Solve for m 1/(2m^2)=1/m-1/2

$\frac{1}{2 m^{2}} = \frac{1}{m} - \frac{1}{2}$

Find the LCD of the terms in the equation.

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

$2 m^{2}, m, 2$

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

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

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

$2$ is a prime number

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

$2$

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

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

$m$ occurs $2$ times.

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

$m^{1} = m$

$m$ occurs $1$ time.

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

$m \cdot m$

Multiply $m$ by $m$ .

$m^{2}$

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

$2 m^{2}$

Multiply each term by $2 m^{2}$ and simplify.

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Multiply each term in $\frac{1}{2 m^{2}} = \frac{1}{m} - \frac{1}{2}$ by $2 m^{2}$ in order to remove all the denominators from the equation.

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

Simplify $\frac{1}{2 m^{2}} \cdot (2 m^{2})$ .

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Rewrite using the commutative property of multiplication.

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

Cancel the common factor of $2$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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Cancel the common factor.

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

Rewrite the expression.

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

Simplify each term.

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Rewrite using the commutative property of multiplication.

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

Combine $2$ and $\frac{1}{m}$ .

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

Cancel the common factor of $m$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

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

$1 = 2 m + \frac{- 1}{2} \cdot (2 m^{2})$

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

$1 = 2 m + \frac{- 1}{2} \cdot (2 (m^{2}))$

Cancel the common factor.

$1 = 2 m + \frac{- 1}{2} \cdot (2 m^{2})$

Rewrite the expression.

$1 = 2 m - 1 \cdot m^{2}$

Rewrite $- 1 m^{2}$ as $- m^{2}$ .

$1 = 2 m - m^{2}$

Solve the equation.

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Rewrite the equation as $2 m - m^{2} = 1$ .

$2 m - m^{2} = 1$

Move $1$ to the left side of the equation by subtracting it from both sides.

$2 m - m^{2} - 1 = 0$

Factor the left side of the equation.

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Let $u = m$ . Substitute $u$ for all occurrences of $m$ .

$2 u - u^{2} - 1$

Factor by grouping.

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Reorder terms.

$- u^{2} + 2 u - 1$

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 $a \cdot c = - 1 \cdot - 1 = 1$ and whose sum is $b = 2$ .

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Factor $2$ out of $2 u$ .

$- u^{2} + 2 (u) - 1$

Rewrite $2$ as $1$ plus $1$

$- u^{2} + (1 + 1) u - 1$

Apply the distributive property.

$- u^{2} + 1 u + 1 u - 1$

Multiply $u$ by $1$ .

$- u^{2} + u + 1 u - 1$

Multiply $u$ by $1$ .

$- u^{2} + u + u - 1$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$(- u^{2} + u) + u - 1$

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

$u (- u + 1) - 1 (- u + 1)$

Factor the polynomial by factoring out the greatest common factor, $- u + 1$ .

$(- u + 1) (u - 1)$

Replace all occurrences of $u$ with $m$ .

$(- m + 1) (m - 1)$

Replace the left side with the factored expression.

$(- m + 1) (m - 1) = 0$

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

$- m + 1 = 0$

$m - 1 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$- m + 1 = 0$

Subtract $1$ from both sides of the equation.

$- m = - 1$

Multiply each term in $- m = - 1$ by $- 1$

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Multiply each term in $- m = - 1$ by $- 1$ .

$(- m) \cdot - 1 = (- 1) \cdot - 1$

Multiply $(- m) \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$1 m = (- 1) \cdot - 1$

Multiply $m$ by $1$ .

$m = (- 1) \cdot - 1$

Multiply $- 1$ by $- 1$ .

$m = 1$

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

$m = 1$

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