The first step on this problem is to use the equation that you're provided so that you can solve for . To do so, take and divide both sides by . That leaves you with . Of course, the question doesn't ask you for the value of , but rather the value of . So plug in and the expression looks like: which is . That simplifies to , giving you your final answer.

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

This system of equations provides you an excellent opportunity to use the Elimination Method to isolate a single variable. With the two equations provided, you already have a negative term (in the second equation) and a positive term (in the first). If you multiply the second equation by 2, you can get the coefficients the same and sum the equations to arrive at a single variable, . First, multiply the entire second equation by 2:

Then stack the updated equations and sum them:

Gives you:

This means that but remember to always double check that you've solved for the proper variable. The question wants to know so you can plug back into either equation to solve. Using the original second equation:

So

If you found yourself staring at the initial fraction with no idea how to get started on the algebra, you’re not alone. The first big lesson here is that you should always take a look at the answer choices before you get started. The SAT involves a lot of “algebraic equivalency” – problems that provide you with an algebraic expression and ask you which answer choice is equivalent to it – and as you can see in this case, the answers aren’t necessarily any simpler or cleaner than the original. So an important concept when you’re translating algebra is to see which options they give you for the translation. That way you have a goal in sight and aren’t just casually performing algebra steps in the hopes of arriving at an answer choice.

Then keep in mind: with algebraic equivalency, that equivalency has to hold for all values of the variable. They’re not asking you “what is y?” but rather “which algebraic expression equals this one?” So algebraic equivalency problems – those with variables in the answer choices – are fantastic opportunities to just pick numbers and see which answer choice holds true.

For example, here if you decided to try , then the initial equation would be . Now your job would be to plug in to the other answer choices to see if you get a match at .

For , clearly you won't get a fraction by plugging in so that is incorrect. For you should also see quickly that the answer is not . That leaves the two similar-looking fractions, and . If you plug in to you'll get . Since , this choice works out to exactly , proving that you have the right answer.

The first step on this problem is to use the equation that you're provided so that you can solve for . To do so, take and divide both sides by . That leaves you with . Of course, the question doesn't ask you for the value of , but rather the value of . So plug in and the expression looks like: which is . That simplifies to , giving you your final answer.

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

This system of equations provides you an excellent opportunity to use the Elimination Method to isolate a single variable. With the two equations provided, you already have a negative term (in the second equation) and a positive term (in the first). If you multiply the second equation by 2, you can get the coefficients the same and sum the equations to arrive at a single variable, . First, multiply the entire second equation by 2:

Then stack the updated equations and sum them:

Gives you:

This means that but remember to always double check that you've solved for the proper variable. The question wants to know so you can plug back into either equation to solve. Using the original second equation:

So

If you found yourself staring at the initial fraction with no idea how to get started on the algebra, you’re not alone. The first big lesson here is that you should always take a look at the answer choices before you get started. The SAT involves a lot of “algebraic equivalency” – problems that provide you with an algebraic expression and ask you which answer choice is equivalent to it – and as you can see in this case, the answers aren’t necessarily any simpler or cleaner than the original. So an important concept when you’re translating algebra is to see which options they give you for the translation. That way you have a goal in sight and aren’t just casually performing algebra steps in the hopes of arriving at an answer choice.

Then keep in mind: with algebraic equivalency, that equivalency has to hold for all values of the variable. They’re not asking you “what is y?” but rather “which algebraic expression equals this one?” So algebraic equivalency problems – those with variables in the answer choices – are fantastic opportunities to just pick numbers and see which answer choice holds true.

For example, here if you decided to try , then the initial equation would be . Now your job would be to plug in to the other answer choices to see if you get a match at .

For , clearly you won't get a fraction by plugging in so that is incorrect. For you should also see quickly that the answer is not . That leaves the two similar-looking fractions, and . If you plug in to you'll get . Since , this choice works out to exactly , proving that you have the right answer.

To solve this problem, first perform algebra on the given equation to isolate the terms on one side of the equation and the numeric terms on the other. That means subtracting from both sides and adding to both sides to get:

You can then divide both sides by to realize that .

Now, notice that the question did not ask you for the value of , but rather for the value of . So you now need to plug in for to finish the job:

To solve this problem, first perform algebra on the given equation to isolate the terms on one side of the equation and the numeric terms on the other. That means subtracting from both sides and adding to both sides to get:

You can then divide both sides by to realize that .

Now, notice that the question did not ask you for the value of , but rather for the value of . So you now need to plug in for to finish the job: