Solve x^2-4/x^2+3x-10=x+2/x+5

$$\frac { x ^ { 2 } - 4 } { x ^ { 2 } + 3 x - 10 } = \frac { x + 2 } { x + 5 }$$

Solve for x (complex solution)

$$x\in \mathrm{C}\setminus -5,2$$

Solution Steps

Variable x cannot be equal to any of the values -5,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+5\right), the least common multiple of x^{2}+3x-10,x+5.

$$x^{2}-4=\left(x-2\right)\left(x+2\right)$$

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.

$$x^{2}-4=x^{2}-4$$

Subtract x^{2} from both sides.

$$x^{2}-4-x^{2}=-4$$

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

$$-4=-4$$

Compare -4 and -4.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{C}$$

Variable x cannot be equal to any of the values -5,2.

$$x\in \mathrm{C}\setminus -5,2$$

Solve for x

$$x\in \mathrm{R}\setminus 2,-5$$

Solution Steps

Variable x cannot be equal to any of the values -5,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+5\right), the least common multiple of x^{2}+3x-10,x+5.

$$x^{2}-4=\left(x-2\right)\left(x+2\right)$$

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.

$$x^{2}-4=x^{2}-4$$

Subtract x^{2} from both sides.

$$x^{2}-4-x^{2}=-4$$

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

$$-4=-4$$

Compare -4 and -4.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{R}$$

Variable x cannot be equal to any of the values -5,2.

$$x\in \mathrm{R}\setminus -5,2$$