If $d$ is a metric on $M$, show that $\left|d(x,z) - d(y,z)\right|\leq d(x,y)$ for any $x,y,z\in M$.

Question

I have to solve the following exercise:

Let $M$ be an arbitrary set. If $d$ is a metric on $M$, show that $\left|d(x,z) - d(y,z)\right|\leq d(x,y)$ for any $x,y,z\in M$.

Question: How do I solve this exercise?

I've tried to use the triangle inequality (using the fact that $d(x,y)\in\mathbb{R}_+$) to arrive at $$\left|d(x,z) - d(y,z)\right|\leq \left|d(x,z) + d(z,y)\right| + \left|-d(z,y) - d(y,z)\right|$$ where I know that $\left|d(x,z) + d(z,y)\right| = d(x,z) + d(z,y)\geq d(x,y)$ and $\left|-d(z,y) - d(y,z)\right| = d(z,y) + d(y,z)$. This means that $$\left|d(x,z) - d(y,z)\right|\leq d(x,z) + d(z,y) + d(z,y) + d(y,z) = d(x,z) + 3d(z,y)$$ but this is not the result I need..

Answer 1

$d(x,z)\leq d(x,y)+d(y,z)$ implies $d(x,z)-d(y,z)\leq d(x,y)$

$d(y,z)\leq d(y,x)+d(x,z)$ implies $d(y,z)-d(x,z)\leq d(x,y)$

This implies that $|d(x,z)-d(y,z)|\leq d(x,y)$.