The discussion revolves around simplifying the expression (6x+12y)/(10a +5) times (100a^2-25)/(9x^2-81y^2), which involves algebraic fractions and factoring techniques. The subject area includes algebra and polynomial manipulation.
Some participants provide insights on factoring techniques and suggest looking at individual factors for further simplification. There is acknowledgment of potential mistakes in previous attempts, and a focus on ensuring that all factors are accounted for correctly. Multiple interpretations of the factoring process are being explored.
Participants note the importance of not dropping factors without justification and emphasize the need to apply the difference of squares correctly in their simplifications. There is a recognition of the complexity of the problem, which may require careful consideration of each step in the simplification process.
If you learn a little bit of LaTeX, you can make this look nicer.science_rules said:Homework Statement
Simplify (6x+12y)/(10a +5) times (100a2 - 25)/(9x2 - 81y2)
Homework Equations
(6x+12y)/(10a +5) times (100a2 - 25)/(9x2 - 81y2)
The Attempt at a Solution
(6x+12y)/(10a +5) times (100a2 - 25)/(9x2 - 81y2) =
3(2x+4y)/(10a+5) times (10a - 5)/(3x-9y) = (3(2x+4y))(5(2a-1))/(5(2a-1))(3(x-3y))
I do not know how to simplify this any further.
Let me try this again and see what I come up withSammyS said:If you learn a little bit of LaTeX, you can make this look nicer.
How did ##\displaystyle \ \frac{(100a^2-25)}{(9x^2-81y^2)} \ ## become ##\displaystyle \ \frac{(10a-5)}{(3x-9y)} \ ## ?
(6x+12y)/(10a+5) times (100a2 - 25)/(9x2 - 81y2) = (3(x+4y))/(5(2a+1)) times (5(20a2 - 5))/(3(3x2 - 27y2))SammyS said:If you learn a little bit of LaTeX, you can make this look nicer.
How did ##\displaystyle \ \frac{(100a^2-25)}{(9x^2-81y^2)} \ ## become ##\displaystyle \ \frac{(10a-5)}{(3x-9y)} \ ## ?
Let's look at the individual factors, (individual numerators & denominators).science_rules said:(6x+12y)/(10a+5) times (100a2 - 25)/(9x2 - 81y2) = (3(x+4y))/(5(2a+1)) times (5(20a2 - 5))/(3(3x2 - 27y2))
Right.science_rules said:I think i made a mistake: (3(x+4y))/(5(2a+1)) times (5(20a2 - 5))/(3(3x2 - 27y2))
(3(x+4y)) should have been: (3(2x+4y))
SammyS said:Let's look at the individual factors, (individual numerators & denominators).
The first fraction is fine, both numerator & denominator are totally factored.
You can factor out an additional 5 from 100a2 - 25 = 5(20a2 - 5), which will get you back to a difference of squares for one of the factors, and thus can be factored even further.
Similarly, you can factor out an additional 3 from 9x2 - 81y2 = 3(3x2 - 27y2), which will get you back to a difference of squares for one of the factors, and thus can be factored even further.
Then put all into one grand rational expression ("fraction") & do some further cancelling.
You can't just drop factors for no reason.science_rules said:The only factoring I can see to do is: (5(20a2 + 5))/(3(3x2 - 27y2)) = (4a2 +1)/(x2 - 9y2) and then put these together: (3(2x + 4y))(4a2 + 1)/(5(2a +1))(x2-9y2)