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The question was longer than 150 characters, so I couldn't fit it all in the title. Here's the full question:

Let $A$ and $B$ be nonempty sets of real numbers and let $L ∈ \Bbb R$. Suppose that there exist positive real numbers $x$ and $y$ such that $|a − L| < x$ for all $a ∈ A$ and $|2b − L| < y$ for all $b ∈ B$. Use the Triangle Inequality to prove that: $$|a − 4b| < x + 2y + |L|$$ for all $a ∈ A$ and $b ∈ B$.

(HINT: $|a − 4b| = |a − 2L + 2L − 4b|$.)

I know you can use the triangle inequality theorem to separate the hint out to where

$$|a-2L+2L-4b| ≤ |a-2L| + |2L-4b|$$ but I'm not sure where to go after that.

The triangle inequality theorem states: $$|x+y|≤|x|+|y|,$$ $$|x-y|≥||x|-|y||,$$ $$|x-y|≥|x|-|y|,$$ and $$|x-y|≥|y|-|x|.$$

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    Hint: $|a-2L| \le |a-L|+L$ and $|2L-4b|= 2\,|2b-L|$. – 2017-01-31
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    I understand how you can take the constant out of the second absolute value, but how can you take the L out of the first one? Sorry, just need some reasoning behind it so that I can understand it. – 2017-01-31
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    $|a-2L|=\left|(a-L)-L\right|\le |a-L|+|L|$. – 2017-01-31

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Here's a basic outline:

We start off with what you've done, $$\begin{align}|a-4b| &= |a-2L + 2L - 4b| \tag{From the hint}\\&\le |a-2L| + |2L - 4b| \tag{Triangle Inequality}\\&= |a-2L| + 2|L-2b| \tag{Factor out $2$}\\&= |a-2L| + 2|2b - L|\tag{Absolute Value removes $-1$}\\ &= |(a -L) - L| + 2|2b-L| \tag{Rewrite left quantity}\\&\le|a-L| + |L| + 2|2b-L| \tag{Triangle Inequality}\\& < x + |L| +2y. \tag{Hypothesis}\end{align}$$

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    That's a nice write-up, +1. – 2017-01-31
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    Okay. I understand it until you go from the 5th line to the 6th line. Which part of the Triangle Inequality are you using? From the theorem, |x−y|≥|x|−|y|. How do you get |a-L|+|L|? – 2017-01-31
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    @johnie4usc, we may look at the problem like this using the standard form of the triangle inequality: $$\begin{align}|a-2L| &= |(a - L) - L| \\&= |(a-L) + (0 - L)| \\&\le |a-L| + |0 - L| \tag{Triangle Inequality}\\&= |a-L| + |-L| \\&= |a-L| + |L|.\end{align}$$ This relies on the fact that $|u| = |-u|$ for any real number $u$, because the absolute value makes negative numbers nonnegative. – 2017-01-31
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    Hm. Alright, that makes sense. I'm just wondering if I should use that since we haven't covered it. Thanks a lot for the help. I really appreciate it. – 2017-01-31
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    @johnie4usc, covered which part? The fact that $|u| = |-u|$ for any real $u$? You can also look at that step this way: $$|-L| = |-1\cdot L| = |-1||L| = 1\cdot|L|,$$ which uses the same logic from factoring out the $2$ earlier. – 2017-01-31
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    Yeah it all makes sense. I appreciate it. – 2017-01-31