If you can write $4x^2 - 9y^2$ in the form $a^2 - b^2$, use the identity $a^2 - b^2 = (a - b)(a + b)$ to factor it.

Once the numerator is factored, compare it to the denominator $2x - 3y$. Is there a common factor you can cancel, given that $2x \ne 3y$?

In Desmos, type x = 1 and y = 1 on two separate lines. Turn them into sliders so you can change the values of $x$ and $y$ easily.

On a new line, type (4x^2 - 9y^2)/(2x - 3y). Desmos will show a numeric value using the current slider values of $x$ and $y$ (avoid values where 2x - 3y = 0).

On separate lines, enter each option:

• 2x - 3y
• (4x^2 + 9y^2)/(2x + 3y)
• (2x + 3y)/2
• 2x + 3y Desmos will compute a value for each using the same $x$ and $y$ sliders.

Move the $x$ and $y$ sliders to several different pairs of values where 2x - 3y ≠ 0. For each pair, compare the value of the original expression to each of the four options. The equivalent expression is the one whose value always matches the original expression for every tested pair.

Look at the numerator $4x^2 - 9y^2$.

• Notice that $4x^2 = (2x)^2$ and $9y^2 = (3y)^2$.
• So the numerator has the form $a^2 - b^2$, where $a = 2x$ and $b = 3y$.

This is the difference of squares pattern.

Use the identity $a^2 - b^2 = (a - b)(a + b)$ with $a = 2x$ and $b = 3y$. So

• $4x^2 - 9y^2 = (2x)^2 - (3y)^2 = (2x - 3y)(2x + 3y)$.

Now the whole fraction becomes

• $\dfrac{4x^2 - 9y^2}{2x - 3y} = \dfrac{(2x - 3y)(2x + 3y)}{2x - 3y}$.

In $\dfrac{(2x - 3y)(2x + 3y)}{2x - 3y}$, the factor $2x - 3y$ appears in both numerator and denominator.

• Because the problem states $2x \ne 3y$, we know $2x - 3y \ne 0$, so it is safe to divide both numerator and denominator by $2x - 3y$.
• After canceling this common factor, only one factor remains in the numerator.

After canceling $2x - 3y$, the simplified expression is $2x + 3y$.

Therefore, the expression equivalent to $\dfrac{4x^2 - 9y^2}{2x - 3y}$ (for $2x \ne 3y$) is $2x + 3y$, which corresponds to choice D.