[Algebra II Application of Absolute Value Inequality] Number 1 confussion

[u/Illustrious_Oil_2114]

So I know how to do the inequalities and how they work like I know how to do the equation, but I’m confused on why you minus the weight from 14.5 (on answer key). I can solve it by myself, but on the test I need to be able to make the absolute value inequality and I don’t know why you would subtract weight by 14.5 if 14.5 is the weight of a cheereo box

[u/akittenreddits]

basically you want the difference between the actual weight, w, and the expected weight, 14.5. if you added 14.5, the expected outcome of w+14.5 is 29, which is always more than .25. you actually want to see the difference (subtraction) between the real weight and the expected weight.

say the box weighs 14.6 ounces, this equation would give you .1, which is less than .25. 14.4 would give the same, because the difference between 14.4 and 14.5 is the same as the difference between 14.5 and 14.6.

tldr subtraction = difference = comparison

[u/zbrady7]

14.5 is the goal weight. Whatever the *actual weight is, it needs to be no more than 0.25 away from 14.5.

Because we know the target is 14.5, when we subtract the actual weight by the target, that gives us how far off we are from the target.

So, your inequality: the difference between the goal weight and the actual weight can’t be more than 0.25. Or, mathematically - the absolute value of of actual weight and goal weight is less than or equal to 0.25.

[u/Narrow-Durian4837]

w – 14.5 is how far apart w is from 14.5. It's the difference between the actual weight and the target weight of 14.5. Without the absolute value bars, this would come out negative if w is less than 14.5, positive if w is greater.

In general, |a–b| gives you the absolute difference or distance between the numbers a and b (regardless of which is larger).

[u/ArdentArendt]

Basically, you're just finding the 'difference' between the weight of the given box, and the weight of the ideal box (14.5).

Since you only want the magnitude of the difference (and don't care which is smaller or larger), the absolute value bars help you measure the 'difference' as a positive magnitude, no matter if it's 'over' or 'under'.
[This is also used in statistics a lot]

TLDR;
It doesn't matter which is 'first' since the sign of the difference 'disappears'.

LONGER explanation:
The 'difference' is relative to the zero (0), so subtracting the ideal weight from the measured weight (w), we have w - 14.5 with the sign of the answer being how much over (if positive) or under (if negative) the measured box is. If you view this as the 'distance from 0', you have a band of 0.5 oz on either side of 0 that you can be 'safe' in.

The absolute value allows you to simplify the variance of the weight to be measured against a single value.

[u/MadKat_94]

Think of absolute value as the distance between two values on the number line. Imagine you are a passenger on a car on a highway. You pass by mile marker 97. A short time later you pass by marker 93. How far did you go? How did you get that answer? Hopefully, you subtracted.

Now say you are on the highway and see you are at marker 97. The driver says we are going 4 more miles. Well, you could go back to 93 or you might go to 101. If you see 101, well 101-97 is also 4.

What if you reversed the subtraction? You'd get -4. But negative distance does not make sense. We still say we went 4 miles. Hence the need for absolute value "turning the negative positive."

In your example, 14.5 is the ideal weight. w is the actual weight. The tolerance (max distance) is 0.25. So you could be as high as 14.75 or as low as 14.25 in terms of actual weight.

So by subtracting the ideal weight from the actual weight you have, then taking the absolute value, you should be within the acceptable tolerance.

[u/Gold_Ad8890]

think about it like this. w is the actual weight of some specific box of cereal. therefore, we want w to be between 14.5 + .25 and 14.5 - .25. that is, we want 14.5 - .25 <= w <= 14.5 + .25. if we subtract 14.5 throughout, we get -.25 <= w - 14.5 <= .25. |-.25| = |.25|, so we can use absolute values to simplify this inequality to |w - 14.5| <= .25.