Solve x-3/x+2+x-2/x-3=2x^2-5x-6/x^2-x-6

$$\frac { x - 3 } { x + 2 } + \frac { x - 2 } { x - 3 } = \frac { 2 x ^ { 2 } - 5 x - 6 } { x ^ { 2 } - x - 6 }$$

Solve for x

$$x=11$$

Steps for Solving Linear Equation

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

$$\left(x-3\right)\left(x-3\right)+\left(x+2\right)\left(x-2\right)=2x^{2}-5x-6$$

Multiply x-3 and x-3 to get \left(x-3\right)^{2}.

$$\left(x-3\right)^{2}+\left(x+2\right)\left(x-2\right)=2x^{2}-5x-6$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9+\left(x+2\right)\left(x-2\right)=2x^{2}-5x-6$$

Consider \left(x+2\right)\left(x-2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

$$x^{2}-6x+9+x^{2}-4=2x^{2}-5x-6$$

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

$$2x^{2}-6x+9-4=2x^{2}-5x-6$$

Subtract 4 from 9 to get 5.

$$2x^{2}-6x+5=2x^{2}-5x-6$$

Subtract 2x^{2} from both sides.

$$2x^{2}-6x+5-2x^{2}=-5x-6$$

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

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

Add 5x to both sides.

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

Combine -6x and 5x to get -x.

$$-x+5=-6$$

Subtract 5 from both sides.

$$-x=-6-5$$

Subtract 5 from -6 to get -11.

$$-x=-11$$

Multiply both sides by -1.

$$x=11$$