Solve for x -4-1/x=-5x

$- 4 - \frac{1}{x} = - 5 x$

Add $4$ to both sides of the equation.

$- \frac{1}{x} = - 5 x + 4$

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 $- \frac{1}{x} = - 5 x + 4$ by $x$ to eliminate the fractions.

Multiply each term in $- \frac{1}{x} = - 5 x + 4$ by $x$ .

$- \frac{1}{x} x = - 5 x \cdot x + 4 x$

Simplify the left side.

Cancel the common factor of $x$ .

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

$\frac{- 1}{x} x = - 5 x \cdot x + 4 x$

Cancel the common factor.

$\frac{- 1}{x} x = - 5 x \cdot x + 4 x$

Rewrite the expression.

$- 1 = - 5 x \cdot x + 4 x$

Simplify the right side.

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

Move $x$ .

$- 1 = - 5 (x \cdot x) + 4 x$

Multiply $x$ by $x$ .

$- 1 = - 5 x^{2} + 4 x$

Solve the equation.

Rewrite the equation as $- 5 x^{2} + 4 x = - 1$ .

$- 5 x^{2} + 4 x = - 1$

Add $1$ to both sides of the equation.

$- 5 x^{2} + 4 x + 1 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- 5 x^{2} + 4 x + 1$ .

Factor $- 1$ out of $- 5 x^{2}$ .

$- (5 x^{2}) + 4 x + 1 = 0$

Factor $- 1$ out of $4 x$ .

$- (5 x^{2}) - (- 4 x) + 1 = 0$

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

$- (5 x^{2}) - (- 4 x) - 1 \cdot - 1 = 0$

Factor $- 1$ out of $- (5 x^{2}) - (- 4 x)$ .

$- (5 x^{2} - 4 x) - 1 \cdot - 1 = 0$

Factor $- 1$ out of $- (5 x^{2} - 4 x) - 1 (- 1)$ .

$- (5 x^{2} - 4 x - 1) = 0$

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 $a \cdot c = 5 \cdot - 1 = - 5$ and whose sum is $b = - 4$ .

Factor $- 4$ out of $- 4 x$ .

$- (5 x^{2} - 4 x - 1) = 0$

Rewrite $- 4$ as $1$ plus $- 5$

$- (5 x^{2} + (1 - 5) x - 1) = 0$

Apply the distributive property.

$- (5 x^{2} + 1 x - 5 x - 1) = 0$

Multiply $x$ by $1$ .

$- (5 x^{2} + x - 5 x - 1) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- ((5 x^{2} + x) - 5 x - 1) = 0$

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

$- (x (5 x + 1) - (5 x + 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $5 x + 1$ .

$- ((5 x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$- (5 x + 1) (x - 1) = 0$

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

$5 x + 1 = 0$

$x - 1 = 0$

Set $5 x + 1$ equal to $0$ and solve for $x$ .

Set $5 x + 1$ equal to $0$ .

$5 x + 1 = 0$

Solve $5 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$5 x = - 1$

Divide each term in $5 x = - 1$ by $5$ and simplify.

Divide each term in $5 x = - 1$ by $5$ .

$\frac{5 x}{5} = \frac{- 1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{- 1}{5}$

Divide $x$ by $1$ .

$x = \frac{- 1}{5}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{5}$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $- (5 x + 1) (x - 1) = 0$ true.

$x = - \frac{1}{5}, 1$

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1}{5}, 1$

Decimal Form:

$x = - 0.2, 1$

Solve for x -4-1/x=-5x

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