Solve 1/x^2-x+x/x^2+2x-3=2x+9/2x^2+6x

$$\frac{ 1 }{ { x }^{ 2 } -x } + \frac{ x }{ { x }^{ 2 } +2x-3 } = \frac{ 2x+9 }{ 2 { x }^{ 2 } +6x }$$

Solve for x

$$x=3$$

Steps for Solving Linear Equation

Variable x cannot be equal to any of the values -3,0,1 since division by zero is not defined. Multiply both sides of the equation by 2x\left(x-1\right)\left(x+3\right), the least common multiple of x^{2}-x,x^{2}+2x-3,2x^{2}+6x.

$$2x+6+2xx=\left(x-1\right)\left(2x+9\right)$$

Multiply x and x to get x^{2}.

$$2x+6+2x^{2}=\left(x-1\right)\left(2x+9\right)$$

Use the distributive property to multiply x-1 by 2x+9 and combine like terms.

$$2x+6+2x^{2}=2x^{2}+7x-9$$

Subtract 2x^{2} from both sides.

$$2x+6+2x^{2}-2x^{2}=7x-9$$

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

$$2x+6=7x-9$$

Subtract 7x from both sides.

$$2x+6-7x=-9$$

Combine 2x and -7x to get -5x.

$$-5x+6=-9$$

Subtract 6 from both sides.

$$-5x=-9-6$$

Subtract 6 from -9 to get -15.

$$-5x=-15$$

Divide both sides by -5.

$$x=\frac{-15}{-5}$$

Divide -15 by -5 to get 3.

$$x=3$$