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I thought absolute values are supposed to be a number's distance from 0, which is always positive. So I had this equation:

| 7 – y | = 12

According to practice tests they say this,

This equation means we need to solve two equations: 7–y=12, and 7–y=-12 which means y=-5,19.

Even if the answer inside the vertical bars was negative, wouldn't the vertical bars make it positive, resulting in just 1 positive answer?

Thanks for any help you can give!

Edit

I found out what happened! I didn't notice when they said the equations could have both 12, and -12 they didn't include absolute value bars, so they were just referring to how the variable can be both positive, or negative.

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Not quite.

Recall that $$|y| = \begin{cases}y,&\text{if } y\geq 0\\-y,&\text{if } y<0\end{cases}$$

Notice that the problem asks for all values $y$ such that $$|7-y| = 12.\tag 1$$ If $y = -5$, then $$|7-(-5)| = |7+5| = |12| = 12.$$ Further, if $y = 19$, then $$|7-(19)| = |-12| = -(-12) = 12$$ using the rule above.

Hence the values of $y$ that satisfy equation $(1)$ are $-5$ and $19$.

So although $y = 19$ gives $|-12|$, it still satisfies equation $(1)$ since we take the absolute value: since $-12<0$, we have that $|-12| = -(-12) = 12.$

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  • $\begingroup$ Thank you so much! Yes I didn't notice when the question said the answer could have both 12, and -12 it was not using absolute value bars, and was just referring to how the variable can be both positive, and negative. This answer my question! $\endgroup$ – Christie Apr 11 '16 at 7:26
  • $\begingroup$ I'm glad you got it. Good luck! $\endgroup$ – Em. Apr 11 '16 at 7:27
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$7-y=12$ or $7-y=-12$

$y=-5$ or $y=19$

If $y=-5$ then $|7-(-5)|=|12|=12$ and if $y=19$, then $|7-19|=|-12|=12$

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$|7-y|=12$ means that the distance (as you point out, this distance can only be nonnegative) between 7 and y is 12. Now, in $\mathbb{R}$ (the space of real numbers), there are two values of $y$ that are 12 units distant from 7. One of them is 12 + 7 = 19, while the other is -12 + 7 = -5.

Even though you have not asked for this in your OP, if we are talking about points in $\mathbb{R}^2$ and consider the "7" to be (7,0), the equation corresponds to the points on a circle centered at (7,0) with radius 12. In this case, there will be infinite number of solutions to this equation.

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Even if the answer inside the vertical bars was negative, wouldn't the vertical bars make it positive, resulting in just 1 positive answer?

The vertical bars are there to express a property you want your $y$ to have. They don't influence what $y$ itself is.

A solution to the equation means the value of the $y$ you start your computation with, before subtracting it from $7$ and positivising the result. Both of those things happen during the calculation you want to work, but doing them doesn't change the fact that the $y$ you started out with might have been $-5$.

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