Simplify completely the quantity 14 times x to the 5th power times y to the 4th power plus 21 times x to the third power times y to the 2nd power all over 7 times x to the third power times y. 2x8y5 + 3x6y3 2x2y4 + 3xy 2x2y3 + 3y 2x2y3 + 3y2

We have to simplify the given expression:[tex]\frac{14x^5y^4+21x^3y^2}{7x^3y}[/tex]Dividing the terms of the numerator by the given term of denominator individually, we get= [tex]\frac{14x^5y^4}{7x^3y}+\frac{21x^3y^2}{7x^3y}[/tex]By using the laws of exponent, [tex]a^m \div a^n = a^{m-n}[/tex], we get= [tex]\frac{7 \times 2 x^5y^4} {7x^3y} + \frac{7 \times 3 x^3y^2}{7x^3y}[/tex]= [tex]\frac{2 x^5y^4} {x^3y} + \frac{3 x^3y^2}{x^3y}[/tex]= [tex]{2 x^{5-3}y^{4-1}} + 3 y^{2-1}[/tex]= [tex]{2 x^{2}y^{3}} + 3 y[/tex]Therefore, the simplification of the given expression is  [tex]{2 x^{2}y^{3}} + 3 y[/tex].So, Option 3 is the correct answer.