Solve for x 3x+5/(x-6)=2x

$3 x + \frac{5}{x - 6} = 2 x$

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.

$1, x - 6, 1$

Remove parentheses.

$1, x - 6, 1$

The LCM of one and any expression is the expression.

$x - 6$

Multiply each term in $3 x + \frac{5}{x - 6} = 2 x$ by $x - 6$ to eliminate the fractions.

Multiply each term in $3 x + \frac{5}{x - 6} = 2 x$ by $x - 6$ .

$3 x (x - 6) + \frac{5}{x - 6} (x - 6) = 2 x (x - 6)$

Simplify the left side.

Simplify each term.

Apply the distributive property.

$3 x \cdot x + 3 x \cdot - 6 + \frac{5}{x - 6} (x - 6) = 2 x (x - 6)$

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

Move $x$ .

$3 (x \cdot x) + 3 x \cdot - 6 + \frac{5}{x - 6} (x - 6) = 2 x (x - 6)$

Multiply $x$ by $x$ .

$3 x^{2} + 3 x \cdot - 6 + \frac{5}{x - 6} (x - 6) = 2 x (x - 6)$

Multiply $- 6$ by $3$ .

$3 x^{2} - 18 x + \frac{5}{x - 6} (x - 6) = 2 x (x - 6)$

Cancel the common factor of $x - 6$ .

Cancel the common factor.

$3 x^{2} - 18 x + \frac{5}{x - 6} (x - 6) = 2 x (x - 6)$

Rewrite the expression.

$3 x^{2} - 18 x + 5 = 2 x (x - 6)$

Simplify the right side.

Apply the distributive property.

$3 x^{2} - 18 x + 5 = 2 x \cdot x + 2 x \cdot - 6$

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

Move $x$ .

$3 x^{2} - 18 x + 5 = 2 (x \cdot x) + 2 x \cdot - 6$

Multiply $x$ by $x$ .

$3 x^{2} - 18 x + 5 = 2 x^{2} + 2 x \cdot - 6$

Multiply $- 6$ by $2$ .

$3 x^{2} - 18 x + 5 = 2 x^{2} - 12 x$

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.

$3 x^{2} - 18 x + 5 - 2 x^{2} = - 12 x$

Add $12 x$ to both sides of the equation.

$3 x^{2} - 18 x + 5 - 2 x^{2} + 12 x = 0$

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

$x^{2} - 18 x + 5 + 12 x = 0$

Add $- 18 x$ and $12 x$ .

$x^{2} - 6 x + 5 = 0$

Factor $x^{2} - 6 x + 5$ 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 $- 6$ .

$- 5, - 1$

Write the factored form using these integers.

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

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

$x - 5 = 0$

$x - 1 = 0$

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

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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.

$x = 1$

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

$x = 5, 1$

Solve for x 3x+5/(x-6)=2x

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