Precalculus Examples

$\frac{- 4}{5 x} + \frac{1}{25} = 0$

Subtract $\frac{1}{25}$ from both sides of the equation.

$\frac{- 4}{5 x} = - \frac{1}{25}$

Move the negative in front of the fraction.

$- \frac{4}{5 x} = - \frac{1}{25}$

Set up the rational expression with the same denominator over the entire equation.

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Multiply each term by a factor of $1$ that will equate all the denominators. In this case, all terms need a denominator of $25 x$ . The $- \frac{4}{5 x}$ expression needs to be multiplied by $\frac{5}{5}$ to make the denominator $25 x$ . The $- \frac{1}{25}$ expression needs to be multiplied by $\frac{x}{x}$ to make the denominator $25 x$ .

$- \frac{4}{5 x} \cdot \frac{5}{5} = - \frac{1}{25} \cdot \frac{x}{x}$

Multiply the expression by a factor of $1$ to create the least common denominator (LCD) of $25 x$ .

$4 (5)$

Multiply $4$ by $5$ .

$- \frac{20}{25 x} = - \frac{1}{25} \cdot \frac{x}{x}$

Multiply the expression by a factor of $1$ to create the least common denominator (LCD) of $25 x$ .

$1 (x)$

Multiply $x$ by $1$ .

$- \frac{20}{25 x} = - \frac{x}{25 x}$

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$- (20) = - (x)$

Rewrite the equation as $- (x) = - (20)$ .

$- (x) = - (20)$

Multiply each term in $- (x) = - (20)$ by $- 1$

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Multiply each term in $- (x) = - (20)$ by $- 1$ .

$- x \cdot - 1 = - 20 \cdot - 1$

Multiply $- x \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$1 x = - 20 \cdot - 1$

Multiply $x$ by $1$ .

$x = - 20 \cdot - 1$

Multiply $- 20$ by $- 1$ .

$x = 20$

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