Solve {l}{3*x+x^2=(x+1)^2}{y=1}{z=y}{a=z}{text{Solvefor}btext{where}}{b=a}

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

Solve for x, y, z, a, b

$$b=1$$

Solution Steps

Consider the first equation. Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

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

Subtract x^{2} from both sides.

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

Combine x^{2} and -x^{2} to get 0.

$$3x=2x+1$$

Subtract 2x from both sides.

$$3x-2x=1$$

Combine 3x and -2x to get x.

$$x=1$$

The system is now solved.

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