Solve for x 1-1/x=-6x

$1 - \frac{1}{x} = - 6 x$

Subtract $1$ from both sides of the equation.

$- \frac{1}{x} = - 6 x - 1$

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$x, 1, 1$

The LCM of one and any expression is the expression.

$x$

Multiply each term in $- \frac{1}{x} = - 6 x - 1$ by $x$ to eliminate the fractions.

Multiply each term in $- \frac{1}{x} = - 6 x - 1$ by $x$ .

$- \frac{1}{x} x = - 6 x \cdot x - x$

Simplify the left side.

Cancel the common factor of $x$ .

Move the leading negative in $- \frac{1}{x}$ into the numerator.

$\frac{- 1}{x} x = - 6 x \cdot x - x$

Cancel the common factor.

$\frac{- 1}{x} x = - 6 x \cdot x - x$

Rewrite the expression.

$- 1 = - 6 x \cdot x - x$

Simplify the right side.

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$- 1 = - 6 (x \cdot x) - x$

Multiply $x$ by $x$ .

$- 1 = - 6 x^{2} - x$

Solve the equation.

Rewrite the equation as $- 6 x^{2} - x = - 1$ .

$- 6 x^{2} - x = - 1$

Add $1$ to both sides of the equation.

$- 6 x^{2} - x + 1 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- 6 x^{2} - x + 1$ .

Factor $- 1$ out of $- 6 x^{2}$ .

$- (6 x^{2}) - x + 1 = 0$

Factor $- 1$ out of $- x$ .

$- (6 x^{2}) - (x) + 1 = 0$

Rewrite $1$ as $- 1 (- 1)$ .

$- (6 x^{2}) - (x) - 1 \cdot - 1 = 0$

Factor $- 1$ out of $- (6 x^{2}) - (x)$ .

$- (6 x^{2} + x) - 1 \cdot - 1 = 0$

Factor $- 1$ out of $- (6 x^{2} + x) - 1 (- 1)$ .

$- (6 x^{2} + x - 1) = 0$

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 6 \cdot - 1 = - 6$ and whose sum is $b = 1$ .

Multiply by $1$ .

$- (6 x^{2} + 1 x - 1) = 0$

Rewrite $1$ as $- 2$ plus $3$

$- (6 x^{2} + (- 2 + 3) x - 1) = 0$

Apply the distributive property.

$- (6 x^{2} - 2 x + 3 x - 1) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- ((6 x^{2} - 2 x) + 3 x - 1) = 0$

Factor out the greatest common factor (GCF) from each group.

$- (2 x (3 x - 1) + 1 (3 x - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x - 1$ .

$- ((3 x - 1) (2 x + 1)) = 0$

Remove unnecessary parentheses.

$- (3 x - 1) (2 x + 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$3 x - 1 = 0$

$2 x + 1 = 0$

Set $3 x - 1$ equal to $0$ and solve for $x$ .

Set $3 x - 1$ equal to $0$ .

$3 x - 1 = 0$

Solve $3 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$3 x = 1$

Divide each term in $3 x = 1$ by $3$ and simplify.

Divide each term in $3 x = 1$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3}$

Divide $x$ by $1$ .

$x = \frac{1}{3}$

Set $2 x + 1$ equal to $0$ and solve for $x$ .

Set $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Solve $2 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Divide each term in $2 x = - 1$ by $2$ and simplify.

Divide each term in $2 x = - 1$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2}$

Divide $x$ by $1$ .

$x = \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

The final solution is all the values that make $- (3 x - 1) (2 x + 1) = 0$ true.

$x = \frac{1}{3}, - \frac{1}{2}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{3}, - \frac{1}{2}$

Decimal Form:

$x = 0.\overline{3}, - 0.5$

Solve for x 1-1/x=-6x

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