Solve by Factoring 10/x-12/(x-3)+4=0

$\frac{10}{x} - \frac{12}{x - 3} + 4 = 0$

Step 1

To write $\frac{10}{x}$ as a fraction with a common denominator, multiply by $\frac{x - 3}{x - 3}$ .

$\frac{10}{x} \cdot \frac{x - 3}{x - 3} - \frac{12}{x - 3} + 4 = 0$

Step 2

To write $- \frac{12}{x - 3}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$\frac{10}{x} \cdot \frac{x - 3}{x - 3} - \frac{12}{x - 3} \cdot \frac{x}{x} + 4 = 0$

Step 3

Write each expression with a common denominator of $x (x - 3)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{10}{x}$ by $\frac{x - 3}{x - 3}$ .

$\frac{10 (x - 3)}{x (x - 3)} - \frac{12}{x - 3} \cdot \frac{x}{x} + 4 = 0$

Multiply $\frac{12}{x - 3}$ by $\frac{x}{x}$ .

$\frac{10 (x - 3)}{x (x - 3)} - \frac{12 x}{(x - 3) x} + 4 = 0$

Reorder the factors of $(x - 3) x$ .

$\frac{10 (x - 3)}{x (x - 3)} - \frac{12 x}{x (x - 3)} + 4 = 0$

Step 4

Combine the numerators over the common denominator.

$\frac{10 (x - 3) - 12 x}{x (x - 3)} + 4 = 0$

Step 5

Simplify the numerator.

Factor $2$ out of $10 (x - 3) - 12 x$ .

Factor $2$ out of $10 (x - 3)$ .

$\frac{2 (5 (x - 3)) - 12 x}{x (x - 3)} + 4 = 0$

Factor $2$ out of $- 12 x$ .

$\frac{2 (5 (x - 3)) + 2 (- 6 x)}{x (x - 3)} + 4 = 0$

Factor $2$ out of $2 (5 (x - 3)) + 2 (- 6 x)$ .

$\frac{2 (5 (x - 3) - 6 x)}{x (x - 3)} + 4 = 0$

Apply the distributive property.

$\frac{2 (5 x + 5 \cdot - 3 - 6 x)}{x (x - 3)} + 4 = 0$

Multiply $5$ by $- 3$ .

$\frac{2 (5 x - 15 - 6 x)}{x (x - 3)} + 4 = 0$

Subtract $6 x$ from $5 x$ .

$\frac{2 (- x - 15)}{x (x - 3)} + 4 = 0$

Step 6

To write $4$ as a fraction with a common denominator, multiply by $\frac{x (x - 3)}{x (x - 3)}$ .

$\frac{2 (- x - 15)}{x (x - 3)} + 4 \cdot \frac{x (x - 3)}{x (x - 3)} = 0$

Step 7

Combine $4$ and $\frac{x (x - 3)}{x (x - 3)}$ .

$\frac{2 (- x - 15)}{x (x - 3)} + \frac{4 (x (x - 3))}{x (x - 3)} = 0$

Step 8

Combine the numerators over the common denominator.

$\frac{2 (- x - 15) + 4 (x (x - 3))}{x (x - 3)} = 0$

Step 9

Remove parentheses.

$\frac{2 (- x - 15) + 4 x (x - 3)}{x (x - 3)} = 0$

Step 10

Simplify the numerator.

Factor $2$ out of $2 (- x - 15) + 4 x (x - 3)$ .

Factor $2$ out of $4 x (x - 3)$ .

$\frac{2 (- x - 15) + 2 (2 x (x - 3))}{x (x - 3)} = 0$

Factor $2$ out of $2 (- x - 15) + 2 (2 x (x - 3))$ .

$\frac{2 (- x - 15 + 2 x (x - 3))}{x (x - 3)} = 0$

Apply the distributive property.

$\frac{2 (- x - 15 + 2 x \cdot x + 2 x \cdot - 3)}{x (x - 3)} = 0$

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

Move $x$ .

$\frac{2 (- x - 15 + 2 (x \cdot x) + 2 x \cdot - 3)}{x (x - 3)} = 0$

Multiply $x$ by $x$ .

$\frac{2 (- x - 15 + 2 x^{2} + 2 x \cdot - 3)}{x (x - 3)} = 0$

Multiply $- 3$ by $2$ .

$\frac{2 (- x - 15 + 2 x^{2} - 6 x)}{x (x - 3)} = 0$

Subtract $6 x$ from $- x$ .

$\frac{2 (- 7 x - 15 + 2 x^{2})}{x (x - 3)} = 0$

Reorder terms.

$\frac{2 (2 x^{2} - 7 x - 15)}{x (x - 3)} = 0$

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 15 = - 30$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 x$ .

$\frac{2 (2 x^{2} - 7 x - 15)}{x (x - 3)} = 0$

Rewrite $- 7$ as $3$ plus $- 10$

$\frac{2 (2 x^{2} + (3 - 10) x - 15)}{x (x - 3)} = 0$

Apply the distributive property.

$\frac{2 (2 x^{2} + 3 x - 10 x - 15)}{x (x - 3)} = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{2 ((2 x^{2} + 3 x) - 10 x - 15)}{x (x - 3)} = 0$

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

$\frac{2 (x (2 x + 3) - 5 (2 x + 3))}{x (x - 3)} = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x + 3$ .

$\frac{2 ((2 x + 3) (x - 5))}{x (x - 3)} = 0$

Remove unnecessary parentheses.

$\frac{2 (2 x + 3) (x - 5)}{x (x - 3)} = 0$

Step 11

Set the numerator equal to zero.

$2 (2 x + 3) (x - 5) = 0$

Step 12

Solve the equation for $x$ .

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

$2 x + 3 = 0$

$x - 5 = 0$

Set $2 x + 3$ equal to $0$ and solve for $x$ .

Set $2 x + 3$ equal to $0$ .

$2 x + 3 = 0$

Solve $2 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Divide each term in $2 x = - 3$ by $2$ and simplify.

Divide each term in $2 x = - 3$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

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.

$x = 5$

The final solution is all the values that make $2 (2 x + 3) (x - 5) = 0$ true.

$x = - \frac{3}{2}, 5$

Solve by Factoring 10/x-12/(x-3)+4=0

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