Question

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

Answer 1

2x^2y^3 + 3y

Explanation:

Divide the Coefficients Like you would a normal division problem for each term separately. Then subtract the like exponents.

Answer 2

We have to simplify the given expression:

`(14x^5y^4+21x^3y^2)/(7x^3y)`

Dividing the terms of the numerator by the given term of denominator individually, we get

=
`(14x^5y^4)/(7x^3y)+(21x^3y^2)/(7x^3y)`

By using the laws of exponent,
`a^m / a^n = a^(m-n)`, we get

=
`\frac{7 * 2 x^5y^4} {7x^3y} + (7 * 3 x^3y^2)/(7x^3y)`

=
`\frac{2 x^5y^4} {x^3y} + (3 x^3y^2)/(x^3y)`

=
`{2 x^(5-3)y^(4-1)} + 3 y^(2-1)`

=
`{2 x^(2)y^(3)} + 3 y`

Therefore, the simplification of the given expression is
`{2 x^(2)y^(3)} + 3 y`.

So, Option 3 is the correct answer.