Basic Math Examples

$\frac{1}{f} + \frac{1}{t} + \frac{1}{k} = t$

Step 1

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

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Step 1.1

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

$\frac{1}{t} + \frac{1}{k} = t - \frac{1}{f}$

Step 1.2

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

$\frac{1}{t} = t - \frac{1}{f} - \frac{1}{k}$

Step 2

Find the LCD of the terms in the equation.

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Step 2.1

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

$t, 1, f, k$

Step 2.2

Since $t, 1, f, k$ 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 $t^{1}, f^{1}, k^{1}$ .

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 $t^{1}$ is $t$ itself.

$t^{1} = t$

$t$ occurs $1$ time.

Step 2.7

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

$f^{1} = f$

$f$ occurs $1$ time.

Step 2.8

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

$k^{1} = k$

$k$ occurs $1$ time.

Step 2.9

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

$t \cdot f \cdot k$

Step 2.10

Multiply $t f$ by $k$ .

$t f k$

Step 3

Multiply each term in $\frac{1}{t} = t - \frac{1}{f} - \frac{1}{k}$ by $t f k$ to eliminate the fractions.

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Step 3.1

Multiply each term in $\frac{1}{t} = t - \frac{1}{f} - \frac{1}{k}$ by $t f k$ .

$\frac{1}{t} (t f k) = t (t f k) - \frac{1}{f} (t f k) - \frac{1}{k} (t f k)$

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of $t$ .

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Step 3.2.1.1

Factor $t$ out of $t f k$ .

$\frac{1}{t} (t (f k)) = t (t f k) - \frac{1}{f} (t f k) - \frac{1}{k} (t f k)$

Step 3.2.1.2

Cancel the common factor.

$\frac{1}{t} (t (f k)) = t (t f k) - \frac{1}{f} (t f k) - \frac{1}{k} (t f k)$

Step 3.2.1.3

Rewrite the expression.

$f k = t (t f k) - \frac{1}{f} (t f k) - \frac{1}{k} (t f k)$

Step 3.3

Simplify the right side.

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Step 3.3.1

Simplify each term.

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Step 3.3.1.1

Multiply $t$ by $t$ by adding the exponents.

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Step 3.3.1.1.1

Move $t$ .

$f k = t \cdot t (f k) - \frac{1}{f} (t f k) - \frac{1}{k} (t f k)$

Step 3.3.1.1.2

Multiply $t$ by $t$ .

$f k = t^{2} (f k) - \frac{1}{f} (t f k) - \frac{1}{k} (t f k)$

Step 3.3.1.2

Cancel the common factor of $f$ .

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Step 3.3.1.2.1

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

$f k = t^{2} f k + \frac{- 1}{f} (t f k) - \frac{1}{k} (t f k)$

Step 3.3.1.2.2

Factor $f$ out of $t f k$ .

$f k = t^{2} f k + \frac{- 1}{f} (f (t k)) - \frac{1}{k} (t f k)$

Step 3.3.1.2.3

Cancel the common factor.

$f k = t^{2} f k + \frac{- 1}{f} (f (t k)) - \frac{1}{k} (t f k)$

Step 3.3.1.2.4

Rewrite the expression.

$f k = t^{2} f k - (t k) - \frac{1}{k} (t f k)$

Step 3.3.1.3

Cancel the common factor of $k$ .

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Step 3.3.1.3.1

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

$f k = t^{2} f k - t k + \frac{- 1}{k} (t f k)$

Step 3.3.1.3.2

Factor $k$ out of $t f k$ .

$f k = t^{2} f k - t k + \frac{- 1}{k} (k (t f))$

Step 3.3.1.3.3

Cancel the common factor.

$f k = t^{2} f k - t k + \frac{- 1}{k} (k (t f))$

Step 3.3.1.3.4

Rewrite the expression.

$f k = t^{2} f k - t k - (t f)$

$f k = t^{2} f k - t k - t f$

Step 4

Solve the equation.

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Step 4.1

Rewrite the equation as $t^{2} f k - t k - t f = f k$ .

$t^{2} f k - t k - t f = f k$

Step 4.2

Subtract $f k$ from both sides of the equation.

$t^{2} f k - t k - t f - f k = 0$

Step 4.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 4.4

Substitute the values $a = f k$ , $b = - k - f$ , and $c = - f k$ into the quadratic formula and solve for $t$ .

$\frac{- (- k - f) \pm \sqrt{{(- k - f)}^{2} - 4 \cdot (f k \cdot (- f k))}}{2 (f k)}$

Step 4.5

Simplify the numerator.

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Step 4.5.1

Apply the distributive property.

$t = \frac{k + f \pm \sqrt{{(- k - f)}^{2} - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.2

Multiply $- - k$ .

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Step 4.5.2.1

Multiply $- 1$ by $- 1$ .

$t = \frac{1 k + f \pm \sqrt{{(- k - f)}^{2} - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.2.2

Multiply $k$ by $1$ .

$t = \frac{k + f \pm \sqrt{{(- k - f)}^{2} - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.3

Multiply $- - f$ .

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Step 4.5.3.1

Multiply $- 1$ by $- 1$ .

$t = \frac{k + 1 f \pm \sqrt{{(- k - f)}^{2} - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.3.2

Multiply $f$ by $1$ .

$t = \frac{k + f \pm \sqrt{{(- k - f)}^{2} - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.4

Rewrite ${(- k - f)}^{2}$ as $(- k - f) (- k - f)$ .

$t = \frac{k + f \pm \sqrt{(- k - f) (- k - f) - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.5

Expand $(- k - f) (- k - f)$ using the FOIL Method.

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Step 4.5.5.1

Apply the distributive property.

$t = \frac{k + f \pm \sqrt{- k (- k - f) - f (- k - f) - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.5.2

Apply the distributive property.

$t = \frac{k + f \pm \sqrt{- k (- k) - k (- f) - f (- k - f) - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.5.3

Apply the distributive property.

$t = \frac{k + f \pm \sqrt{- k (- k) - k (- f) - f (- k) - f (- f) - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.6

Simplify and combine like terms.

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Step 4.5.6.1

Simplify each term.

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Step 4.5.6.1.1

Rewrite using the commutative property of multiplication.

$t = \frac{k + f \pm \sqrt{- 1 \cdot (- 1 k \cdot k) - k (- f) - f (- k) - f (- f) - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.6.1.2

Multiply $k$ by $k$ by adding the exponents.

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Step 4.5.6.1.2.1

Move $k$ .

$t = \frac{k + f \pm \sqrt{- 1 \cdot (- 1 (k \cdot k)) - k (- f) - f (- k) - f (- f) - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.6.1.2.2

Multiply $k$ by $k$ .

$t = \frac{k + f \pm \sqrt{- 1 \cdot (- 1 k^{2}) - k (- f) - f (- k) - f (- f) - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.6.1.3

Multiply $- 1$ by $- 1$ .

$t = \frac{k + f \pm \sqrt{1 k^{2} - k (- f) - f (- k) - f (- f) - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.6.1.4

Multiply $k^{2}$ by $1$ .

$t = \frac{k + f \pm \sqrt{k^{2} - k (- f) - f (- k) - f (- f) - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.6.1.5

Rewrite using the commutative property of multiplication.

$t = \frac{k + f \pm \sqrt{k^{2} - 1 \cdot (- 1 k f) - f (- k) - f (- f) - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.6.1.6

Multiply $- 1$ by $- 1$ .

$t = \frac{k + f \pm \sqrt{k^{2} + 1 k f - f (- k) - f (- f) - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.6.1.7

Multiply $k$ by $1$ .

$t = \frac{k + f \pm \sqrt{k^{2} + k f - f (- k) - f (- f) - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.6.1.8

Rewrite using the commutative property of multiplication.

$t = \frac{k + f \pm \sqrt{k^{2} + k f - 1 \cdot (- 1 f k) - f (- f) - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.6.1.9

Multiply $- 1$ by $- 1$ .

$t = \frac{k + f \pm \sqrt{k^{2} + k f + 1 f k - f (- f) - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.6.1.10

Multiply $f$ by $1$ .

$t = \frac{k + f \pm \sqrt{k^{2} + k f + f k - f (- f) - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.6.1.11

Rewrite using the commutative property of multiplication.

$t = \frac{k + f \pm \sqrt{k^{2} + k f + f k - 1 \cdot (- 1 f \cdot f) - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.6.1.12

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

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Step 4.5.6.1.12.1

Move $f$ .

$t = \frac{k + f \pm \sqrt{k^{2} + k f + f k - 1 \cdot (- 1 (f \cdot f)) - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.6.1.12.2

Multiply $f$ by $f$ .

$t = \frac{k + f \pm \sqrt{k^{2} + k f + f k - 1 \cdot (- 1 f^{2}) - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.6.1.13

Multiply $- 1$ by $- 1$ .

$t = \frac{k + f \pm \sqrt{k^{2} + k f + f k + 1 f^{2} - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.6.1.14

Multiply $f^{2}$ by $1$ .

$t = \frac{k + f \pm \sqrt{k^{2} + k f + f k + f^{2} - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.6.2

Add $k f$ and $f k$ .

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Step 4.5.6.2.1

Reorder $k$ and $f$ .

$t = \frac{k + f \pm \sqrt{k^{2} + f k + f k + f^{2} - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.6.2.2

Add $f k$ and $f k$ .

$t = \frac{k + f \pm \sqrt{k^{2} + 2 f k + f^{2} - 4 \cdot (f k) \cdot (- f k)}}{2 f k}$

Step 4.5.7

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

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Step 4.5.7.1

Move $f$ .

$t = \frac{k + f \pm \sqrt{k^{2} + 2 f k + f^{2} - 4 \cdot ((f \cdot f) k) \cdot (- 1 k)}}{2 f k}$

Step 4.5.7.2

Multiply $f$ by $f$ .

$t = \frac{k + f \pm \sqrt{k^{2} + 2 f k + f^{2} - 4 \cdot (f^{2} k) \cdot (- 1 k)}}{2 f k}$

Step 4.5.8

Multiply $k$ by $k$ by adding the exponents.

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Step 4.5.8.1

Move $k$ .

$t = \frac{k + f \pm \sqrt{k^{2} + 2 f k + f^{2} - 4 \cdot (f^{2} (k \cdot k)) \cdot - 1}}{2 f k}$

Step 4.5.8.2

Multiply $k$ by $k$ .

$t = \frac{k + f \pm \sqrt{k^{2} + 2 f k + f^{2} - 4 \cdot (f^{2} k^{2}) \cdot - 1}}{2 f k}$

Step 4.5.9

Multiply $- 1$ by $- 4$ .

$t = \frac{k + f \pm \sqrt{k^{2} + 2 f k + f^{2} + 4 f^{2} k^{2}}}{2 f k}$

Step 4.6

The final answer is the combination of both solutions.

$t = \frac{k + f + \sqrt{k^{2} + 2 f k + f^{2} + 4 f^{2} k^{2}}}{2 f k}$

$t = \frac{k + f - \sqrt{k^{2} + 2 f k + f^{2} + 4 f^{2} k^{2}}}{2 f k}$

$t = \frac{k + f + \sqrt{k^{2} + 2 f k + f^{2} + 4 f^{2} k^{2}}}{2 f k}$

$t = \frac{k + f - \sqrt{k^{2} + 2 f k + f^{2} + 4 f^{2} k^{2}}}{2 f k}$