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 pickso that you can easily set some of thedenominators 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 usingto see which answer choice produces the value. And in doing this work, you should find that onlygives you that 2 you're looking for. When you plug inthis 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.

这里的方法是选择一个容易计算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 ofandis simply. With that multiplication you see that you are left with:

Next you should factor thein 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 ofIf you create a fraction ofto 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:

现在,虽然这答案不匹配选择t, 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 commonin each term that can be factored. So you can make your fraction look like:

Note that theterms 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 thatcannot 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 pluginto 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, withyou know you won't end up with an improper fraction when multiplyingby,所以你没有穿孔orm 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 pluginto the answer choices, you'll see that onlyreturns the target value of:

当你面对“等效表达式”的问题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答案的选择,你会发现只有一个choice,, returns the same target value of:

Thereforeis correct.

当你面对“等效表达式”的问题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 useas 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 sameinto 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. Givenand, how would you make one look like the other? Multiplying both numerator and denominator bywould 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: