Precalculus Examples

12 x - 7 = \frac{12}{x}

$12 x - 7 = \frac{12}{x}$

Step 1

Add

7

$7$ to both sides of the equation.

12 x = \frac{12}{x} + 7

$12 x = \frac{12}{x} + 7$

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.

1, x, 1

$1, x, 1$

Step 2.2

The LCM of one and any expression is the expression.

x

$x$

x

$x$

Step 3

Multiply each term in

12 x = \frac{12}{x} + 7

$12 x = \frac{12}{x} + 7$ by

x

$x$ to eliminate the fractions.

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Step 3.1

Multiply each term in

12 x = \frac{12}{x} + 7

$12 x = \frac{12}{x} + 7$ by

x

$x$ .

12 x \cdot x = \frac{12}{x} x + 7 x

$12 x \cdot x = \frac{12}{x} x + 7 x$

Step 3.2

Simplify the left side.

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Step 3.2.1

Multiply

x

$x$ by

x

$x$ by adding the exponents.

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Step 3.2.1.1

Move

x

$x$ .

12 (x \cdot x) = \frac{12}{x} x + 7 x

$12 (x \cdot x) = \frac{12}{x} x + 7 x$

Step 3.2.1.2

Multiply

x

$x$ by

x

$x$ .

12 x^{2} = \frac{12}{x} x + 7 x

$12 x^{2} = \frac{12}{x} x + 7 x$

12 x^{2} = \frac{12}{x} x + 7 x

$12 x^{2} = \frac{12}{x} x + 7 x$

12 x^{2} = \frac{12}{x} x + 7 x

$12 x^{2} = \frac{12}{x} x + 7 x$

Step 3.3

Simplify the right side.

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Step 3.3.1

Cancel the common factor of

x

$x$ .

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Step 3.3.1.1

Cancel the common factor.

12 x^{2} = \frac{12}{x} x + 7 x

$12 x^{2} = \frac{12}{x} x + 7 x$

Step 3.3.1.2

Rewrite the expression.

12 x^{2} = 12 + 7 x

$12 x^{2} = 12 + 7 x$

12 x^{2} = 12 + 7 x

$12 x^{2} = 12 + 7 x$

12 x^{2} = 12 + 7 x

$12 x^{2} = 12 + 7 x$

12 x^{2} = 12 + 7 x

$12 x^{2} = 12 + 7 x$

Step 4

Solve the equation.

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Step 4.1

Subtract

7 x

$7 x$ from both sides of the equation.

12 x^{2} - 7 x = 12

$12 x^{2} - 7 x = 12$

Step 4.2

Subtract

12

$12$ from both sides of the equation.

12 x^{2} - 7 x - 12 = 0

$12 x^{2} - 7 x - 12 = 0$

Step 4.3

Factor by grouping.

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Step 4.3.1

For a polynomial of the form

a x^{2} + b x + c

$a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is

a \cdot c = 12 \cdot - 12 = - 144

$a \cdot c = 12 \cdot - 12 = - 144$ and whose sum is

b = - 7

$b = - 7$ .

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Step 4.3.1.1

Factor

- 7

$- 7$ out of

- 7 x

$- 7 x$ .

12 x^{2} - 7 x - 12 = 0

$12 x^{2} - 7 x - 12 = 0$

Step 4.3.1.2

Rewrite

- 7

$- 7$ as

9

$9$ plus

- 16

$- 16$

12 x^{2} + (9 - 16) x - 12 = 0

$12 x^{2} + (9 - 16) x - 12 = 0$

Step 4.3.1.3

Apply the distributive property.

12 x^{2} + 9 x - 16 x - 12 = 0

$12 x^{2} + 9 x - 16 x - 12 = 0$

12 x^{2} + 9 x - 16 x - 12 = 0

$12 x^{2} + 9 x - 16 x - 12 = 0$

Step 4.3.2

Factor out the greatest common factor from each group.

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Step 4.3.2.1

Group the first two terms and the last two terms.

(12 x^{2} + 9 x) - 16 x - 12 = 0

$(12 x^{2} + 9 x) - 16 x - 12 = 0$

Step 4.3.2.2

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

3 x (4 x + 3) - 4 (4 x + 3) = 0

$3 x (4 x + 3) - 4 (4 x + 3) = 0$

3 x (4 x + 3) - 4 (4 x + 3) = 0

$3 x (4 x + 3) - 4 (4 x + 3) = 0$

Step 4.3.3

Factor the polynomial by factoring out the greatest common factor,

4 x + 3

$4 x + 3$ .

(4 x + 3) (3 x - 4) = 0

$(4 x + 3) (3 x - 4) = 0$

(4 x + 3) (3 x - 4) = 0

$(4 x + 3) (3 x - 4) = 0$

Step 4.4

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

4 x + 3 = 0

$4 x + 3 = 0$

3 x - 4 = 0

$3 x - 4 = 0$

Step 4.5

Set

4 x + 3

$4 x + 3$ equal to

0

$0$ and solve for

x

$x$ .

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Step 4.5.1

Set

4 x + 3

$4 x + 3$ equal to

0

$0$ .

4 x + 3 = 0

$4 x + 3 = 0$

Step 4.5.2

Solve

4 x + 3 = 0

$4 x + 3 = 0$ for

x

$x$ .

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Step 4.5.2.1

Subtract

3

$3$ from both sides of the equation.

4 x = - 3

$4 x = - 3$

Step 4.5.2.2

Divide each term in

4 x = - 3

$4 x = - 3$ by

4

$4$ and simplify.

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Step 4.5.2.2.1

Divide each term in

4 x = - 3

$4 x = - 3$ by

4

$4$ .

\frac{4 x}{4} = \frac{- 3}{4}

$\frac{4 x}{4} = \frac{- 3}{4}$

Step 4.5.2.2.2

Simplify the left side.

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Step 4.5.2.2.2.1

Cancel the common factor of

4

$4$ .

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Step 4.5.2.2.2.1.1

Cancel the common factor.

\frac{4 x}{4} = \frac{- 3}{4}

$\frac{4 x}{4} = \frac{- 3}{4}$

Step 4.5.2.2.2.1.2

Divide

x

$x$ by

1

$1$ .

x = \frac{- 3}{4}

$x = \frac{- 3}{4}$

x = \frac{- 3}{4}

$x = \frac{- 3}{4}$

x = \frac{- 3}{4}

$x = \frac{- 3}{4}$

Step 4.5.2.2.3

Simplify the right side.

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Step 4.5.2.2.3.1

Move the negative in front of the fraction.

x = - \frac{3}{4}

$x = - \frac{3}{4}$

x = - \frac{3}{4}

$x = - \frac{3}{4}$

x = - \frac{3}{4}

$x = - \frac{3}{4}$

x = - \frac{3}{4}

$x = - \frac{3}{4}$

x = - \frac{3}{4}

$x = - \frac{3}{4}$

Step 4.6

Set

3 x - 4

$3 x - 4$ equal to

0

$0$ and solve for

x

$x$ .

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Step 4.6.1

Set

3 x - 4

$3 x - 4$ equal to

0

$0$ .

3 x - 4 = 0

$3 x - 4 = 0$

Step 4.6.2

Solve

3 x - 4 = 0

$3 x - 4 = 0$ for

x

$x$ .

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Step 4.6.2.1

Add

4

$4$ to both sides of the equation.

3 x = 4

$3 x = 4$

Step 4.6.2.2

Divide each term in

3 x = 4

$3 x = 4$ by

3

$3$ and simplify.

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Step 4.6.2.2.1

Divide each term in

3 x = 4

$3 x = 4$ by

3

$3$ .

\frac{3 x}{3} = \frac{4}{3}

$\frac{3 x}{3} = \frac{4}{3}$

Step 4.6.2.2.2

Simplify the left side.

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Step 4.6.2.2.2.1

Cancel the common factor of

3

$3$ .

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Step 4.6.2.2.2.1.1

Cancel the common factor.

\frac{3 x}{3} = \frac{4}{3}

$\frac{3 x}{3} = \frac{4}{3}$

Step 4.6.2.2.2.1.2

Divide

x

$x$ by

1

$1$ .

x = \frac{4}{3}

$x = \frac{4}{3}$

x = \frac{4}{3}

$x = \frac{4}{3}$

x = \frac{4}{3}

$x = \frac{4}{3}$

x = \frac{4}{3}

$x = \frac{4}{3}$

x = \frac{4}{3}

$x = \frac{4}{3}$

x = \frac{4}{3}

$x = \frac{4}{3}$

Step 4.7

The final solution is all the values that make

(4 x + 3) (3 x - 4) = 0

$(4 x + 3) (3 x - 4) = 0$ true.

x = - \frac{3}{4}, \frac{4}{3}

$x = - \frac{3}{4}, \frac{4}{3}$

x = - \frac{3}{4}, \frac{4}{3}

$x = - \frac{3}{4}, \frac{4}{3}$

Step 5

The result can be shown in multiple forms.

Exact Form:

x = - \frac{3}{4}, \frac{4}{3}

$x = - \frac{3}{4}, \frac{4}{3}$

Decimal Form:

x = - 0.75, 1.\overline{3}

$x = - 0.75, 1.\overline{3}$

Mixed Number Form:

x = - \frac{3}{4}, 1 \frac{1}{3}

$x = - \frac{3}{4}, 1 \frac{1}{3}$