}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { 3 \cdot x + x ^ {2} = {(x + 1)} ^ {2} }\\ { y = 1 }\\ { z = y }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Whakaoti mō x, y, z, a

$$a=1$$

Ngā Upane Otinga

Whakaarohia te whārite tuatahi. Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+1\right)^{2}.

$$3x+x^{2}=x^{2}+2x+1$$

Tangohia te x^{2} mai i ngā taha e rua.

$$3x+x^{2}-x^{2}=2x+1$$

Pahekotia te x^{2} me -x^{2}, ka 0.

$$3x=2x+1$$

Tangohia te 2x mai i ngā taha e rua.

$$3x-2x=1$$

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

$$x=1$$

Kua oti te pūnaha te whakatau.

$$x=1 y=1 z=1 a=1$$