After you factor the numerator, see if any factor in the numerator matches the denominator $2x - 5$. If so, what happens to that factor in a fraction?

Once the fraction is simplified, rewrite the whole expression without the fraction and then combine like terms involving $x$.

When you cancel a factor like $(2x - 5)$ from the numerator and denominator, the result is not defined where $2x - 5 = 0$. Keep in mind you only need equivalence where the original expression is defined.

In Desmos, type f(x) = (4x^2 - 25)/(2x - 5) - 2x to graph the original expression. Notice there will be a missing point (a hole) where the denominator is zero.

Enter additional functions:

• g(x) = 5
• h(x) = 2x + 5
• p(x) = 4x - 5
• q(x) = x + 5 Compare each of these graphs to the graph of $f(x)$.

Look for the function whose graph lies exactly on top of the graph of $f(x)$ everywhere that $f(x)$ is defined (except at the hole where $f(x)$ is undefined). The function that perfectly overlaps in this way is the equivalent expression.

Notice that $4x^2 - 25$ is a difference of squares:

So the expression becomes

In the fraction, $(2x - 5)$ appears in both the numerator and the denominator, so for $2x - 5 \ne 0$ (that is, $x \ne 2.5$), we can cancel it:

Now the whole expression becomes

Now simplify $2x + 5 - 2x$ by combining like terms. The $2x$ and $-2x$ are like terms and add to $0$:

Since $2x - 2x = 0$, we have

So, for all values of $x$ where the original expression is defined (all real $x$ except $x = 2.5$), the expression is equal to $5$. The equivalent expression is $5$, which corresponds to choice A.