Solve for x x/(x-2)=(x-3)/(2x)

$\frac{x}{x - 2} = \frac{x - 3}{2 x}$

Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.

$x (2 x) = (x - 2) (x - 3)$

Solve the equation for $x$ .

Simplify $x (2 x)$ .

Rewrite using the commutative property of multiplication.

$2 x \cdot x = (x - 2) (x - 3)$

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

Move $x$ .

$2 (x \cdot x) = (x - 2) (x - 3)$

Multiply $x$ by $x$ .

$2 x^{2} = (x - 2) (x - 3)$

Simplify $(x - 2) (x - 3)$ .

Expand $(x - 2) (x - 3)$ using the FOIL Method.

Apply the distributive property.

$2 x^{2} = x (x - 3) - 2 (x - 3)$

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Move $- 3$ to the left of $x$ .

$2 x^{2} = x^{2} - 3 \cdot x - 2 x - 2 \cdot - 3$

Multiply $- 2$ by $- 3$ .

$2 x^{2} = x^{2} - 3 x - 2 x + 6$

Subtract $2 x$ from $- 3 x$ .

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

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

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

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

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

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

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

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

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} - 5 x + 6$ .

Factor $- 1$ out of $- x^{2}$ .

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

Factor $- 1$ out of $- 5 x$ .

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

Rewrite $6$ as $- 1 (- 6)$ .

$- (x^{2}) - (5 x) - 1 \cdot - 6 = 0$

Factor $- 1$ out of $- (x^{2}) - (5 x)$ .

$- (x^{2} + 5 x) - 1 \cdot - 6 = 0$

Factor $- 1$ out of $- (x^{2} + 5 x) - 1 (- 6)$ .

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

Factor.

Factor $x^{2} + 5 x - 6$ 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 $- 6$ and whose sum is $5$ .

$- 1, 6$

Write the factored form using these integers.

$- ((x - 1) (x + 6)) = 0$

Remove unnecessary parentheses.

$- (x - 1) (x + 6) = 0$

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

$x - 1 = 0$

$x + 6 = 0$

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 $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $- (x - 1) (x + 6) = 0$ true.

$x = 1, - 6$

Solve for x x/(x-2)=(x-3)/(2x)

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