Question 341465
It's easier to use the laws of exponents rather than expand the expressions.
Further simplification is possible in your answer.
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{{{(70x^2y^4)/(4x^3y) =(70/4)x^(2-3)y^(4-1)}}}
{{{(70x^2y^4)/(4x^3y) =(15)x^(-1)y^(3)}}}
{{{(70x^2y^4)/(4x^3y) =(15y^3)/x}}}