Absolute Value proof involving epsilon and delta

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The question is:
If $x,z$ $ϵ$ $R$, show that for every $ϵ > 0$ there is a $ δ > 0$ such that if $y$ $ϵ$ $R$ satisfies $|y-x| < δ$ then $|zy-zx| < ϵ.$


I honestly have no idea how to go about this problem, any hints or help will be greatly appreciated!

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Hint: $|zy-zx| = |z(y-x)| = |z||y-x|$. If you want to have $|y-x|<\delta$, what relationship should you have between $\varepsilon$ and $\delta$?