Algebra Examples

$- 1 = \frac{15}{x} - \frac{11 x + 5}{x^{2}}$

Step 1

Move all terms to the left side of the equation and simplify.

Tap for more steps...

Step 1.1

Simplify the right side.

Tap for more steps...

Step 1.1.1

Simplify $\frac{15}{x} - \frac{11 x + 5}{x^{2}}$ .

Tap for more steps...

Step 1.1.1.1

Simplify each term.

Tap for more steps...

Step 1.1.1.1.1

Split the fraction $\frac{11 x + 5}{x^{2}}$ into two fractions.

$- 1 = \frac{15}{x} - (\frac{11 x}{x^{2}} + \frac{5}{x^{2}})$

Step 1.1.1.1.2

Cancel the common factor of $x$ and $x^{2}$ .

Tap for more steps...

Step 1.1.1.1.2.1

Factor $x$ out of $11 x$ .

$- 1 = \frac{15}{x} - (\frac{x \cdot 11}{x^{2}} + \frac{5}{x^{2}})$

Step 1.1.1.1.2.2

Cancel the common factors.

Tap for more steps...

Step 1.1.1.1.2.2.1

Factor $x$ out of $x^{2}$ .

$- 1 = \frac{15}{x} - (\frac{x \cdot 11}{x \cdot x} + \frac{5}{x^{2}})$

Step 1.1.1.1.2.2.2

Cancel the common factor.

$- 1 = \frac{15}{x} - (\frac{x \cdot 11}{x \cdot x} + \frac{5}{x^{2}})$

Step 1.1.1.1.2.2.3

Rewrite the expression.

$- 1 = \frac{15}{x} - (\frac{11}{x} + \frac{5}{x^{2}})$

Step 1.1.1.1.3

Apply the distributive property.

$- 1 = \frac{15}{x} - \frac{11}{x} - \frac{5}{x^{2}}$

Step 1.1.1.2

Combine fractions.

Tap for more steps...

Step 1.1.1.2.1

Combine the numerators over the common denominator.

$- 1 = \frac{15 - 11}{x} + \frac{- 5}{x^{2}}$

Step 1.1.1.2.2

Simplify the expression.

Tap for more steps...

Step 1.1.1.2.2.1

Subtract $11$ from $15$ .

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

Step 1.1.1.2.2.2

Move the negative in front of the fraction.

$- 1 = \frac{4}{x} - \frac{5}{x^{2}}$

Step 1.2

Move all the expressions to the left side of the equation.

Tap for more steps...

Step 1.2.1

Subtract $\frac{4}{x}$ from both sides of the equation.

$- 1 - \frac{4}{x} = - \frac{5}{x^{2}}$

Step 1.2.2

Add $\frac{5}{x^{2}}$ to both sides of the equation.

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

Step 2

Find the LCD of the terms in the equation.

Tap for more steps...

Step 2.1

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

$1, x, x^{2}, 1$

Step 2.2

Since $1, x, x^{2}, 1$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $1, 1, 1, 1$ then find LCM for the variable part $x^{1}, x^{2}$ .

Step 2.3

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 2.4

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 2.5

The LCM of $1, 1, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

Step 2.6

The factor for $x^{1}$ is $x$ itself.

$x^{1} = x$

$x$ occurs $1$ time.

Step 2.7

The factors for $x^{2}$ are $x \cdot x$ , which is $x$ multiplied by each other $2$ times.

$x^{2} = x \cdot x$

$x$ occurs $2$ times.

Step 2.8

The LCM of $x^{1}, x^{2}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$x \cdot x$

Step 2.9

Multiply $x$ by $x$ .

$x^{2}$

Step 3

Multiply each term in $- 1 - \frac{4}{x} + \frac{5}{x^{2}} = 0$ by $x^{2}$ to eliminate the fractions.

Tap for more steps...

Step 3.1

Multiply each term in $- 1 - \frac{4}{x} + \frac{5}{x^{2}} = 0$ by $x^{2}$ .

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

Step 3.2

Simplify the left side.

Tap for more steps...

Step 3.2.1

Simplify each term.

Tap for more steps...

Step 3.2.1.1

Cancel the common factor of $x$ .

Tap for more steps...

Step 3.2.1.1.1

Move the leading negative in $- \frac{4}{x}$ into the numerator.

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

Step 3.2.1.1.2

Factor $x$ out of $x^{2}$ .

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

Step 3.2.1.1.3

Cancel the common factor.

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

Step 3.2.1.1.4

Rewrite the expression.

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

Step 3.2.1.2

Cancel the common factor of $x^{2}$ .

Tap for more steps...

Step 3.2.1.2.1

Cancel the common factor.

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

Step 3.2.1.2.2

Rewrite the expression.

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

Step 3.3

Simplify the right side.

Tap for more steps...

Step 3.3.1

Multiply $0$ by $x^{2}$ .

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

Step 4

Solve the equation.

Tap for more steps...

Step 4.1

Factor the left side of the equation.

Tap for more steps...

Step 4.1.1

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

Tap for more steps...

Step 4.1.1.1

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

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

Step 4.1.1.2

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

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

Step 4.1.1.3

Rewrite $5$ as $- 1 (- 5)$ .

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

Step 4.1.1.4

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

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

Step 4.1.1.5

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

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

Step 4.1.2

Factor.

Tap for more steps...

Step 4.1.2.1

Factor $x^{2} + 4 x - 5$ using the AC method.

Tap for more steps...

Step 4.1.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 5$ and whose sum is $4$ .

$- 1, 5$

Step 4.1.2.1.2

Write the factored form using these integers.

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

Step 4.1.2.2

Remove unnecessary parentheses.

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

Step 4.2

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

$x - 1 = 0$

$x + 5 = 0$

Step 4.3

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

Tap for more steps...

Step 4.3.1

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

$x - 1 = 0$

Step 4.3.2

Add $1$ to both sides of the equation.

$x = 1$

Step 4.4

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

Tap for more steps...

Step 4.4.1

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

$x + 5 = 0$

Step 4.4.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 4.5

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

$x = 1, - 5$