Solve f(x)=2x-1h(x)=-5/x-1

$$f ( x ) = 2 x - 1 h ( x ) = - \frac { 5 } { x - 1 }$$

Solve for h

$$h=-\frac{2x^{2}-2x+5}{x\left(1-x\right)}$$

$$x\neq 1\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x-1.

$$2x\left(x-1\right)-hx\left(x-1\right)=-5$$

Use the distributive property to multiply 2x by x-1.

$$2x^{2}-2x-hx\left(x-1\right)=-5$$

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

$$2x^{2}-2x-hx^{2}+xh=-5$$

Subtract 2x^{2} from both sides.

$$-2x-hx^{2}+xh=-5-2x^{2}$$

Add 2x to both sides.

$$-hx^{2}+xh=-5-2x^{2}+2x$$

Combine all terms containing h.

$$\left(-x^{2}+x\right)h=-5-2x^{2}+2x$$

The equation is in standard form.

$$\left(x-x^{2}\right)h=-2x^{2}+2x-5$$

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

$$\frac{\left(x-x^{2}\right)h}{x-x^{2}}=\frac{-2x^{2}+2x-5}{x-x^{2}}$$

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

$$h=\frac{-2x^{2}+2x-5}{x-x^{2}}$$

Divide -5-2x^{2}+2x by -x^{2}+x.

$$h=\frac{-2x^{2}+2x-5}{x\left(1-x\right)}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{-\left(2-h\right)\left(h+18\right)}-h+2}{2\left(2-h\right)}$$

$$x=\frac{-\sqrt{-\left(2-h\right)\left(h+18\right)}-h+2}{2\left(2-h\right)}\text{, }h\neq 2$$

Solve for x

$$x=\frac{\sqrt{-\left(2-h\right)\left(h+18\right)}-h+2}{2\left(2-h\right)}$$

$$x=\frac{-\sqrt{-\left(2-h\right)\left(h+18\right)}-h+2}{2\left(2-h\right)}\text{, }h>2\text{ or }h\leq -18$$