MHB Algebraic Rational Expressions

[jjlittle00]

I am attempting to find the solution to the following question.

Simplify and state the restrictions on the variables$$\frac{5a^5b^6}{10a^2b^3}\div\frac{2a^4b^2}{20a^3b^5}$$

Not really understanding how to find the restrictions with these set variables.

 

[MarkFL]

Hello and welcome to MHB, jjlittle00! (Wave)

I am assuming the expression is as follows:

$$\frac{5a^5b^6}{10a^2b^3}\div\frac{2a^2b^3}{20a^2b^3}$$

Before we proceed, is this correct?

 

[jjlittle00]

Hello and welcome to MHB, jjlittle00! (Wave)

I am assuming the expression is as follows:

$$\frac{5a^5b^6}{10a^2b^3}\div\frac{2a^4b^2}{20a^3b^5}$$

Before we proceed, is this correct?

Yes this is correct. Just made one small correction.

 

[MarkFL]

Yes this is correct. Just made one small correction.

Okay, we now have:

$$\frac{5a^5b^6}{10a^2b^3}\div\frac{2a^4b^2}{20a^3b^5}$$

We have one rational expression being divided by another. In order for these expressions to be defined, we cannot have either denominator being equal to zero. What values of $a$ and/or $b$ will cause either denominator to be zero?

 

[HallsofIvy]

Also, when we divide by a fraction, we "invert and multiply": \frac{5a^5b^6}{10a^2b^3}\frac{20a^3b^5}{2a^4b^2}. So that, in addition to the requirement that the denominators of the original fractions not being 0, 2a^4b^2 cannot be 0. That is effectively saying that a and b cannot be 0.

Of course, to "simplify" you cancel as many "a"s and "b"s, in numerator and denominator, as you can.

 

[MarkFL]

Also, when we divide by a fraction, we "invert and multiply": \frac{5a^5b^6}{10a^2b^3}\frac{20a^3b^5}{2a^4b^2}. So that, in addition to the requirement that the denominators of the original fractions not being 0, 2a^4b^2 cannot be 0. That is effectively saying that a and b cannot be 0.

Of course, to "simplify" you cancel as many "a"s and "b"s, in numerator and denominator, as you can.

I was going to get to all that eventually...honest I was...:p