multiplying and factorizing with algebra

[kopite_123]

this is my first ever post and am not sure if this is the right thread for this question but here it is anyway:

multiply, factorize and cancel down the following:

12ab
----------
25ca squared

x

5dc
------
6b squared

 

[stapel]

I'm sorry, but I cannot follow what you have written. Please use standard web-safe formatting to clarify your meaning.

Thank you.

Eliz.

 

[Trebor]

I'm assuming that this is what your trying to solve:

\(\displaystyle \frac{12ab}{25ca^2} \times \frac{5dc}{6b^2}\)

Simply multiply the top part (numerators) and bottom parts (denominators) together into a new fraction:

\(\displaystyle = \frac{60abcd}{150a^2b^2c}\)

Then, it's just a case of simplifying.

 

[soroban]

Hello, kopite_123!

Welcome aboard!

Multiply, factorize and cancel down the following:

. . \(\displaystyle \L\frac{12ab}{25a^2c}\,\times\,\frac{5cd}{6b^2}\)

Multiply: \(\displaystyle \L\:\frac{60abcd}{150a^2b^2c}\)

Factor: \(\displaystyle \L\:\frac{2\cdot2\cdot3\cdot5\cdot a\cdot b\cdot c\cdot d}{2\cdot3\cdot5\cdot5\cdot a\cdot a\cdot b\cdot b\cdot c}\)

Cancel: \(\displaystyle \L\:\frac{2\cdot\not2\cdot\not3\cdot\not5\cdot\not{a}\cdot\not{b}\cdot \not{c}\cdot d}{\not2\cdot\not3\cdot5\cdot\not{5}\cdot\not{a}\cdot a\cdot\not{b}\cdot b\cdot\not{c}} \;= \;\frac{2d}{5ab}\)