Uniform continuity of a function: Is it correct?

Question

I'm currently attempting to prove this statement:

Below is the proof I drew up; I am hoping someone can help me make this more rigorous and/or see if there is any flaws on it proving uniform continuity on the set of Real Numbers.

f(x) = 3x+7 is uniformly continuous on R if for every Epsilon > 0, there is
a Delta > 0 such that for all x,y the inequality |x-y| < Delta holds,
implying |f(x) - f(y)| < Epsilon.

Proof

Set Delta = (3 * Epsilon). Let Epsilon > 0. 
|(3x+7) - (3y+7)| = |3x - 3y + 7 - 7| = |(3x - 3y)| = (|x-y|) * |3|
< Delta/3 < 1/3 * (3 Epsilon) <= Epsilon

Is this all legal? I feel the absolute value part I may have made some mistake, and I don't know how to specifically show this is true for all numbers in R.

Any ideas?

Answer 1

You actually proved that, for all $x,y\in \mathbb R$ that satisfy |f(x)-f(y)|<\varepsilon, then (I'm just rewriting your calculations backward)

$$|f(x)-f(y)|=|3x+7-3y-7|=|3x+3y|=3|x-y|<\varepsilon$$

so $|x-y|<\frac\varepsilon 3$. That means $\delta=\frac\varepsilon3$ satisfy the definition of uniform continuity for all $x,y$. Everything is correct to me.