some one please help me.

[blueZ]

can some one teach me how to do math questions when trinomials, variables, and fractions and all mixed together?
i have the WORST math teacher, she only spends 5 minute with every lesson, and she only gives us notes, and DOES NOT TEACH.
she only gone over briefly on how to factoring trinomials, when there's only 1 variable. like x[2] + 4x - 6( "[2]" means squared by the way, [3]means to the power of 3)

and she gives us homework like:


and i was like WHAT THE?!!!!
and she never even taught us to factor before we calculate.

can some one please land me a helping hand, my assignment is due tomorrow

[whhattisthiss]

I'm actually doing that in school right now too, and it makes no sense to me either

[Uncle Xyzzy]

On the first one, set it equal to "a", and take the reciprocol, eliminating the fractions, then combine like terms. Then, just take the reciprocol again.
Second one, remember that you're multiplying by the reciprocol, and multiply by the bottom terms, eliminating denominators. Then reduce it.

Factoring trinomials:
1. Look for something common: 2x-2=2(x-2)
2. Look for difference of terms: 4x^2-9=(2x+3)(2x-3)
3. Look for perfect trinomial squares: x^2+12x+36=(x+6)^2
4. Guess and check, and look for two numbers that multiply to C and add to B in Y=AX^2+BX+C: x^2+5x+6=(x+2)(x+3)

[pipsqueeek]

I'm not sure I quite understand what the question is, but what I'd do-

First factor what you can so it's easier to work with. So the y^2-y-6 = (y-3)(y+2) and y^2+3y = y(y+3), etc.

Then for the first get it so you've got a common denominator, for the second divide it, and cancel stuff out.

Does that help at all? I'm not sure if I'm even answering the right thing...

Edit//

On the first one, set it equal to "a", and take the reciprocol, eliminating the fractions, then combine like terms. Then, just take the reciprocol again.

If I'm understanding you right, that doesn't work. That'd be like saying 1/2+1/2+1/2 = 1/(2+2+2). 3/2 =/= 1/6

[blueZ]

but cant you only eliminated when its multiplication of division?
i dont think you can elinminate when its addition or subtraction, can you?
ARRRG I AM SO CONFUSED on how to do additions and subtractions when there are whole bunch of trinomials and variables and exponents.

[pipsqueeek]

You can get it so everything has the same denominator by mulitplying 6xy^3 by ((5x^2y^2)(2x^3y))/((5x^2y^2)(2x^3y)), and do the same for the others. You end up with (10x^5y^3+12x^4y^4+30x^3y^5)/(60x^6y^6). You can then factor out/cancel out a little bit, but what you end up with is still kind of messy. (5x^2+6xy+15y^2)/(30x^3y^3). Does that look like what you should be getting or an I doing the wrong problem?

[blueZ]

what do you do when denominators are the same, but one of the numerators is a variable? like:
1/4(x-1) - x/4(x-1)

[pipsqueeek]

what do you do when denominators are the same, but one of the numerators is a variable? like:
1/4(x-1) - x/4(x-1)

If the denominators are the same, you can combine the numerators. So, (1-x)/(4(x-1)). (Just like, for example, 1/5+3/5=4/5). 1-x = -(x-1), so those can cancel and you're left with -1/4

[blueZ]

If the denominators are the same, you can combine the numerators. So, (1-x)/(4(x-1)). (Just like, for example, 1/5+3/5=4/5). 1-x = -(x-1), so those can cancel and you're left with -1/4

thats so smart.........
but how are you suppose know to write 1-x = -(x-1), i would have never thought of that in a million years

You can get it so everything has the same denominator by mulitplying 6xy^3 by ((5x^2y^2)(2x^3y))/((5x^2y^2)(2x^3y)), and do the same for the others. You end up with (10x^5y^3+12x^4y^4+30x^3y^5)/(60x^6y^6). You can then factor out/cancel out a little bit, but what you end up with is still kind of messy. (5x^2+6xy+15y^2)/(30x^3y^3). Does that look like what you should be getting or an I doing the wrong problem?

can you explain further, i kind of get it, but not completely
why do you times 6xy^3 by ((5x^2y^2)(2x^3y))/((5x^2y^2)(2x^3y)),
dont you times 1/6xy^3 by ((5x^2y^2)(2x^3y))/((5x^2y^2)(2x^3y))?

and what do you do with the other 2 fractions?

[pipsqueeek]

You can get it so everything has the same denominator by mulitplying 6xy^3 by ((5x^2y^2)(2x^3y))/((5x^2y^2)(2x^3y)), and do the same for the others. You end up with (10x^5y^3+12x^4y^4+30x^3y^5)/(60x^6y^6). You can then factor out/cancel out a little bit, but what you end up with is still kind of messy. (5x^2+6xy+15y^2)/(30x^3y^3). Does that look like what you should be getting or an I doing the wrong problem?

can you explain further, i kind of get it, but not completely
why do you times 6xy^3 by ((5x^2y^2)(2x^3y))/((5x^2y^2)(2x^3y)),
dont you times 1/6xy^3 by ((5x^2y^2)(2x^3y))/((5x^2y^2)(2x^3y))?

and what do you do with the other 2 fractions?

er, yeah. I meant 1/6xy^3

with the others, you do the same thing, but you use the other numbers. What you're trying to do is get them so they all have the same denominator. Take the example of 1/3+1/5+1/7. The common denominator for these would be 3*5*7. So the 3 would have to be multiplied by 5*7, the 5 by 3*7, and the 7 by 3*5. You have to keep the values the same though, or you'd change the problem, so you muliply them each by one. These ones come in the form of (5*7)/(5*7), (3*7)/(3*7), and (3*5)/(3*5). They're equivalent to one, so you can multiply them without changing your values, but you end up with the same denominator for everything, so you can then combine them.

[blueZ]

yeah thats what i thought, but i am ending up have to be doing 30 calculations,which seems a bit ridiculous, am i on the right track? because i have to multiply everything again when i am trying to find the numerator for all 3 of the fractions

[pipsqueeek]

How are you getting 30? Finding the numerator should only take one calculation for each. (They might look scary with all the variables and exponents and all, but they're actually really easy. All you have to do is mulitply the coefficients and add the exponents(ie, x^2*x^3=x^5))

[blueZ]

well 30 was just an adjective.

do i do
5x[2]y[2] times 2x[3]y for the numerator of the first fration
then
6xy[3] times 2x[3]y for the numerator for the second fraction
then
6xy[3] times 5x[2]y[2] for the numerator for the 3rd fraction

and i do
6xy[3] times 5x[2]y[2] times 2x[3]y for the denominator for every fraction

right?

[pipsqueeek]

yup, just like that

[Runevalkyrie]

This seems to be solved, however, this belongs in the Homework Help thread here next time.