Question

Algebra 2 Help please! @mathmale Will attach problem.

Answer 1

Remember that when you divide fractions, you end up multiplying the reciprocal:
Your question:
\[\frac{24y^5}{15x^8} \div \frac{8y^2}{4x^4}\]
Simplifies to
\[\frac{24y^5}{15x^8} \times \frac{4x^4}{8y^2}\]
Now simplify
\[\frac{24y^5 \times 4x^4}{15x^8 \times 8y^2}\]
And you should be able to carry out the rest (just remember to simplify the exponents and the fraction)

Answer 2

See, we got that far, @DisplayError, but my grandfather and I are stuck past that. We can't figure out how to work it to get one of the given answers.

Answer 3

Hello, Hedgie,
While I can't spend much time with you right now (due to other things on my mind and agenda), I can vouch for what DisplayError has just told y ou. To divide by a fraction, invert the divisor fraction and multiply. That's what DisplayError is doing. All you have to do then is to reduce the equation to lowest terms. For example, (24/15) reduces to (8/5), and
(x^4) / (x^8) to 1 / (x^4).

Answer 4

Nice of your grandfather to offer to help. Don't look now, but I am probably older than he is.

Answer 5

Separate it out. First evaluate the numerical part, so
\[\frac{24 \times 4}{15 \times 8} = \frac{4}{5}\]
Then simplify the exponents. Remember your exponent rules:
\[\frac{x^a}{x^b} = x^{a-b}\]
Applying that to the problem here, you would get a negative exponent for the base x and a positive exponent for the base y. Here is another exponent rule:
\[x^{-a} = \frac{1}{x^a}\]
\[\frac{1}{x^{-a}} = x^a\]

Answer 6

Hedgie: Take my word for it: Our friend Display Error really knows his stuff!

Answer 7

Okay, thank you for helping, guys! My grandfather is trying to figure how to do it based of what you guys are saying. He says "How do we reduce the exponents?"

He's really good at this stuff, and even he's having issues trying to help me work on the problems they give me.

Answer 8

DisplayError has done a fine job of introducing/reviewing rules of exponents. You might try translating each of his rules into English; doing that just might help you to understand what is happenign in each case.

Answer 9

You can only simply when the base is the same. We can't simplify
\[\frac{x^a}{y^b}\]
because the numerator is base x and the denominator is base y. However, if our fraction was this:
\[\frac{x^a}{x^b}\]
Then because the numerator and denominator are both base x, we can simplify, as follows:
\[\frac{x^a}{x^b} = x^{a-b}\]
Here are some additional exponent rules (not required for this problem, but to jog your memory)
\[x^a \times x^b = x^{a+b}\]
\[(x^a)^b = x^{a \times b}\]

Answer 10

Thank you, DisplayError. We got it figured out, we think. :)