Solve f(x)=x+3/x+2+h(x)=sqrt{x-10}

$$f ( x ) = \frac { x + 3 } { x + 2 } + h ( x ) = \sqrt { x - 10 }$$

Solve for h

$$h=-\frac{-\sqrt{x-10}x+x-2\sqrt{x-10}+3}{x\left(x+2\right)}$$

$$x\geq 10$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x+2.

$$x+3+hx\left(x+2\right)=\left(x+2\right)\sqrt{x-10}$$

Use the distributive property to multiply hx by x+2.

$$x+3+hx^{2}+2hx=\left(x+2\right)\sqrt{x-10}$$

Use the distributive property to multiply x+2 by \sqrt{x-10}.

$$x+3+hx^{2}+2hx=x\sqrt{x-10}+2\sqrt{x-10}$$

Subtract x from both sides.

$$3+hx^{2}+2hx=x\sqrt{x-10}+2\sqrt{x-10}-x$$

Subtract 3 from both sides.

$$hx^{2}+2hx=x\sqrt{x-10}+2\sqrt{x-10}-x-3$$

Combine all terms containing h.

$$\left(x^{2}+2x\right)h=x\sqrt{x-10}+2\sqrt{x-10}-x-3$$

The equation is in standard form.

$$\left(x^{2}+2x\right)h=\sqrt{x-10}x-x+2\sqrt{x-10}-3$$

Divide both sides by x^{2}+2x.

$$\frac{\left(x^{2}+2x\right)h}{x^{2}+2x}=\frac{\sqrt{x-10}x-x+2\sqrt{x-10}-3}{x^{2}+2x}$$

Dividing by x^{2}+2x undoes the multiplication by x^{2}+2x.

$$h=\frac{\sqrt{x-10}x-x+2\sqrt{x-10}-3}{x^{2}+2x}$$

Divide x\sqrt{x-10}+2\sqrt{x-10}-x-3 by x^{2}+2x.

$$h=\frac{\sqrt{x-10}x-x+2\sqrt{x-10}-3}{x\left(x+2\right)}$$