24X^5Y^8/30X^3Y^10

## 24X^5Y^8/30X^3Y^10

# 1 Answer

It's not very clear, but I'm assuming that problem looks like the following:

(24·x^{5}·y^{8})/(30·x^{3}·y^{10})

First, notice that 6 is a factor of both the coefficients in the numerator and the denominator. So the expression can be reduced as follows:

(6·4·x^{5}·y^{8})/(6·5·x^{3}·y^{10}) = (4·x^{5}·y^{8})/(5·x^{3}·y^{10})

When dividing exponents with like bases, we keep the same base and subtract the smaller exponent from the larger exponent. For instance,

x^{5}/x^{3} = x^{5-3}/x^{3-3} = x^{2}/x^{0} = x^{2}/1 = x^{2}

y^{8}/y^{10} = y^{8-8}/y^{10-8} = y^{0}/y^{2} = 1/y^{2}

Therefore,

(4·x^{5}·y^{8})/(5·x^{3}·y^{10})
= (4·x^{5-3}·y^{8-8})/(5·x^{3-3}·y^{10-8})

= (4·x^{2}·y^{0})/(5·x^{0}·y^{2})

= (4·x^{2}·1)/(5·1·y^{2})

= (4·x^{2})/(5·y^{2})

Thus,

(24x^{5}y^{8})/(30x^{3}y^{10}) = (4x^{2})/(5y^{2})