Solve for x 3x-5/(x+4)=2x

$3 x - \frac{5}{x + 4} = 2 x$

Step 1

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.

$1, x + 4, 1$

Remove parentheses.

$1, x + 4, 1$

The LCM of one and any expression is the expression.

$x + 4$

Step 2

Multiply each term in $3 x - \frac{5}{x + 4} = 2 x$ by $x + 4$ to eliminate the fractions.

Multiply each term in $3 x - \frac{5}{x + 4} = 2 x$ by $x + 4$ .

$3 x (x + 4) - \frac{5}{x + 4} (x + 4) = 2 x (x + 4)$

Simplify the left side.

Simplify each term.

Apply the distributive property.

$3 x \cdot x + 3 x \cdot 4 - \frac{5}{x + 4} (x + 4) = 2 x (x + 4)$

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

Move $x$ .

$3 (x \cdot x) + 3 x \cdot 4 - \frac{5}{x + 4} (x + 4) = 2 x (x + 4)$

Multiply $x$ by $x$ .

$3 x^{2} + 3 x \cdot 4 - \frac{5}{x + 4} (x + 4) = 2 x (x + 4)$

Multiply $4$ by $3$ .

$3 x^{2} + 12 x - \frac{5}{x + 4} (x + 4) = 2 x (x + 4)$

Cancel the common factor of $x + 4$ .

Move the leading negative in $- \frac{5}{x + 4}$ into the numerator.

$3 x^{2} + 12 x + \frac{- 5}{x + 4} (x + 4) = 2 x (x + 4)$

Cancel the common factor.

$3 x^{2} + 12 x + \frac{- 5}{x + 4} (x + 4) = 2 x (x + 4)$

Rewrite the expression.

$3 x^{2} + 12 x - 5 = 2 x (x + 4)$

Simplify the right side.

Apply the distributive property.

$3 x^{2} + 12 x - 5 = 2 x \cdot x + 2 x \cdot 4$

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

Move $x$ .

$3 x^{2} + 12 x - 5 = 2 (x \cdot x) + 2 x \cdot 4$

Multiply $x$ by $x$ .

$3 x^{2} + 12 x - 5 = 2 x^{2} + 2 x \cdot 4$

Multiply $4$ by $2$ .

$3 x^{2} + 12 x - 5 = 2 x^{2} + 8 x$

Step 3

Solve the equation.

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

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

$3 x^{2} + 12 x - 5 - 2 x^{2} = 8 x$

Subtract $8 x$ from both sides of the equation.

$3 x^{2} + 12 x - 5 - 2 x^{2} - 8 x = 0$

Subtract $2 x^{2}$ from $3 x^{2}$ .

$x^{2} + 12 x - 5 - 8 x = 0$

Subtract $8 x$ from $12 x$ .

$x^{2} + 4 x - 5 = 0$

Factor $x^{2} + 4 x - 5$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 5$ and whose sum is $4$ .

$- 1, 5$

Write the factored form using these integers.

$(x - 1) (x + 5) = 0$

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

$x - 1 = 0$

$x + 5 = 0$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

The final solution is all the values that make $(x - 1) (x + 5) = 0$ true.

$x = 1, - 5$

Solve for x 3x-5/(x+4)=2x

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