Multiplying Fractions: Simplifying with GCF

[science_rules]

Homework Statement

I would just like to know if I got this right. ((x₂ + 2x + 1)/(18cw₃) times ((12c₃w)/(x₂ - 1)) = (12c₃wx₂ + 2x12c₃w + 12c₃w)/(18cw₃)(x₂ - 18cw₃

Homework Equations

((x₂ + 2x + 1)/(18cw₃) times ((12c₃w)(x₂ - 1))

The Attempt at a Solution

(12c₃wx₂ + 2x12c₃w + 12c₃w)/((18cw₃)(x₂ - 1)) = (12c₃wx₂ + 2x12c₃w + 12c₃w)/(18cw₃x₂ - 18cw₃

 

[Samy_A]

Removed, as apparently there is an error in the formulation of the question.

 

[science_rules]

I accidentally used a subscript all the way through instead of a power, argh!

 

[Mark44]

$(12c_3wx_2 + 2x12c_3w + 12c_3w)/(18cw_3x_2 - 18cw_3$

On the right side one of the terms is $2x12c_3w$. Should that have been $2x^{12}c^3$? Also, you're missing a right parenthesis.

 

[haruspex]

((x2 + 2x + 1)/(18cw3) times ((12c3w)(x2 - 1))

Do you mean $\frac{x^2+2x+1}{18cw^3}\frac{12c^3w}{x^2-1}$?
If so, there is a lot of cancellation available. Factorise the two expressions involving x.

 

[science_rules]

On the right side one of the terms is $2x12c_3w$. Should that have been $2x^{12}c^3$? Also, you're missing a right parenthesis.

It should have been as haruspex said: (x² + 2x + 1)/(18cw³) times (12c³w)/(x²-1)

 

[science_rules]

Do you mean $\frac{x^2+2x+1}{18cw^3}\frac{12c^3w}{x^2-1}$?
If so, there is a lot of cancellation available. Factorise the two expressions involving x.

Yes, that is what I meant.

 

[science_rules]

Yes, that is what I meant.

Do you mean $\frac{x^2+2x+1}{18cw^3}\frac{12c^3w}{x^2-1}$?
If so, there is a lot of cancellation available. Factorise the two expressions involving x.

(x+1)(x+1)/(18cw³) times (12c³w)/(x-1)(x+1)
The x+1 cancels, leaving: (x+1)/(18cw³) times (12c³w)/(x-1)

 

[haruspex]

(x+1)(x+1)/(18cw³) times (12c³w)/(x-1)(x+1)
The x+1 cancels, leaving: (x+1)/(18cw³) times (12c³w)/(x-1)

More cancellation to go. Look at the constants, look at the powers of c and w.

 

[science_rules]

More cancellation to go. Look at the constants, look at the powers of c and w.

(x+1)/((3w)(6cw²)) times ((3w)(4c³))/(x-1) = (x+1)/(6cw²) times (4c³)/(x-1)

 

[science_rules]

(x+1)/((3w)(6cw²)) times ((3w)(4c³))/(x-1) = (x+1)/(6cw²) times (4c³)/(x-1)

The 3w is canceled

 

[science_rules]

The 3w is canceled

I meant that 3wc is canceled, leaving: (x+1)/(w²) times (4c²)/(x-1)

 

[haruspex]

I meant that 3wc is canceled, leaving: (x+1)/(w²) times (4c²)/(x-1)

You made a mistake with the constants.

 

[science_rules]

You made a mistake with the constants.

Should it be: 3cw(w²) and 3cw(4c²)??

 

[science_rules]

Should it be: 3cw(w²) and 3cw(4c²)??

Oops I meant: 3cw(6w²) and 3cw(4c²)

 

[haruspex]

Oops I meant: 3cw(6w²) and 3cw(4c²)

Yes.

 

[science_rules]

Yes.

Thank you for your help

 

[science_rules]

Yes.

So the answer is: (x+1)/(6w²) times (4c²)/(x-1)

 

[Mark44]

(x+1)/((3w)(6cw²)) times ((3w)(4c³))/(x-1) = (x+1)/(6cw²) times (4c³)/(x-1)

The 3w is canceled

I meant that 3wc is canceled, leaving: (x+1)/(w²) times (4c²)/(x-1)

Should it be: 3cw(w²) and 3cw(4c²)??

Oops I meant: 3cw(6w²) and 3cw(4c²)

Instead of adding new posts each time you discover something you should have said, you can edit the post you want to change.

 

[Thewindyfan]

So the answer is: (x+1)/(6w²) times (4c²)/(x-1)

The constants can be further simplified, and then you should bring it together so you have one fraction.

 

[science_rules]

The constants can be further simplified, and then you should bring it together so you have one fraction.

Like this?: (x+1)/(6w²) times (4c²)/(x-1) = 4c²x + 4c²/6w²x-6w² = 2c(2cx+2c)/2w(3wx-3w)<------(Answer)?

 

[Samy_A]

Like this?: (x+1)/(6w²) times (4c²)/(x-1) = 4c²x + 4c²/6w²x-6w² = 2c(2cx+2c)/2w(3wx-3w)<------(Answer)?

1)

The constants can be further simplified

2) I think you can leave the nominator as $2c²(x+1)$, and similarly for the denominator. Certainly would give a better readable result.

 

[science_rules]

1)

2) I think you can leave the nominator as $2c²(x+1)$, and similarly for the denominator. Certainly would give a better readable result.

How did you get 2c²(x+1) when it was 4c²/(x+1)?? Isn't the numerator supposed to be 2c²(2x+2)??
And the denominator is supposed to be: 2w²(3x-3)??

 

[haruspex]

How did you get 2c²(x+1) when it was 4c²/(x-1)?? Isn't the numerator supposed to be 2c²(2x+2)??
And the denominator is supposed to be: 2w²(3x-3)??

You are still not doing all the simplification you could. In what you posted above, each parenthetic term has a factor that can be taken outside. Having done that, one of the primes occurs above and below the line.

 

[science_rules]

You are still not doing all the simplification you could. In what you posted above, each parenthetic term has a factor that can be taken outside. Having done that, one of the primes occurs above and below the line.

Okay, I think I have it now: (x² + 2x + 1)/(18cw³) times (12c³w)/(x²-1) = (x+1)²(12c³w)/(18cw²)(x+1)(x-1) Cancel the (x+1) = (12c³w)(x+1)/(18cw²)(x-1) Cancel the c and w = (12c²)(x+1)/(18w)(x-1) divided by 3 (numerator and denominator) = (4c²)(x+1)/(6w)(x-1) divided by 2 (numerator and denominator) = (2c²)(x+1)/(3w)(x-1) Answer

 

[SammyS]

Okay, I think I have it now: (x² + 2x + 1)/(18cw³) times (12c³w)/(x²-1) = (x+1)²(12c³w)/(18cw²)(x+1)(x-1) Cancel the (x+1) = (12c³w)(x+1)/(18cw²)(x-1) Cancel the c and w = (12c²)(x+1)/(18w)(x-1) divided by 3 (numerator and denominator) = (4c²)(x+1)/(6w)(x-1) divided by 2 (numerator and denominator) = (2c²)(x+1)/(3w)(x-1) Answer

Enclose the entire denominator in parentheses.
(2c²)(x+1)/((3w)(x-1)) .

 

[Thewindyfan]

Okay, I think I have it now: (x² + 2x + 1)/(18cw³) times (12c³w)/(x²-1) = (x+1)²(12c³w)/(18cw²)(x+1)(x-1) Cancel the (x+1) = (12c³w)(x+1)/(18cw²)(x-1) Cancel the c and w = (12c²)(x+1)/(18w)(x-1) divided by 3 (numerator and denominator) = (4c²)(x+1)/(6w)(x-1) divided by 2 (numerator and denominator) = (2c²)(x+1)/(3w)(x-1) Answer

You could've reached the last step by recognizing that 6 is the GCF for the numerator and denominator.