When the SAT asks "equivalent expression" questions like you see here, it is almost always faster and easier to pick numbers to test the answer choices against the original; the algebra can be time-consuming and a bit abstract, but since two equivalent expressions will produce the same number in both forms, you can get away with testing numbers.

When you do pick numbers, it's best to pick numbers that are easy to calculate. Here you might pick  so that you can easily set some of the  denominators in the answer choices equal to 1, making for quicker arithmetic. If you do that, you'll find that the original expression works out to:

Now your job is to test the answer choices using   to see which answer choice produces the value .  And in doing this work, you should find that only   gives you that 2 you're looking for.  When you plug in  this expression becomes:

 which works out to .   

When the SAT asks you "equivalent expression" questions, it is often much easier to plug in numbers than it is to try to recreate the abstract algebra. And this strategy works because if two expressions are truly equivalent, then when you plug in numbers for variables you'll get the same answer.  

The technique here is to pick an easy-to-calculate number to plug in for the variable, and then to get a numerical value for the given expression. Then you can plug in the same number as the variable for each answers, and see which choice(s) match the output value. 

Here you might pick , making for an easy number to calculate with. That makes the value of the expression 

Whenever you encounter a multi-denominator expression, simplify that expression by multiplying the top and bottom by the least common multiple of the different denominators. Here the least common multiple of  and  is simply . With that multiplication you see that you are left with:

Next you should factor the  in the numerators so that you can leverage the fact that . With this factoring and then substituting , you see that:

When you approach algebra that features multiple denominators - as you see in this problem, you begin with four "levels" of fraction - a strong algebraic first step is to "multiply by one." This means that you create a fraction with the same numerator and denominator, and use it to cancel all the smaller denominators and greatly reduce the number of fraction "levels" you're working with.

Here, for example, note that the inner fractions include denominators of  If you create a fraction of  to multiply by, that's the same thing as multiplying by 1 (and therefore keeping the value the same), and it will cancel several of the denominators that are making the given fraction complicated:

Now, while this does not match an answer choice yet, it's vastly streamlined compared to the original. And it also lends itself to factoring.  The numerator is a classic Difference of Squares setup: .  And the denominator has a common  in each term that can be factored. So you can make your fraction look like:

Note that the  terms will cancel, leaving you with , the correct answer.

When the SAT asks you to find an equivalent expression, it is often fastest to pick numbers. To do so, choose a number for each variable in the given expression, making sure to choose easy numbers to work with for the situation. Here you know that  cannot be 1 or -1, so you might pick a small number like . Once you've identified your numbers for the given expression, plug those in and get a target value. Here that's:

Now you can plug  into all the answer choices, looking for a match; after all, if the expressions truly are equivalent, then they will produce the same value given the same input for .  Note that you can stop doing any calculation as soon as you realize you won't get your target value. For example, with  you know you won't end up with an improper fraction when multiplying  by , so you don't actually have to perform the math if you know it won't match.

In doing so here, you'll find that the only match is , as .

Often when you're dealing with equivalent expression questions on the SAT, the fastest way to an answer is to pick numbers. That means choosing a number for each variable in the given expression, plugging that number in, and establishing a target value. Then you can plug in that same number into each answer choice and see which choice(s) match the original. If the expressions are equivalent, the target value will match.

Here you might try , which would make the target value calculation look like:

Which simplifies quickly to 

If you then plug  into the answer choices, you'll see that only  returns the target value of :

When you face "equivalent expression" questions on the SAT, often the quickest way to get an answer is by picking numbers. That means choosing a number for each variable in the given expression, then plugging those numbers in to elicit a target value. When you then plug your numbers in for the variables in the answer choices, the correct answer will produce the same target value. This is because if the expressions really are equivalent, they'll produce the same outcome for the same input.

Here you might choose to use , as your number-picking goal should always be to choose a number that's easy to work with. If you do so, you would get an initial expression of:

When you then apply  to the answer choices, you'll find that only one choice, , returns the same target value of :

Therefore  is correct.

When you face "equivalent expression" questions on the SAT, often the quickest way to get an answer is by picking numbers. That means choosing a number for each variable in the given expression, then plugging those numbers in to elicit a target value. When you then plug your numbers in for the variables in the answer choices, the correct answer will produce the same target value. This is because if the expressions really are equivalent, they'll produce the same outcome for the same input.

Here you might choose to use  as your number-picking goal should always be to choose a number that's easy to work with. If you do so, you would get an initial expression of:

Now your job is to plug in that same  into each answer choice to see which one produces the target value of . In doing so, you'll find that the only one that does is , as that produces .

This problem offers an incredible shortcut for those who check the answer choices before performing algebra. The given expression is a proper fraction, in which the numerator ( is between 4 and 5) is smaller than than the denominator (which is 6).  Of the answer choices, ONLY ONE follows that same format, . The others all have a numerator which is greater than the denominator, so none can be correct.

To work on these choices algebraically, a good strategy is to try to make each answer choice look like the given expression.  You can see this with the correct answer. Given  and , how would you make one look like the other?  Multiplying both numerator and denominator by  would mean that you take the answer choice numerator,  and convert it to . And since if you multiply by the same numerator as denominator you're multiplying by one, you're allowed to do that. So you could do:

And since , you can simplify the denominator and see that you've arrived at the given expression: