$$\frac { - x - 1 } { x + 1 } = \frac { 9 x } { x - 5 }$$
Solve for x
$x=\frac{1}{2}=0.5$
Steps Using Factoring By Grouping
Steps Using the Quadratic Formula
Steps for Completing the Square
Steps Using Factoring By Grouping
Variable $x$ cannot be equal to any of the values $-1,5$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-5\right)\left(x+1\right)$, the least common multiple of $x+1,x-5$.
Use the distributive property to multiply $x+1$ by $9$.
$$x\left(-x\right)+4x+5=\left(9x+9\right)x$$
Use the distributive property to multiply $9x+9$ by $x$.
$$x\left(-x\right)+4x+5=9x^{2}+9x$$
Subtract $9x^{2}$ from both sides.
$$x\left(-x\right)+4x+5-9x^{2}=9x$$
Subtract $9x$ from both sides.
$$x\left(-x\right)+4x+5-9x^{2}-9x=0$$
Combine $4x$ and $-9x$ to get $-5x$.
$$x\left(-x\right)-5x+5-9x^{2}=0$$
Multiply $x$ and $x$ to get $x^{2}$.
$$x^{2}\left(-1\right)-5x+5-9x^{2}=0$$
Combine $x^{2}\left(-1\right)$ and $-9x^{2}$ to get $-10x^{2}$.
$$-10x^{2}-5x+5=0$$
Divide both sides by $5$.
$$-2x^{2}-x+1=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-2x^{2}+ax+bx+1$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-1$$ $$ab=-2=-2$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.
$$a=1$$ $$b=-2$$
Rewrite $-2x^{2}-x+1$ as $\left(-2x^{2}+x\right)+\left(-2x+1\right)$.
$$\left(-2x^{2}+x\right)+\left(-2x+1\right)$$
Factor out $-x$ in the first and $-1$ in the second group.
$$-x\left(2x-1\right)-\left(2x-1\right)$$
Factor out common term $2x-1$ by using distributive property.
$$\left(2x-1\right)\left(-x-1\right)$$
To find equation solutions, solve $2x-1=0$ and $-x-1=0$.
$$x=\frac{1}{2}$$ $$x=-1$$
Variable $x$ cannot be equal to $-1$.
$$x=\frac{1}{2}$$
Steps Using the Quadratic Formula
Variable $x$ cannot be equal to any of the values $-1,5$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-5\right)\left(x+1\right)$, the least common multiple of $x+1,x-5$.
Use the distributive property to multiply $x+1$ by $9$.
$$x\left(-x\right)+4x+5=\left(9x+9\right)x$$
Use the distributive property to multiply $9x+9$ by $x$.
$$x\left(-x\right)+4x+5=9x^{2}+9x$$
Subtract $9x^{2}$ from both sides.
$$x\left(-x\right)+4x+5-9x^{2}=9x$$
Subtract $9x$ from both sides.
$$x\left(-x\right)+4x+5-9x^{2}-9x=0$$
Combine $4x$ and $-9x$ to get $-5x$.
$$x\left(-x\right)-5x+5-9x^{2}=0$$
Multiply $x$ and $x$ to get $x^{2}$.
$$x^{2}\left(-1\right)-5x+5-9x^{2}=0$$
Combine $x^{2}\left(-1\right)$ and $-9x^{2}$ to get $-10x^{2}$.
$$-10x^{2}-5x+5=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-10$ for $a$, $-5$ for $b$, and $5$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Now solve the equation $x=\frac{5±15}{-20}$ when $±$ is plus. Add $5$ to $15$.
$$x=\frac{20}{-20}$$
Divide $20$ by $-20$.
$$x=-1$$
Now solve the equation $x=\frac{5±15}{-20}$ when $±$ is minus. Subtract $15$ from $5$.
$$x=-\frac{10}{-20}$$
Reduce the fraction $\frac{-10}{-20}$ to lowest terms by extracting and canceling out $10$.
$$x=\frac{1}{2}$$
The equation is now solved.
$$x=-1$$ $$x=\frac{1}{2}$$
Variable $x$ cannot be equal to $-1$.
$$x=\frac{1}{2}$$
Steps for Completing the Square
Variable $x$ cannot be equal to any of the values $-1,5$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-5\right)\left(x+1\right)$, the least common multiple of $x+1,x-5$.
Reduce the fraction $\frac{-5}{-10}$ to lowest terms by extracting and canceling out $5$.
$$x^{2}+\frac{1}{2}x=-\frac{5}{-10}$$
Reduce the fraction $\frac{-5}{-10}$ to lowest terms by extracting and canceling out $5$.
$$x^{2}+\frac{1}{2}x=\frac{1}{2}$$
Divide $\frac{1}{2}$, the coefficient of the $x$ term, by $2$ to get $\frac{1}{4}$. Then add the square of $\frac{1}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
Add $\frac{1}{2}$ to $\frac{1}{16}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{9}{16}$$
Factor $x^{2}+\frac{1}{2}x+\frac{1}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(x+\frac{1}{4}\right)^{2}=\frac{9}{16}$$
Take the square root of both sides of the equation.
