Question 1082260
Look at it like this instead. 
You're not multiplying by {{{10}}} but by {{{2*5}}} instead.
You're trying to get rid of every denominator in your equation and make for simpler addition and subtraction.
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{{{(2*5)((3x)/2)=(2*5)(x/5)-(2*5)(39/5)}}}
{{{(cross(2)*5)((3x)/cross(2))=(2*cross(5))(x/cross(5))-(2*cross(5))(39/cross(5))}}}
{{{5(3x)=(2)x-2*39}}}
{{{15x=2x-78}}}
{{{13x=-78}}}
{{{x=-6}}}