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Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3|

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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3|  [#permalink]

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New post 29 Dec 2019, 10:56
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Quote:
Hi

I am not clear why are we taking ONLY x+3 as x+3>=0 and not x+3<=0?


Whenever you translate an absolute value to 'go both ways' like in that explanation, you need to take an additional step and double check to make sure you haven't come up with an invalid solution. You're correct that you need to consider both x + 3 > 0 and x + 3 < 0. But once you've figured out what that implies about the value of x, double check by plugging x back in. For instance, if x + 3 < 0, we ended up figuring out that x has to equal 0. But that's impossible, because 0 + 3 isn't less than 0. So, we didn't get a valid solution at all, and only the x + 3 > 0 solution worked.
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3|  [#permalink]

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New post 29 Dec 2019, 18:13
ccooley wrote:
Quote:
Hi

I am not clear why are we taking ONLY x+3 as x+3>=0 and not x+3<=0?


Whenever you translate an absolute value to 'go both ways' like in that explanation, you need to take an additional step and double check to make sure you haven't come up with an invalid solution. You're correct that you need to consider both x + 3 > 0 and x + 3 < 0. But once you've figured out what that implies about the value of x, double check by plugging x back in. For instance, if x + 3 < 0, we ended up figuring out that x has to equal 0. But that's impossible, because 0 + 3 isn't less than 0. So, we didn't get a valid solution at all, and only the x + 3 > 0 solution worked.


Thank you...noted! :)
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3|   [#permalink] 29 Dec 2019, 18:13

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