Solve f(x)=x-1/3+xquadyquadg(x)=sqrt{x}

$$f ( x ) = \frac { x - 1 } { 3 + x } \quad y \quad g ( x ) = \sqrt { x }$$

Solve for g

$$\left\{\begin{matrix}g=\frac{x+3}{\sqrt{x}y\left(x-1\right)}\text{, }&x\neq 1\text{ and }y\neq 0\text{ and }x>0\\g\in \mathrm{R}\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x+3.

$$\left(x-1\right)ygx=\left(x+3\right)\sqrt{x}$$

Use the distributive property to multiply x-1 by y.

$$\left(xy-y\right)gx=\left(x+3\right)\sqrt{x}$$

Use the distributive property to multiply xy-y by g.

$$\left(xyg-yg\right)x=\left(x+3\right)\sqrt{x}$$

Use the distributive property to multiply xyg-yg by x.

$$ygx^{2}-ygx=\left(x+3\right)\sqrt{x}$$

Use the distributive property to multiply x+3 by \sqrt{x}.

$$ygx^{2}-ygx=x\sqrt{x}+3\sqrt{x}$$

Combine all terms containing g.

$$\left(yx^{2}-yx\right)g=x\sqrt{x}+3\sqrt{x}$$

The equation is in standard form.

$$\left(yx^{2}-xy\right)g=\sqrt{x}x+3\sqrt{x}$$

Divide both sides by yx^{2}-yx.

$$\frac{\left(yx^{2}-xy\right)g}{yx^{2}-xy}=\frac{\sqrt{x}\left(x+3\right)}{yx^{2}-xy}$$

Dividing by yx^{2}-yx undoes the multiplication by yx^{2}-yx.

$$g=\frac{\sqrt{x}\left(x+3\right)}{yx^{2}-xy}$$

Divide \left(3+x\right)\sqrt{x} by yx^{2}-yx.

$$g=\frac{x+3}{\sqrt{x}y\left(x-1\right)}$$