Trigonometry Examples

$4 (x) = \frac{12 (x)}{4 + x}$

Step 1

Find the LCD of the terms in the equation.

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Step 1.1

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

$1, 4 + x$

Step 1.2

Remove parentheses.

$1, 4 + x$

Step 1.3

The LCM of one and any expression is the expression.

$4 + x$

Step 2

Multiply each term in $4 x = \frac{12 x}{4 + x}$ by $4 + x$ to eliminate the fractions.

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Step 2.1

Multiply each term in $4 x = \frac{12 x}{4 + x}$ by $4 + x$ .

$4 x (4 + x) = \frac{12 x}{4 + x} (4 + x)$

Step 2.2

Simplify the left side.

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Step 2.2.1

Apply the distributive property.

$4 x \cdot 4 + 4 x \cdot x = \frac{12 x}{4 + x} (4 + x)$

Step 2.2.2

Multiply $4$ by $4$ .

$16 x + 4 x \cdot x = \frac{12 x}{4 + x} (4 + x)$

Step 2.2.3

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

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Step 2.2.3.1

Move $x$ .

$16 x + 4 (x \cdot x) = \frac{12 x}{4 + x} (4 + x)$

Step 2.2.3.2

Multiply $x$ by $x$ .

$16 x + 4 x^{2} = \frac{12 x}{4 + x} (4 + x)$

Step 2.3

Simplify the right side.

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Step 2.3.1

Cancel the common factor of $4 + x$ .

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Step 2.3.1.1

Cancel the common factor.

$16 x + 4 x^{2} = \frac{12 x}{4 + x} (4 + x)$

Step 2.3.1.2

Rewrite the expression.

$16 x + 4 x^{2} = 12 x$

Step 3

Solve the equation.

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Step 3.1

Move all terms containing $x$ to the left side of the equation.

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Step 3.1.1

Subtract $12 x$ from both sides of the equation.

$16 x + 4 x^{2} - 12 x = 0$

Step 3.1.2

Subtract $12 x$ from $16 x$ .

$4 x^{2} + 4 x = 0$

Step 3.2

Factor $4 x$ out of $4 x^{2} + 4 x$ .

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Step 3.2.1

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

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

Step 3.2.2

Factor $4 x$ out of $4 x$ .

$4 x (x) + 4 x (1) = 0$

Step 3.2.3

Factor $4 x$ out of $4 x (x) + 4 x (1)$ .

$4 x (x + 1) = 0$

Step 3.3

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

$x = 0$

$x + 1 = 0$

Step 3.4

Set $x$ equal to $0$ .

$x = 0$

Step 3.5

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 3.5.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 3.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3.6

The final solution is all the values that make $4 x (x + 1) = 0$ true.

$x = 0, - 1$