Solve for f 3/2-1/f=(f-2)/3

$\frac{3}{2} - \frac{1}{f} = \frac{f - 2}{3}$

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

Subtract $\frac{3}{2}$ from both sides of the equation.

$- \frac{1}{f} = \frac{f - 2}{3} - \frac{3}{2}$

Simplify each term.

Split the fraction $\frac{f - 2}{3}$ into two fractions.

$- \frac{1}{f} = \frac{f}{3} + \frac{- 2}{3} - \frac{3}{2}$

Move the negative in front of the fraction.

$- \frac{1}{f} = \frac{f}{3} - \frac{2}{3} - \frac{3}{2}$

To write $- \frac{2}{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- \frac{1}{f} = \frac{f}{3} - \frac{2}{3} \cdot \frac{2}{2} - \frac{3}{2}$

To write $- \frac{3}{2}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- \frac{1}{f} = \frac{f}{3} - \frac{2}{3} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{3}{3}$

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{2}{3}$ by $\frac{2}{2}$ .

$- \frac{1}{f} = \frac{f}{3} - \frac{2 \cdot 2}{3 \cdot 2} - \frac{3}{2} \cdot \frac{3}{3}$

Multiply $3$ by $2$ .

$- \frac{1}{f} = \frac{f}{3} - \frac{2 \cdot 2}{6} - \frac{3}{2} \cdot \frac{3}{3}$

Multiply $\frac{3}{2}$ by $\frac{3}{3}$ .

$- \frac{1}{f} = \frac{f}{3} - \frac{2 \cdot 2}{6} - \frac{3 \cdot 3}{2 \cdot 3}$

Multiply $2$ by $3$ .

$- \frac{1}{f} = \frac{f}{3} - \frac{2 \cdot 2}{6} - \frac{3 \cdot 3}{6}$

Combine the numerators over the common denominator.

$- \frac{1}{f} = \frac{f}{3} + \frac{- 2 \cdot 2 - 3 \cdot 3}{6}$

Simplify the numerator.

Multiply $- 2$ by $2$ .

$- \frac{1}{f} = \frac{f}{3} + \frac{- 4 - 3 \cdot 3}{6}$

Multiply $- 3$ by $3$ .

$- \frac{1}{f} = \frac{f}{3} + \frac{- 4 - 9}{6}$

Subtract $9$ from $- 4$ .

$- \frac{1}{f} = \frac{f}{3} + \frac{- 13}{6}$

Move the negative in front of the fraction.

$- \frac{1}{f} = \frac{f}{3} - \frac{13}{6}$

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.

$f, 3, 6$

Since $f, 3, 6$ contain both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $1, 3, 6$ then find LCM for the variable part $f^{1}$ .

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.

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

Not prime

Since $3$ has no factors besides $1$ and $3$ .

$3$ is a prime number

$6$ has factors of $2$ and $3$ .

$2 \cdot 3$

Multiply $2$ by $3$ .

$6$

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

$f^{1} = f$

$f$ occurs $1$ time.

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

$f$

The LCM for $f, 3, 6$ is the numeric part $6$ multiplied by the variable part.

$6 f$

Multiply each term in $- \frac{1}{f} = \frac{f}{3} - \frac{13}{6}$ by $6 f$ to eliminate the fractions.

Multiply each term in $- \frac{1}{f} = \frac{f}{3} - \frac{13}{6}$ by $6 f$ .

$- \frac{1}{f} (6 f) = \frac{f}{3} (6 f) - \frac{13}{6} (6 f)$

Simplify the left side.

Cancel the common factor of $f$ .

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

$\frac{- 1}{f} (6 f) = \frac{f}{3} (6 f) - \frac{13}{6} (6 f)$

Factor $f$ out of $6 f$ .

$\frac{- 1}{f} (f \cdot 6) = \frac{f}{3} (6 f) - \frac{13}{6} (6 f)$

Cancel the common factor.

$\frac{- 1}{f} (f \cdot 6) = \frac{f}{3} (6 f) - \frac{13}{6} (6 f)$

Rewrite the expression.

$- 1 \cdot 6 = \frac{f}{3} (6 f) - \frac{13}{6} (6 f)$

Multiply $- 1$ by $6$ .

$- 6 = \frac{f}{3} (6 f) - \frac{13}{6} (6 f)$

Simplify the right side.

Simplify each term.

Rewrite using the commutative property of multiplication.

$- 6 = 6 \frac{f}{3} f - \frac{13}{6} (6 f)$

Cancel the common factor of $3$ .

Factor $3$ out of $6$ .

$- 6 = 3 (2) \frac{f}{3} f - \frac{13}{6} (6 f)$

Cancel the common factor.

$- 6 = 3 \cdot 2 \frac{f}{3} f - \frac{13}{6} (6 f)$

Rewrite the expression.

$- 6 = 2 f \cdot f - \frac{13}{6} (6 f)$

Multiply $f$ by $f$ by adding the exponents.

Move $f$ .

$- 6 = 2 (f \cdot f) - \frac{13}{6} (6 f)$

Multiply $f$ by $f$ .

$- 6 = 2 f^{2} - \frac{13}{6} (6 f)$

Cancel the common factor of $6$ .

Move the leading negative in $- \frac{13}{6}$ into the numerator.

$- 6 = 2 f^{2} + \frac{- 13}{6} (6 f)$

Factor $6$ out of $6 f$ .

$- 6 = 2 f^{2} + \frac{- 13}{6} (6 (f))$

Cancel the common factor.

$- 6 = 2 f^{2} + \frac{- 13}{6} (6 f)$

Rewrite the expression.

$- 6 = 2 f^{2} - 13 f$

Solve the equation.

Rewrite the equation as $2 f^{2} - 13 f = - 6$ .

$2 f^{2} - 13 f = - 6$

Add $6$ to both sides of the equation.

$2 f^{2} - 13 f + 6 = 0$

Factor by grouping.

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 $a \cdot c = 2 \cdot 6 = 12$ and whose sum is $b = - 13$ .

Factor $- 13$ out of $- 13 f$ .

$2 f^{2} - 13 f + 6 = 0$

Rewrite $- 13$ as $- 1$ plus $- 12$

$2 f^{2} + (- 1 - 12) f + 6 = 0$

Apply the distributive property.

$2 f^{2} - 1 f - 12 f + 6 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 f^{2} - 1 f) - 12 f + 6 = 0$

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

$f (2 f - 1) - 6 (2 f - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 f - 1$ .

$(2 f - 1) (f - 6) = 0$

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

$2 f - 1 = 0$

$f - 6 = 0$

Set $2 f - 1$ equal to $0$ and solve for $f$ .

Set $2 f - 1$ equal to $0$ .

$2 f - 1 = 0$

Solve $2 f - 1 = 0$ for $f$ .

Add $1$ to both sides of the equation.

$2 f = 1$

Divide each term in $2 f = 1$ by $2$ and simplify.

Divide each term in $2 f = 1$ by $2$ .

$\frac{2 f}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 f}{2} = \frac{1}{2}$

Divide $f$ by $1$ .

$f = \frac{1}{2}$

Set $f - 6$ equal to $0$ and solve for $f$ .

Set $f - 6$ equal to $0$ .

$f - 6 = 0$

Add $6$ to both sides of the equation.

$f = 6$

The final solution is all the values that make $(2 f - 1) (f - 6) = 0$ true.

$f = \frac{1}{2}, 6$

The result can be shown in multiple forms.

Exact Form:

$f = \frac{1}{2}, 6$

Decimal Form:

$f = 0.5, 6$

Solve for f 3/2-1/f=(f-2)/3

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