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How do you solve absolute values? - Answers
If, for example, you had a problem looking something like this:l3-4l= xFirst you would completely forget that there is an absolute value sign. Just solve 3-4. You should get -1. Absolute value is just the distance on a number line something is from 0. Since -1 is 1 digit away from 0, the answer would be 1.Another trick is to just solve the problem inside the absolute value sign. If you get a positive number, that's your answer. If you get a negative number, just take away the negative sign.If you're solving equations or inequalities, then it can help to break it into two parts:Absolute value of a variable u is: {u if u > 0}, {-u if u < 0}, so you can treat it as two different things to solve.Example, say you have y > |x+2|, where the vertical bars are absolute value.So you have y > x +2, when (x+2)>0, and you have y > -(x+2), when (x+2) x + 2, when x > -2} and {y > -x - 2, when x < -2}. When x = 2, you can use either one, because they are both equal to zero.==========The definition of "absolute value" for a number x (written as |x| ) is:|x| = x for x > 0|x| = 0 for x = 0|x| = -x for x < 0... where x is replaced by some function - for example f(x):|f(x)| = f(x) for f(x)>0|f(x)| = 0 for f(x)=0|f(x)| = -f(x) for f(x) |x+2| which yields:y > x+2 for x > -2y > 0 for x = -2y > -(x+2) for x < -2 ... or alternatively, -y < x+2 for x < -2
Bing
How do you solve absolute values? - Answers
If, for example, you had a problem looking something like this:l3-4l= xFirst you would completely forget that there is an absolute value sign. Just solve 3-4. You should get -1. Absolute value is just the distance on a number line something is from 0. Since -1 is 1 digit away from 0, the answer would be 1.Another trick is to just solve the problem inside the absolute value sign. If you get a positive number, that's your answer. If you get a negative number, just take away the negative sign.If you're solving equations or inequalities, then it can help to break it into two parts:Absolute value of a variable u is: {u if u > 0}, {-u if u < 0}, so you can treat it as two different things to solve.Example, say you have y > |x+2|, where the vertical bars are absolute value.So you have y > x +2, when (x+2)>0, and you have y > -(x+2), when (x+2) x + 2, when x > -2} and {y > -x - 2, when x < -2}. When x = 2, you can use either one, because they are both equal to zero.==========The definition of "absolute value" for a number x (written as |x| ) is:|x| = x for x > 0|x| = 0 for x = 0|x| = -x for x < 0... where x is replaced by some function - for example f(x):|f(x)| = f(x) for f(x)>0|f(x)| = 0 for f(x)=0|f(x)| = -f(x) for f(x) |x+2| which yields:y > x+2 for x > -2y > 0 for x = -2y > -(x+2) for x < -2 ... or alternatively, -y < x+2 for x < -2
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How do you solve absolute values? - Answers
If, for example, you had a problem looking something like this:l3-4l= xFirst you would completely forget that there is an absolute value sign. Just solve 3-4. You should get -1. Absolute value is just the distance on a number line something is from 0. Since -1 is 1 digit away from 0, the answer would be 1.Another trick is to just solve the problem inside the absolute value sign. If you get a positive number, that's your answer. If you get a negative number, just take away the negative sign.If you're solving equations or inequalities, then it can help to break it into two parts:Absolute value of a variable u is: {u if u > 0}, {-u if u < 0}, so you can treat it as two different things to solve.Example, say you have y > |x+2|, where the vertical bars are absolute value.So you have y > x +2, when (x+2)>0, and you have y > -(x+2), when (x+2) x + 2, when x > -2} and {y > -x - 2, when x < -2}. When x = 2, you can use either one, because they are both equal to zero.==========The definition of "absolute value" for a number x (written as |x| ) is:|x| = x for x > 0|x| = 0 for x = 0|x| = -x for x < 0... where x is replaced by some function - for example f(x):|f(x)| = f(x) for f(x)>0|f(x)| = 0 for f(x)=0|f(x)| = -f(x) for f(x) |x+2| which yields:y > x+2 for x > -2y > 0 for x = -2y > -(x+2) for x < -2 ... or alternatively, -y < x+2 for x < -2
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