Solve by Factoring 4/(2x-3)=x/5

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

Step 1

Subtract $\frac{x}{5}$ from both sides of the equation.

$\frac{4}{2 x - 3} - \frac{x}{5} = 0$

Step 2

Simplify $\frac{4}{2 x - 3} - \frac{x}{5}$ .

To write $\frac{4}{2 x - 3}$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$\frac{4}{2 x - 3} \cdot \frac{5}{5} - \frac{x}{5} = 0$

To write $- \frac{x}{5}$ as a fraction with a common denominator, multiply by $\frac{2 x - 3}{2 x - 3}$ .

$\frac{4}{2 x - 3} \cdot \frac{5}{5} - \frac{x}{5} \cdot \frac{2 x - 3}{2 x - 3} = 0$

Write each expression with a common denominator of $(2 x - 3) \cdot 5$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{4}{2 x - 3}$ by $\frac{5}{5}$ .

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

Multiply $\frac{x}{5}$ by $\frac{2 x - 3}{2 x - 3}$ .

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

Reorder the factors of $(2 x - 3) \cdot 5$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $4$ by $5$ .

$\frac{20 - x (2 x - 3)}{5 (2 x - 3)} = 0$

Apply the distributive property.

$\frac{20 - x (2 x) - x \cdot - 3}{5 (2 x - 3)} = 0$

Rewrite using the commutative property of multiplication.

$\frac{20 - 1 \cdot 2 x \cdot x - x \cdot - 3}{5 (2 x - 3)} = 0$

Multiply $- 3$ by $- 1$ .

$\frac{20 - 1 \cdot 2 x \cdot x + 3 x}{5 (2 x - 3)} = 0$

Simplify each term.

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

Move $x$ .

$\frac{20 - 1 \cdot 2 (x \cdot x) + 3 x}{5 (2 x - 3)} = 0$

Multiply $x$ by $x$ .

$\frac{20 - 1 \cdot 2 x^{2} + 3 x}{5 (2 x - 3)} = 0$

Multiply $- 1$ by $2$ .

$\frac{20 - 2 x^{2} + 3 x}{5 (2 x - 3)} = 0$

Reorder terms.

$\frac{- 2 x^{2} + 3 x + 20}{5 (2 x - 3)} = 0$

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 $a \cdot c = - 2 \cdot 20 = - 40$ and whose sum is $b = 3$ .

Factor $3$ out of $3 x$ .

$\frac{- 2 x^{2} + 3 (x) + 20}{5 (2 x - 3)} = 0$

Rewrite $3$ as $- 5$ plus $8$

$\frac{- 2 x^{2} + (- 5 + 8) x + 20}{5 (2 x - 3)} = 0$

Apply the distributive property.

$\frac{- 2 x^{2} - 5 x + 8 x + 20}{5 (2 x - 3)} = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(- 2 x^{2} - 5 x) + 8 x + 20}{5 (2 x - 3)} = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $- 2 x - 5$ .

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

Factor $- 1$ out of $- 2 x$ .

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

Rewrite $- 5$ as $- 1 (5)$ .

$\frac{(- (2 x) - 1 (5)) (x - 4)}{5 (2 x - 3)} = 0$

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

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

Rewrite $- (2 x + 5)$ as $- 1 (2 x + 5)$ .

$\frac{- 1 (2 x + 5) (x - 4)}{5 (2 x - 3)} = 0$

Move the negative in front of the fraction.

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

Step 3

Set the numerator equal to zero.

$(2 x + 5) (x - 4) = 0$

Step 4

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 $0$ .

$2 x + 5 = 0$

$x - 4 = 0$

Set $2 x + 5$ equal to $0$ and solve for $x$ .

Set $2 x + 5$ equal to $0$ .

$2 x + 5 = 0$

Solve $2 x + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Divide each term in $2 x = - 5$ by $2$ and simplify.

Divide each term in $2 x = - 5$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

The final solution is all the values that make $(2 x + 5) (x - 4) = 0$ true.

$x = - \frac{5}{2}, 4$

Solve by Factoring 4/(2x-3)=x/5

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