Finite Math Examples

$f {(x)}^{- 1} = \frac{5 x}{2 x - 7}$

Step 1

Multiply the equation by $2 x - 7$ .

$(2 x - 7) (f x^{- 1}) = (\frac{5 x}{2 x - 7}) (2 x - 7)$

Step 2

Simplify the left side.

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Step 2.1

Simplify $(2 x - 7) (f x^{- 1})$ .

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Step 2.1.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$(2 x - 7) (f \frac{1}{x}) = (\frac{5 x}{2 x - 7}) (2 x - 7)$

Step 2.1.2

Combine $f$ and $\frac{1}{x}$ .

$(2 x - 7) \frac{f}{x} = (\frac{5 x}{2 x - 7}) (2 x - 7)$

Step 2.1.3

Multiply $2 x - 7$ by $\frac{f}{x}$ .

$\frac{(2 x - 7) f}{x} = (\frac{5 x}{2 x - 7}) (2 x - 7)$

Step 2.1.4

Reorder factors in $\frac{(2 x - 7) f}{x}$ .

$\frac{f (2 x - 7)}{x} = (\frac{5 x}{2 x - 7}) (2 x - 7)$

Step 3

Simplify the right side.

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Step 3.1

Cancel the common factor of $2 x - 7$ .

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Step 3.1.1

Cancel the common factor.

$\frac{f (2 x - 7)}{x} = \frac{5 x}{2 x - 7} (2 x - 7)$

Step 3.1.2

Rewrite the expression.

$\frac{f (2 x - 7)}{x} = 5 x$

Step 4

Solve for $x$ .

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Step 4.1

Find the LCD of the terms in the equation.

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Step 4.1.1

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

$x, 1$

Step 4.1.2

The LCM of one and any expression is the expression.

$x$

Step 4.2

Multiply each term in $\frac{f (2 x - 7)}{x} = 5 x$ by $x$ to eliminate the fractions.

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Step 4.2.1

Multiply each term in $\frac{f (2 x - 7)}{x} = 5 x$ by $x$ .

$\frac{f (2 x - 7)}{x} x = 5 x \cdot x$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $x$ .

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Step 4.2.2.1.1

Cancel the common factor.

$\frac{f (2 x - 7)}{x} x = 5 x \cdot x$

Step 4.2.2.1.2

Rewrite the expression.

$f (2 x - 7) = 5 x \cdot x$

Step 4.2.2.2

Apply the distributive property.

$f (2 x) + f \cdot - 7 = 5 x \cdot x$

Step 4.2.2.3

Reorder.

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Step 4.2.2.3.1

Rewrite using the commutative property of multiplication.

$2 f x + f \cdot - 7 = 5 x \cdot x$

Step 4.2.2.3.2

Move $- 7$ to the left of $f$ .

$2 f x - 7 f = 5 x \cdot x$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Multiply $x$ by $x$ by adding the exponents.

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Step 4.2.3.1.1

Move $x$ .

$2 f x - 7 f = 5 (x \cdot x)$

Step 4.2.3.1.2

Multiply $x$ by $x$ .

$2 f x - 7 f = 5 x^{2}$

Step 4.3

Solve the equation.

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Step 4.3.1

Subtract $5 x^{2}$ from both sides of the equation.

$2 f x - 7 f - 5 x^{2} = 0$

Step 4.3.2

Use the quadratic formula to find the solutions.

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

Step 4.3.3

Substitute the values $a = - 5$ , $b = 2 f$ , and $c = - 7 f$ into the quadratic formula and solve for $x$ .

$\frac{- 2 f \pm \sqrt{{(2 f)}^{2} - 4 \cdot (- 5 \cdot (- 7 f))}}{2 \cdot - 5}$

Step 4.3.4

Simplify.

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Step 4.3.4.1

Simplify the numerator.

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Step 4.3.4.1.1

Add parentheses.

$x = \frac{- 2 f \pm \sqrt{{(2 f)}^{2} - 4 \cdot (- 5 \cdot (- 7 f))}}{2 \cdot - 5}$

Step 4.3.4.1.2

Let $u = - 5 \cdot (- 7 f)$ . Substitute $u$ for all occurrences of $- 5 \cdot (- 7 f)$ .

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Step 4.3.4.1.2.1

Apply the product rule to $2 f$ .

$x = \frac{- 2 f \pm \sqrt{2^{2} f^{2} - 4 \cdot u}}{2 \cdot - 5}$

Step 4.3.4.1.2.2

Raise $2$ to the power of $2$ .

$x = \frac{- 2 f \pm \sqrt{4 f^{2} - 4 u}}{2 \cdot - 5}$

Step 4.3.4.1.3

Factor $4$ out of $4 f^{2} - 4 u$ .

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Step 4.3.4.1.3.1

Factor $4$ out of $4 f^{2}$ .

$x = \frac{- 2 f \pm \sqrt{4 (f^{2}) - 4 u}}{2 \cdot - 5}$

Step 4.3.4.1.3.2

Factor $4$ out of $- 4 u$ .

$x = \frac{- 2 f \pm \sqrt{4 (f^{2}) + 4 (- u)}}{2 \cdot - 5}$

Step 4.3.4.1.3.3

Factor $4$ out of $4 (f^{2}) + 4 (- u)$ .

$x = \frac{- 2 f \pm \sqrt{4 (f^{2} - u)}}{2 \cdot - 5}$

Step 4.3.4.1.4

Replace all occurrences of $u$ with $- 5 \cdot (- 7 f)$ .

$x = \frac{- 2 f \pm \sqrt{4 (f^{2} - (- 5 \cdot (- 7 f)))}}{2 \cdot - 5}$

Step 4.3.4.1.5

Simplify each term.

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Step 4.3.4.1.5.1

Multiply $- 7$ by $- 5$ .

$x = \frac{- 2 f \pm \sqrt{4 (f^{2} - (35 f))}}{2 \cdot - 5}$

Step 4.3.4.1.5.2

Multiply $35$ by $- 1$ .

$x = \frac{- 2 f \pm \sqrt{4 (f^{2} - 35 f)}}{2 \cdot - 5}$

Step 4.3.4.1.6

Factor $f$ out of $f^{2} - 35 f$ .

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Step 4.3.4.1.6.1

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

$x = \frac{- 2 f \pm \sqrt{4 (f \cdot f - 35 f)}}{2 \cdot - 5}$

Step 4.3.4.1.6.2

Factor $f$ out of $- 35 f$ .

$x = \frac{- 2 f \pm \sqrt{4 (f \cdot f + f \cdot - 35)}}{2 \cdot - 5}$

Step 4.3.4.1.6.3

Factor $f$ out of $f \cdot f + f \cdot - 35$ .

$x = \frac{- 2 f \pm \sqrt{4 (f (f - 35))}}{2 \cdot - 5}$

$x = \frac{- 2 f \pm \sqrt{4 f (f - 35)}}{2 \cdot - 5}$

Step 4.3.4.1.7

Rewrite $4 f (f - 35)$ as $2^{2} (f (f - 35))$ .

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Step 4.3.4.1.7.1

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 f \pm \sqrt{2^{2} f (f - 35)}}{2 \cdot - 5}$

Step 4.3.4.1.7.2

Add parentheses.

$x = \frac{- 2 f \pm \sqrt{2^{2} (f (f - 35))}}{2 \cdot - 5}$

Step 4.3.4.1.8

Pull terms out from under the radical.

$x = \frac{- 2 f \pm 2 \sqrt{f (f - 35)}}{2 \cdot - 5}$

Step 4.3.4.2

Multiply $2$ by $- 5$ .

$x = \frac{- 2 f \pm 2 \sqrt{f (f - 35)}}{- 10}$

Step 4.3.4.3

Simplify $\frac{- 2 f \pm 2 \sqrt{f (f - 35)}}{- 10}$ .

$x = \frac{f \pm \sqrt{f (f - 35)}}{5}$

Step 4.3.5

The final answer is the combination of both solutions.

$x = \frac{f + \sqrt{f (f - 35)}}{5}$

$x = \frac{f - \sqrt{f (f - 35)}}{5}$

$x = \frac{f + \sqrt{f (f - 35)}}{5}$

$x = \frac{f - \sqrt{f (f - 35)}}{5}$

$x = \frac{f + \sqrt{f (f - 35)}}{5}$

$x = \frac{f - \sqrt{f (f - 35)}}{5}$