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

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

Datrys ar gyfer x, y, z, a

$$a=1$$

Camau Ateb

Ystyriwch yr hafaliad cyntaf. Defnyddio'r theorem binomaidd \left(a+b\right)^{2}=a^{2}+2ab+b^{2} i ehangu'r \left(x+1\right)^{2}.

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

Tynnu x^{2} o'r ddwy ochr.

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

Cyfuno x^{2} a -x^{2} i gael 0.

$$3x=2x+1$$

Tynnu 2x o'r ddwy ochr.

$$3x-2x=1$$

Cyfuno 3x a -2x i gael x.

$$x=1$$

Mae’r system wedi’i datrys nawr.

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