Simplifying Fraction: (6x+12y)/(10a+5) * (100a^2-25)/(9x^2-81y^2)

[science_rules]

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 (4a - 1)/(3x-27y) = (3(2x+4y))(4a-1)/(5(2a+1))(3(x-9y))
I do not know how to simplify this any further.

 

[SammyS]

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.

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)} \$ ?

 

[science_rules]

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 me try this again and see what I come up with

 

[science_rules]

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]

(6x+12y)/(10a+5) times (100a² - 25)/(9x² - 81y²) = (3(x+4y))/(5(2a+1)) times (5(20a² - 5))/(3(3x² - 27y²))

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.

 

[science_rules]

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]

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))

Right.

You had that part correct previously.

 

[science_rules]

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.

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)

 

[SammyS]

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)

You can't just drop factors for no reason.

Start with 5(20a² - 5) = 5⋅5(4a² -1) . (Yes, that is subtraction.)

Do you know how to factor a difference of squares?

Similarly, 3(3x² - 27y²) = 3⋅3(x² - 9y²) . Also has a difference of squares.