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

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

Solvi għal x, y, z, a

$$a=1$$

Passi tas-Soluzzjonijiet

Ikkunsidra l-ewwel ekwazzjoni. Uża teorema binomjali \left(a+b\right)^{2}=a^{2}+2ab+b^{2} biex tespandi \left(x+1\right)^{2}.

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

Naqqas x^{2} miż-żewġ naħat.

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

Ikkombina x^{2} u -x^{2} biex tikseb 0.

$$3x=2x+1$$

Naqqas 2x miż-żewġ naħat.

$$3x-2x=1$$

Ikkombina 3x u -2x biex tikseb x.

$$x=1$$

Is-sistema issa solvuta.

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