Algebra Examples

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

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

Step 1

Subtract

\frac{1}{25}

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

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

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

Step 2

Move the negative in front of the fraction.

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

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

Step 3

Find the LCD of the terms in the equation.

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Step 3.1

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

5 x, 25

$5 x, 25$

Step 3.2

Since

5 x, 25

$5 x, 25$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part

5, 25

$5, 25$ then find LCM for the variable part

x^{1}

$x^{1}$ .

Step 3.3

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 3.4

Since

5

$5$ has no factors besides

1

$1$ and

5

$5$ .

5

$5$ is a prime number

Step 3.5

25

$25$ has factors of

5

$5$ and

5

$5$ .

5 \cdot 5

$5 \cdot 5$

Step 3.6

Multiply

5

$5$ by

5

$5$ .

25

$25$

Step 3.7

The factor for

x^{1}

$x^{1}$ is

x

$x$ itself.

x^{1} = x

$x^{1} = x$

x

$x$ occurs

1

$1$ time.

Step 3.8

The LCM of

x^{1}

$x^{1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

x

$x$

Step 3.9

The LCM for

5 x, 25

$5 x, 25$ is the numeric part

25

$25$ multiplied by the variable part.

25 x

$25 x$

25 x

$25 x$

Step 4

Multiply each term in

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

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

25 x

$25 x$ to eliminate the fractions.

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Step 4.1

Multiply each term in

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

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

25 x

$25 x$ .

- \frac{4}{5 x} (25 x) = - \frac{1}{25} (25 x)

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

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of

5 x

$5 x$ .

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Step 4.2.1.1

Move the leading negative in

- \frac{4}{5 x}

$- \frac{4}{5 x}$ into the numerator.

\frac{- 4}{5 x} (25 x) = - \frac{1}{25} (25 x)

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

Step 4.2.1.2

Factor

5 x

$5 x$ out of

25 x

$25 x$ .

\frac{- 4}{5 x} (5 x (5)) = - \frac{1}{25} (25 x)

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

Step 4.2.1.3

Cancel the common factor.

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

Step 4.2.1.4

Rewrite the expression.

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

Step 4.2.2

Multiply

$- 4$ by

$5$ .

$- 20 = - \frac{1}{25} (25 x)$

Step 4.3

Simplify the right side.

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Step 4.3.1

Cancel the common factor of

$25$ .

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Step 4.3.1.1

Move the leading negative in

$- \frac{1}{25}$ into the numerator.

$- 20 = \frac{- 1}{25} (25 x)$

Step 4.3.1.2

Factor

$25$ out of

$25 x$ .

$- 20 = \frac{- 1}{25} (25 (x))$

Step 4.3.1.3

Cancel the common factor.

$- 20 = \frac{- 1}{25} (25 x)$

Step 4.3.1.4

Rewrite the expression.

$- 20 = - x$

Step 5

Solve the equation.

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Step 5.1

Rewrite the equation as

$- x = - 20$ .

$- x = - 20$

Step 5.2

Divide each term in

$- x = - 20$ by

$- 1$ and simplify.

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Step 5.2.1

Divide each term in

$- x = - 20$ by

$- 1$ .

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

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Dividing two negative values results in a positive value.

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

Step 5.2.2.2

Divide

$x$ by

$1$ .

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

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Divide

$- 20$ by

$- 1$ .

$x = 20$

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