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10x=-80, x=?, solve using multiplication principle
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The idea behind this is that every element of the field has a multiplicative inverse.
\[10x\cdot10^{-1} = 10x\cdot\frac{1}{10} = -80 \cdot \frac{1}{10}\]
Now applying commutativity of multiplication we have \[\frac{10}{10} \cdot x = \frac{-80}{10}\]\[x = -8\]
Is that the so called "multiplication principle" math teachers talk about?
Or you could note that 10 *x = 10*-8...
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