# Simplifying a fraction

## 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.

Last edited:

Related Precalculus Mathematics Homework Help News on Phys.org
SammyS
Staff Emeritus
Homework Helper
Gold Member

## 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)} \ ## ?

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

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
Staff Emeritus
Homework Helper
Gold Member
(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
Staff Emeritus
Homework Helper
Gold Member
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
Staff Emeritus
Homework Helper
Gold Member
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.

science_rules