I can't prove the following remark which appears in Matsumura's 'Commutative Ring Theory':
If $B$ is an $A$-algebra and $M$ and $N$ are $B$-modules, $M\otimes_{B}N$ is the quotient of $M\otimes_{A}N$ by the submodule generated by elements of the form $bx\otimes_{A}y-x\otimes_{A}by$.
I think I was able to do it using the universal property (in a completely straightforward fashion), but I am suspicious of this as I couldn't do it by proving directly that $\ker(M\otimes_{A}N\rightarrow M\otimes_{B}N)$ is generated by these elements.
Using the universal property for $\otimes_A$ one gets a map $$ M\otimes_AN\to M\otimes_BN, \quad m\otimes_A n\mapsto m\otimes_B n. $$ This sends the submodule $L$ generated by all $mb\otimes_A n-m\otimes_A bn$ to zero, so we get the induced map $$ (M\otimes_AN)/L\to M\otimes_BN, \quad m\otimes_An+L \mapsto m\otimes_Bn. $$
We can construct the inverse map using the universal property for $\otimes_B$. For, the map $$ M\times N\to (M\otimes_AN)/L, \quad (m,n)\mapsto m\otimes_A n+L $$ is $B$-bilinear, so induces the map $$ M\otimes_BN\to (M\otimes_AN)/L, \quad m\otimes_Bn \mapsto m\otimes_An+L. $$ These two maps are clearly mutually inverse, as required.