Solve the Rational Equation for x 3/(6x)-(5x)/(10x^3)=4/(x^2)

$\frac{3}{6 x} - \frac{5 x}{10 x^{3}} = \frac{4}{x^{2}}$

Factor each term.

Reduce the expression $\frac{3}{6 x}$ by cancelling the common factors.

Factor $3$ out of $3$ .

$\frac{3 (1)}{6 x} - \frac{5 x}{10 x^{3}} = \frac{4}{x^{2}}$

Factor $3$ out of $6 x$ .

$\frac{3 (1)}{3 (2 x)} - \frac{5 x}{10 x^{3}} = \frac{4}{x^{2}}$

Cancel the common factor.

$\frac{3 \cdot 1}{3 (2 x)} - \frac{5 x}{10 x^{3}} = \frac{4}{x^{2}}$

Rewrite the expression.

$\frac{1}{2 x} - \frac{5 x}{10 x^{3}} = \frac{4}{x^{2}}$

Reduce the expression $\frac{5 x}{10 x^{3}}$ by cancelling the common factors.

Factor $5$ out of $5 x$ .

$\frac{1}{2 x} - \frac{5 (x)}{10 x^{3}} = \frac{4}{x^{2}}$

Factor $5$ out of $10 x^{3}$ .

$\frac{1}{2 x} - \frac{5 (x)}{5 (2 x^{3})} = \frac{4}{x^{2}}$

Cancel the common factor.

$\frac{1}{2 x} - \frac{5 x}{5 (2 x^{3})} = \frac{4}{x^{2}}$

Rewrite the expression.

$\frac{1}{2 x} - \frac{x}{2 x^{3}} = \frac{4}{x^{2}}$

Cancel the common factor of $x$ and $x^{3}$ .

Raise $x$ to the power of $1$ .

$\frac{1}{2 x} - \frac{x^{1}}{2 x^{3}} = \frac{4}{x^{2}}$

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

$\frac{1}{2 x} - \frac{x \cdot 1}{2 x^{3}} = \frac{4}{x^{2}}$

Cancel the common factors.

Factor $x$ out of $2 x^{3}$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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.

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

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

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.

Since $2$ has no factors besides $1$ and $2$ .

$2$ is a prime number

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

Not prime

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

$2$

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

$x^{1} = x$

$x$ occurs $1$ time.

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.

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

$x \cdot x$

Multiply $x$ by $x$ .

$x^{2}$

The LCM for $2 x, 2 x^{2}, x^{2}$ is the numeric part $2$ multiplied by the variable part.

$2 x^{2}$

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

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

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

Simplify the left side.

Simplify each term.

Rewrite using the commutative property of multiplication.

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

Cancel the common factor of $2$ .

Factor $2$ out of $2 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Move the leading negative in $- \frac{1}{2 x^{2}}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the right side.

Rewrite using the commutative property of multiplication.

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

Multiply $2 \frac{4}{x^{2}}$ .

Combine $2$ and $\frac{4}{x^{2}}$ .

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

Multiply $2$ by $4$ .

$x - 1 = \frac{8}{x^{2}} x^{2}$

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

Cancel the common factor.

$x - 1 = \frac{8}{x^{2}} x^{2}$

Rewrite the expression.

$x - 1 = 8$

Move all terms not containing $x$ to the right side of the equation.

Add $1$ to both sides of the equation.

$x = 8 + 1$

Add $8$ and $1$ .

$x = 9$

Solve the Rational Equation for x 3/(6x)-(5x)/(10x^3)=4/(x^2)

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