$$\left. \begin{array} { l } { \frac { x - 1 } { x - 3 } - \frac { x ^ { 2 } - x + 2 } { x ^ { 2 } - 2 x - 3 } = \frac { 1 } { x + 1 } } \end{array} \right.$$
Solve for x (complex solution)
$x\in \mathrm{C}\setminus -1,3$
Solution Steps
Variable $x$ cannot be equal to any of the values $-1,3$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-3\right)\left(x+1\right)$, the least common multiple of $x-3,x^{2}-2x-3,x+1$.
Consider $\left(x+1\right)\left(x-1\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $1$.
$$x^{2}-1-\left(x^{2}-x+2\right)=x-3$$
To find the opposite of $x^{2}-x+2$, find the opposite of each term.
$$x^{2}-1-x^{2}+x-2=x-3$$
Combine $x^{2}$ and $-x^{2}$ to get $0$.
$$-1+x-2=x-3$$
Subtract $2$ from $-1$ to get $-3$.
$$-3+x=x-3$$
Subtract $x$ from both sides.
$$-3+x-x=-3$$
Combine $x$ and $-x$ to get $0$.
$$-3=-3$$
Compare $-3$ and $-3$.
$$\text{true}$$
This is true for any $x$.
$$x\in \mathrm{C}$$
Variable $x$ cannot be equal to any of the values $-1,3$.
Variable $x$ cannot be equal to any of the values $-1,3$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-3\right)\left(x+1\right)$, the least common multiple of $x-3,x^{2}-2x-3,x+1$.
Consider $\left(x+1\right)\left(x-1\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $1$.
$$x^{2}-1-\left(x^{2}-x+2\right)=x-3$$
To find the opposite of $x^{2}-x+2$, find the opposite of each term.
$$x^{2}-1-x^{2}+x-2=x-3$$
Combine $x^{2}$ and $-x^{2}$ to get $0$.
$$-1+x-2=x-3$$
Subtract $2$ from $-1$ to get $-3$.
$$-3+x=x-3$$
Subtract $x$ from both sides.
$$-3+x-x=-3$$
Combine $x$ and $-x$ to get $0$.
$$-3=-3$$
Compare $-3$ and $-3$.
$$\text{true}$$
This is true for any $x$.
$$x\in \mathrm{R}$$
Variable $x$ cannot be equal to any of the values $-1,3$.
