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2 bytes added ,  04:25, April 11, 2015
m
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so
 
so
 
::<math>\frac{x^2 - 8 x + 15}{x-3} = x-5\,</math>
 
::<math>\frac{x^2 - 8 x + 15}{x-3} = x-5\,</math>
−
from which one gets ''x''=12.
+
from which one gets ''x'' = 12.
    
This was just a case of the principle that you can perform arithmetic operations even when the expression contain unknowns.  In this case we were actually applying the distributive law 3 times.
 
This was just a case of the principle that you can perform arithmetic operations even when the expression contain unknowns.  In this case we were actually applying the distributive law 3 times.
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