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All SAT Mathematics Resources

137 Practice Tests Question of the Day Flashcards Learn by Concept

Want to review SAT Mathematics but don’t feel like sitting for a whole test at the moment? Varsity Tutors has you covered with thousands of different SAT Mathematics flashcards! Our SAT Mathematics flashcards allow you to practice with as few or as many questions as you like. Get some studying in now with our numerous SAT Mathematics flashcards.

Flashcards: Linear Algebra

The expression $\frac{4y-7}{y+3}$ is equivalent to which of the following?

$4-\frac{7}{3}$

3y-4

$4-\frac{19}{y+3}$

$4+\frac{19}{y+3}$

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Flashcards: Linear Algebra

The expression $\frac{4y-7}{y+3}$ is equivalent to which of the following?

Your answer:

$4-\frac{7}{3}$

Your answer:

3y-4

Your answer:

$4+\frac{19}{y+3}$

Correct answer:

$4-\frac{19}{y+3}$

Explanation:

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 y=2 , then the initial equation would be $\frac{4(2)-7}{2+5}=\frac{1}{5}$ . Now your job would be to plug in y=2 to the other answer choices to see if you get a match at $\frac{1}{5}$ .

For 3y-4 , clearly you won't get a fraction by plugging in y=2 so that is incorrect. For $4-\frac{7}{3}$ you should also see quickly that the answer is not $\frac{1}{5}$ . That leaves the two similar-looking fractions, $4-\frac{19}{y+3}$ and $4+\frac{19}{y+3}$ . If you plug in y=2 to $4-\frac{19}{y+3}$ you'll get $4-\frac{19}{5}$ . Since $4=\frac{20}{5}$ , this choice works out to exactly $\frac{1}{5}$ , proving that you have the right answer.

Flashcards: Linear Algebra

If a>-2 , which of the following is equivalent to a^2+5a+6 ?

$\frac{1}{\frac{1}{a+2}+\frac{1}{a+3}}$

$\frac{1}{\frac{1}{a+2}-\frac{1}{a+3}}$

$\frac{1}{\frac{1}{a+3}-\frac{1}{a+2}}$

$\frac{1}{(a+2)(a+3)}$

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Flashcards: Linear Algebra

If a>-2 , which of the following is equivalent to a^2+5a+6 ?

Your answer:

$\frac{1}{\frac{1}{a+2}+\frac{1}{a+3}}$

Your answer:

$\frac{1}{\frac{1}{a+3}-\frac{1}{a+2}}$

Your answer:

$\frac{1}{(a+2)(a+3)}$

Correct answer:

$\frac{1}{\frac{1}{a+2}-\frac{1}{a+3}}$

Explanation:

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 a=-1 so that you can easily set some of the a+2 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:

(-1)^2+5(-1)+6 = 1-5+6=2

Now your job is to test the answer choices using a=-1 to see which answer choice produces the value . And in doing this work, you should find that only $\frac{1}{\frac{1}{a+2}-\frac{1}{a+3}}$ gives you that 2 you're looking for. When you plug in a=-1 this expression becomes:

$\frac{1}{\frac{1}{-1+2}-\frac{1}{-1+3}}=\frac{1}{\frac{1}{1}-\frac{1}{2}}=\frac{1}{\frac{1}{2}}$ which works out to .

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Flashcards: Linear Algebra

Flashcards: Linear Algebra

Flashcards: Linear Algebra

Flashcards: Linear Algebra

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All SAT Mathematics Resources

Flashcards: Linear Algebra

Flashcards: Linear Algebra

Flashcards: Linear Algebra

Flashcards: Linear Algebra

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DMCA Complaint

Contact Information

Address

Complaint Details

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