The discussion revolves around multiplying and simplifying fractions involving algebraic expressions, specifically focusing on the expression \(\frac{x^2 + 2x + 1}{18cw^3} \times \frac{12c^3w}{x^2 - 1}\). Participants are examining the steps taken to simplify the product and the potential errors in notation and formulation.
The discussion is ongoing, with participants providing guidance on factoring and simplifying the expressions. There is a recognition of multiple interpretations of the original problem, and some participants are actively correcting each other’s notation and assumptions.
There are indications of confusion regarding the notation used, particularly with subscripts and powers. Participants are also addressing the need for clarity in the formulation of the problem and the simplification process.
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.science_rules said:##(12c_3wx_2 + 2x12c_3w + 12c_3w)/(18cw_3x_2 - 18cw_3##
Do you mean ##\frac{x^2+2x+1}{18cw^3}\frac{12c^3w}{x^2-1}##?science_rules said:((x2 + 2x + 1)/(18cw3) times ((12c3w)(x2 - 1))
It should have been as haruspex said: (x2 + 2x + 1)/(18cw3) times (12c3w)/(x2-1)Mark44 said: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.
Yes, that is what I meant.haruspex said: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 said:Yes, that is what I meant.
(x+1)(x+1)/(18cw3) times (12c3w)/(x-1)(x+1)haruspex said: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.
More cancellation to go. Look at the constants, look at the powers of c and w.science_rules said:(x+1)(x+1)/(18cw3) times (12c3w)/(x-1)(x+1)
The x+1 cancels, leaving: (x+1)/(18cw3) times (12c3w)/(x-1)
(x+1)/((3w)(6cw2)) times ((3w)(4c3))/(x-1) = (x+1)/(6cw2) times (4c3)/(x-1)haruspex said:More cancellation to go. Look at the constants, look at the powers of c and w.
The 3w is canceledscience_rules said:(x+1)/((3w)(6cw2)) times ((3w)(4c3))/(x-1) = (x+1)/(6cw2) times (4c3)/(x-1)
I meant that 3wc is canceled, leaving: (x+1)/(w2) times (4c2)/(x-1)science_rules said:The 3w is canceled
You made a mistake with the constants.science_rules said:I meant that 3wc is canceled, leaving: (x+1)/(w2) times (4c2)/(x-1)
Should it be: 3cw(w2) and 3cw(4c2)??haruspex said:You made a mistake with the constants.
Oops I meant: 3cw(6w2) and 3cw(4c2)science_rules said:Should it be: 3cw(w2) and 3cw(4c2)??
Yes.science_rules said:Oops I meant: 3cw(6w2) and 3cw(4c2)
Thank you for your helpharuspex said:Yes.
So the answer is: (x+1)/(6w2) times (4c2)/(x-1)haruspex said:Yes.
science_rules said:(x+1)/((3w)(6cw2)) times ((3w)(4c3))/(x-1) = (x+1)/(6cw2) times (4c3)/(x-1)
science_rules said:The 3w is canceled
science_rules said:I meant that 3wc is canceled, leaving: (x+1)/(w2) times (4c2)/(x-1)
science_rules said:Should it be: 3cw(w2) 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.science_rules said:Oops I meant: 3cw(6w2) and 3cw(4c2)
The constants can be further simplified, and then you should bring it together so you have one fraction.science_rules said:So the answer is: (x+1)/(6w2) times (4c2)/(x-1)
Like this?: (x+1)/(6w2) times (4c2)/(x-1) = 4c2x + 4c2/6w2x-6w2 = 2c(2cx+2c)/2w(3wx-3w)<------(Answer)?Thewindyfan said:The constants can be further simplified, and then you should bring it together so you have one fraction.
1)science_rules said:Like this?: (x+1)/(6w2) times (4c2)/(x-1) = 4c2x + 4c2/6w2x-6w2 = 2c(2cx+2c)/2w(3wx-3w)<------(Answer)?
2) I think you can leave the nominator as ##2c²(x+1)##, and similarly for the denominator. Certainly would give a better readable result.Thewindyfan said:The constants can be further simplified
How did you get 2c2(x+1) when it was 4c2/(x+1)?? Isn't the numerator supposed to be 2c2(2x+2)??Samy_A said: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.
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 said: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)??
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) Answerharuspex said: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.
Enclose the entire denominator in parentheses.science_rules said: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.science_rules said: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