Solve {l}{f(a)=2quadf(x)=x^2}{{l}{a+b=5}{a-b=-1}}

$$\left. \begin{array} { l } { f ( a ) = 2 \quad f ( x ) = x ^ { 2 } } \\ { \left\{ \begin{array} { l } { a + b = 5 } \\ { a - b = - 1 } \end{array} \right.} \end{array} \right.$$

Solve for x (complex solution)

$$\left\{\begin{matrix}\\x=\frac{a}{2}\text{, }&\text{unconditionally}\\x\in \mathrm{C}\text{, }&f=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$2fx=fa$$

The equation is in standard form.

$$2fx=af$$

Divide both sides by 2f.

$$\frac{2fx}{2f}=\frac{af}{2f}$$

Dividing by 2f undoes the multiplication by 2f.

$$x=\frac{af}{2f}$$

Divide fa by 2f.

$$x=\frac{a}{2}$$

Solve for x

$$\left\{\begin{matrix}\\x=\frac{a}{2}\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&f=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$2fx=fa$$

The equation is in standard form.

$$2fx=af$$

Divide both sides by 2f.

$$\frac{2fx}{2f}=\frac{af}{2f}$$

Dividing by 2f undoes the multiplication by 2f.

$$x=\frac{af}{2f}$$

Divide fa by 2f.

$$x=\frac{a}{2}$$