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Flashcards: Understanding Absolute Values

Which of the following, if solved, would yield the set of all real numbers that are exactly 15 units away from -3?

$\left | x+3 \right |=15$

$\left | x-15 \right |=3$

$\left | x+15 \right |=3$

$\left | x-3 \right |=15$

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Flashcards: Understanding Absolute Values

Which of the following, if solved, would yield the set of all real numbers that are exactly 15 units away from -3?

Your answer:

$\left | x-15 \right |=3$

Your answer:

$\left | x+15 \right |=3$

Your answer:

$\left | x-3 \right |=15$

Correct answer:

$\left | x+3 \right |=15$

Explanation:

If you're looking for a number exactly units to either side of , you need an absolute value equation that gives you answers of -3+15=12 and -3-15=-18 .

One way forward is to simply solve each equation for using the two-case method, assuming first that the expression within the absolute value signs positive and then that it is negative and solving the resulting equation for each case.

If you do so, choice $\left | x+3 \right |=15$ becomes: x+3=15 , or x=12 and x+3=-15 , or x=-18 , which is correct.

Choice $\left | x-3 \right |=15$ becomes: x-3=15 , or x=18 , which doesn't work.

Choice $\left | x-15 \right |=3$ becomes: x-15=3 , or x=18 , which doesn't match.

Choice $\left | x+15 \right |=3$ becomes: x+15=3 , or x=-12 , which doesn't work.

You could also quickly solve this question if you remember the general way number line absolute value problems like this one work. Each one will be in the general form $\left | x-Center\, Value \right |=Distance$ , where the "Distance" referenced is the distance from the center value to one of the two values for .

So in this case the "center value" would be and the "distance" would be , yielding an equation of $\left | x+3 \right |=15$ .

Flashcards: Understanding Absolute Values

If is an integer, what is the value of such that $\left | -5x+7 \right |$ is minimized?

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Flashcards: Understanding Absolute Values

If is an integer, what is the value of such that $\left | -5x+7 \right |$ is minimized?

Your answer:

Correct answer:

Explanation:

Any time you have an absolute value equation and a question that is asking for the smallest possible value for the expression, remember that absolute value expressions can be zero but can never be less than 0. There are two ways to solve a problem like the one given. The first is by finding the value of such that the expression within the absolute value signs is zero and then finding the closest integer to that number. The second is by inspection (simply using brute force to find the value for x that will minimize the value given).

If you set -5x+7-5x=-7 , and that $x=\frac{7}{5}$ . Since $\frac{7}{5}=1.4$ , the closest integer value is x=1 , which will yield a value of $\left | -5+7 \right |=2$ .

The other way to approach this problem is by inspection, or through brute force. Try plugging in a few values for to see what happens.

If x=0 then the expression is $\left | -5(0)+7 \right |=7$

If x=1 , then the expression is $\left | -5(1)+7 \right |=2$ .

If x=2 then the expression becomes $\left | -5(2)+7 \right |=3$ .

Notice that the function decreases and then increases around x=1 , indicating that (for integer values, at least) x=1 will yield the smallest result for $\left | -5x+7 \right |$ .

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Flashcards: Understanding Absolute Values

Flashcards: Understanding Absolute Values

Flashcards: Understanding Absolute Values

Flashcards: Understanding Absolute Values

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All SAT Mathematics Resources

Flashcards: Understanding Absolute Values

Flashcards: Understanding Absolute Values

Flashcards: Understanding Absolute Values

Flashcards: Understanding Absolute Values

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DMCA Complaint

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Complaint Details

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