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OfflineGrav
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more algebra questions
    #2100793 - 11/13/03 09:51 AM (13 years, 11 months ago)

merggg......

can you help me with these, Merletto?



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InvisibleLetto
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Re: more algebra questions [Re: Grav]
    #2101099 - 11/13/03 12:09 PM (13 years, 11 months ago)

Thanks! Now I have something to do in Philosophy class rather than sleep! I post what I get around 2:30 EST.


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InvisibleLetto
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Re: more algebra questions [Re: Grav]
    #2101670 - 11/13/03 02:30 PM (13 years, 11 months ago)

For the first one, 1 = (b+3)/(b+3), so

[ (b+2+b+3) / (b+3) ] = [ -7 / (b-5) ]

That simplifies to:

[ (2b + 5) / (b + 3) ] = [ -7 / (b-5) ]

By crossmultiplying, you get:

[ -7b - 21 ] = [ 2b^2 - 5b - 25 ]

Then simplify that, and you get

4 = 2b^2 + 2b

which is

2 = b^2 + b

Turn that into a quadratic equation:

b^2 + b - 2 = 0

The roots to that equation are -2 and 1. So those are your answers.

For your second problem, I moved the terms around to get:

[ (x + 1) / (x - 1) ] = [ 12 / (x^2 - x) ]

Cross multiply and you get:

[ x^3 - x ] = [ 12x - 12 ]

Rearrange your terms:

x^3 - 13x + 12 = 0

The roots to this equation are -4, 1, and 3, so these are your answers. But if you plug 1 back into your original equation, it will cause the denominators to equal 0, therefore it cannot be used. So your final answers are -4 and 3.

Let me know if you see anything wrong with my work, or if you have any questions. Also, if you want to ask for more help in the future, I don't mind helping, but at least say what you're having trouble with or show some of your work. I just want to know that I'm not doing your homework for you, but I'm helping you understand.


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Re: more algebra questions [Re: Letto]
    #2103094 - 11/13/03 09:17 PM (13 years, 11 months ago)

how does the 1 go to (b+3)/(b+3) ?

isnt the LCD: (b+3)(b-5) ?

wouldnt that make the 1 the same thing?


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InvisibleLetto
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Re: more algebra questions [Re: Grav]
    #2103171 - 11/13/03 09:39 PM (13 years, 11 months ago)

1 = 2 / 2 = (x + 3) / (x+3) = anything with the same numerator and denominator. Since the 1 is on the left side of the equation, you just set it equal to (b + 3) / (b + 3) so that it has the same denominator as (b + 2) / (b + 3). At that step, you aren't concerned with the right side of the equation, just simplifying each side as much as possible first.


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Re: more algebra questions [Re: Letto]
    #2108233 - 11/15/03 03:03 AM (13 years, 10 months ago)

I see, I see...

I'll try that, but multiplying every factor by (b+3)(b-5) technically should work too, I thought.


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Re: more algebra questions [Re: Grav]
    #2108274 - 11/15/03 03:25 AM (13 years, 10 months ago)

okay i followed you until you got to

4 = 2b^2 + 2b

then how did you get rid of Both 2's ?


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Re: more algebra questions [Re: Grav]
    #2108282 - 11/15/03 03:29 AM (13 years, 10 months ago)

for the second problem, im not sure how you performed your first step.


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OfflineSeussA
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Re: more algebra questions [Re: Grav]
    #2108658 - 11/15/03 09:37 AM (13 years, 10 months ago)

Second problem:

[ x / (x-1) ] - [ 12 / (x^2 - x) ] = -[ 1 / (x-1) ]

Start by moving the second and third terms around:

[ x / (x-1) ] + [ 1 / (x-1) ] = [ 12 / (x^2 - x) ]

Then add the first and second term (common denom):

[ (x + 1) / (x - 1) ] = [ 12 / (x^2 - x) ]

This is the same as if you had ( A / 5 ) + ( B / 5) = (A+B)/5...


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Re: more algebra questions [Re: Seuss]
    #2109589 - 11/15/03 06:54 PM (13 years, 10 months ago)

ah i see... what about the first problem? i see how you got rid of the 2 in the 2b, but what about the 2 in the 2b^2 ?


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Invisiblekaiowas
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Re: more algebra questions [Re: Grav]
    #2110847 - 11/16/03 05:30 AM (13 years, 10 months ago)

change the 1 into b+3/b+3 so have a common denominator
now the left side looks like

[(b+2)+(b+3)]/(b+3)=-7/(b-5)

simplified is looks like (2b+5)/(b+3)=-7/(b-5)

you want the same denominator on both sides of the equal sign. to do this, mulitply the left side by (b-5)/(b-5) and the right side by (b+3)/(b+3)

after multiplying the left side should look lke

(2b^2 -5b -25)/(b+3)(b-5)

and the right is (-7b-21)/(b+3)(b-5)

since the two sides have the same denominator you can cancel them both out.

you are left with

2b^2-5b-25=-7b-21

make the right side zero by adding 7b to the left side and adding 21 to the left side

so you have 2b^2+2b-4=0

the coeffiecients (or the number infront of the variable) are all divisible by 2 so if you divde both sides by 2 you get

b^2+b-2=o

you can do one of two things here, use the quadratic equation or use your head. what two numbers multiplied give you -2 but add up to get +1?? these are the coefficients in front of the variable btw.

that's 2 and -1
(b+2)(b-1)=0

so b=-2 or 1



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Offlinekyanite
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Re: more algebra questions [Re: kaiowas]
    #2110865 - 11/16/03 05:46 AM (13 years, 10 months ago)

x/(x-1) -12/(x^2-x) = -1(x-1)

therefore x/(x-1) -12/x(x-1) = -1(x-1)

therefore LCD = x(x-1)

therefore x.x/x(x-1) -12/x(x-1) = -1.x/(x-1)

drop LCD therefore x^2 -12 = -1x

then put quadratic in standard form x^2 +x -12 =0

solve quadratic (x+4)(x-3)=0

therefore x+-4 or x=3


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Offlinekyanite
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Re: more algebra questions [Re: kyanite]
    #2110866 - 11/16/03 05:49 AM (13 years, 10 months ago)

sorry last line should read

therefore x=-4 or x=+3


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