Algebra Examples

$\frac{x^{2} + 9 y^{2}}{x - 3 y} + \frac{6 x y}{3 y - x}$

Step 1

Simplify terms.

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Step 1.1

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

$\frac{x^{2} + 9 y^{2}}{x - 3 y} + \frac{6 x y}{- (- 3 y) - x}$

Step 1.2

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

$\frac{x^{2} + 9 y^{2}}{x - 3 y} + \frac{6 x y}{- (- 3 y) - (x)}$

Step 1.3

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

$\frac{x^{2} + 9 y^{2}}{x - 3 y} + \frac{6 x y}{- (- 3 y + x)}$

Step 1.4

Simplify the expression.

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Step 1.4.1

Move a negative from the denominator of $\frac{6 x y}{- (- 3 y + x)}$ to the numerator.

$\frac{x^{2} + 9 y^{2}}{x - 3 y} + \frac{- (6 x y)}{- 3 y + x}$

Step 1.4.2

Reorder terms.

$\frac{x^{2} + 9 y^{2}}{x - 3 y} + \frac{- (6 x y)}{x - 3 y}$

Step 1.5

Combine the numerators over the common denominator.

$\frac{x^{2} + 9 y^{2} - (6 x y)}{x - 3 y}$

Step 2

Factor using the perfect square rule.

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Step 2.1

Rearrange terms.

$\frac{x^{2} - 1 \cdot 6 x y + 9 y^{2}}{x - 3 y}$

Step 2.2

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

$\frac{x^{2} - 1 \cdot 6 x y + {(3 y)}^{2}}{x - 3 y}$

Step 2.3

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

$1 \cdot 6 x y = 2 \cdot x \cdot (3 y)$

Step 2.4

Rewrite the polynomial.

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

Step 2.5

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

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

Step 3

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

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Step 3.1

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

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

Step 3.2

Cancel the common factors.

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Step 3.2.1

Multiply by $1$ .

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

Step 3.2.2

Cancel the common factor.

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

Step 3.2.3

Rewrite the expression.

$\frac{x - 3 y}{1}$

Step 3.2.4

Divide $x - 3 y$ by $1$ .

$x - 3 y$