Simplify (9x^2+4y^2-12xy)/(4y^2-9x^2)

$\frac{9 x^{2} + 4 y^{2} - 12 x y}{4 y^{2} - 9 x^{2}}$

Factor using the perfect square rule.

Rearrange terms.

$\frac{9 x^{2} - 12 x y + 4 y^{2}}{4 y^{2} - 9 x^{2}}$

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

$\frac{{(3 x)}^{2} - 12 x y + 4 y^{2}}{4 y^{2} - 9 x^{2}}$

Rewrite $4 y^{2}$ as ${(2 y)}^{2}$ .

$\frac{{(3 x)}^{2} - 12 x y + {(2 y)}^{2}}{4 y^{2} - 9 x^{2}}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$12 x y = 2 \cdot (3 x) \cdot (2 y)$

Rewrite the polynomial.

$\frac{{(3 x)}^{2} - 2 \cdot (3 x) \cdot (2 y) + {(2 y)}^{2}}{4 y^{2} - 9 x^{2}}$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 3 x$ and $b = 2 y$ .

$\frac{{(3 x - 2 y)}^{2}}{4 y^{2} - 9 x^{2}}$

Simplify the denominator.

Rewrite $4 y^{2}$ as ${(2 y)}^{2}$ .

$\frac{{(3 x - 2 y)}^{2}}{{(2 y)}^{2} - 9 x^{2}}$

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

$\frac{{(3 x - 2 y)}^{2}}{{(2 y)}^{2} - {(3 x)}^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 y$ and $b = 3 x$ .

$\frac{{(3 x - 2 y)}^{2}}{(2 y + 3 x) (2 y - (3 x))}$

Multiply $3$ by $- 1$ .

$\frac{{(3 x - 2 y)}^{2}}{(2 y + 3 x) (2 y - 3 x)}$

Simplify terms.

Cancel the common factor of ${(3 x - 2 y)}^{2}$ and $2 y - 3 x$ .

Factor $- 1$ out of $3 x$ .

$\frac{{(- (- 3 x) - 2 y)}^{2}}{(2 y + 3 x) (2 y - 3 x)}$

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

$\frac{{(- (- 3 x) - (2 y))}^{2}}{(2 y + 3 x) (2 y - 3 x)}$

Factor $- 1$ out of $- (- 3 x) - (2 y)$ .

$\frac{{(- (- 3 x + 2 y))}^{2}}{(2 y + 3 x) (2 y - 3 x)}$

Rewrite $- (- 3 x + 2 y)$ as $- 1 (- 3 x + 2 y)$ .

$\frac{{(- 1 (- 3 x + 2 y))}^{2}}{(2 y + 3 x) (2 y - 3 x)}$

Apply the product rule to $- 1 (- 3 x + 2 y)$ .

$\frac{{(- 1)}^{2} {(- 3 x + 2 y)}^{2}}{(2 y + 3 x) (2 y - 3 x)}$

Raise $- 1$ to the power of $2$ .

$\frac{1 {(- 3 x + 2 y)}^{2}}{(2 y + 3 x) (2 y - 3 x)}$

Multiply ${(- 3 x + 2 y)}^{2}$ by $1$ .

$\frac{{(- 3 x + 2 y)}^{2}}{(2 y + 3 x) (2 y - 3 x)}$

Reorder terms.

$\frac{{(- 3 x + 2 y)}^{2}}{(2 y + 3 x) (- 3 x + 2 y)}$

Factor $- 3 x + 2 y$ out of ${(- 3 x + 2 y)}^{2}$ .

$\frac{(- 3 x + 2 y) (- 3 x + 2 y)}{(2 y + 3 x) (- 3 x + 2 y)}$

Cancel the common factors.

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

$\frac{(- 3 x + 2 y) (- 3 x + 2 y)}{(- 3 x + 2 y) (2 y + 3 x)}$

Cancel the common factor.

$\frac{(- 3 x + 2 y) (- 3 x + 2 y)}{(- 3 x + 2 y) (2 y + 3 x)}$

Rewrite the expression.

$\frac{- 3 x + 2 y}{2 y + 3 x}$

Factor $- 1$ out of $- 3 x$ .

$\frac{- (3 x) + 2 y}{2 y + 3 x}$

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

$\frac{- (3 x) - (- 2 y)}{2 y + 3 x}$

Factor $- 1$ out of $- (3 x) - (- 2 y)$ .

$\frac{- (3 x - 2 y)}{2 y + 3 x}$

Simplify the expression.

Rewrite $- (3 x - 2 y)$ as $- 1 (3 x - 2 y)$ .

$\frac{- 1 (3 x - 2 y)}{2 y + 3 x}$

Move the negative in front of the fraction.

$- \frac{3 x - 2 y}{2 y + 3 x}$

Simplify (9x^2+4y^2-12xy)/(4y^2-9x^2)

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