Solve x^2+14x+49/x^2-49=x+7/x-7

$$\frac { x ^ { 2 } + 14 x + 49 } { x ^ { 2 } - 49 } = \frac { x + 7 } { x - 7 }$$

Solve for x (complex solution)

$$x\in \mathrm{C}\setminus -7,7$$

Solution Steps

Variable x cannot be equal to any of the values -7,7 since division by zero is not defined. Multiply both sides of the equation by \left(x-7\right)\left(x+7\right), the least common multiple of x^{2}-49,x-7.

$$x^{2}+14x+49=\left(x+7\right)\left(x+7\right)$$

Multiply x+7 and x+7 to get \left(x+7\right)^{2}.

$$x^{2}+14x+49=\left(x+7\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+7\right)^{2}.

$$x^{2}+14x+49=x^{2}+14x+49$$

Subtract x^{2} from both sides.

$$x^{2}+14x+49-x^{2}=14x+49$$

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

$$14x+49=14x+49$$

Subtract 14x from both sides.

$$14x+49-14x=49$$

Combine 14x and -14x to get 0.

$$49=49$$

Compare 49 and 49.

$$\text{true}$$

This is true for any x.

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

Variable x cannot be equal to any of the values -7,7.

$$x\in \mathrm{C}\setminus -7,7$$

Solve for x

$$x\in \mathrm{R}\setminus 7,-7$$

Solution Steps

Variable x cannot be equal to any of the values -7,7 since division by zero is not defined. Multiply both sides of the equation by \left(x-7\right)\left(x+7\right), the least common multiple of x^{2}-49,x-7.

$$x^{2}+14x+49=\left(x+7\right)\left(x+7\right)$$

Multiply x+7 and x+7 to get \left(x+7\right)^{2}.

$$x^{2}+14x+49=\left(x+7\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+7\right)^{2}.

$$x^{2}+14x+49=x^{2}+14x+49$$

Subtract x^{2} from both sides.

$$x^{2}+14x+49-x^{2}=14x+49$$

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

$$14x+49=14x+49$$

Subtract 14x from both sides.

$$14x+49-14x=49$$

Combine 14x and -14x to get 0.

$$49=49$$

Compare 49 and 49.

$$\text{true}$$

This is true for any x.

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

Variable x cannot be equal to any of the values -7,7.

$$x\in \mathrm{R}\setminus -7,7$$