ID:276423
 
This is some review work for my first math class in college. I've had no problems with the rest of the problems, but I can't figure this one out.

  1         1        5
----- + ----- = -
x - 1 x - 2 6


I multiplied each side by (x-1)(x-2)(6) to get rid of all the fractions, but once I do that, I get the following equation.

(x-2)(6) + (x-1)(6) = 5(x-1)(x-2)


From here, I'm stuck. No matter how I solve the eqution, x doesn't equate to 4, which is the correct answer (I found that out in my head). What am I missing here? I believe I did something wrong on the right side of the equation.

P_S
Remember order of operations, and how to work with polynomial equations...

Working it out, here are my steps (I'm extremely rusty at this sort of thing, not having done it for nearly 5 years now, and I suspect that I'm taking the long way around, but it still works):

(x-2)(6) + (x-1)(6) = 5(x-1)(x-2)
(6x-12) + (6x-6) = 5x^2-15x+10


Well, since you said you suspected making a mistake on the right side of the equation, I'll walk over the steps in working that out into the polynomial form...

Multiplying out the factors containing the x variable (leavig the factor of 5 until the end, for simplicity's sake, although you can just do it all at once if you can keep track of things) you'll get:

5(x^2-2x-1x+2) //You basically have to multiply each part
// by each of the others, and list the results... The first
// x*the second x, the first x*the -2, the -1*the second x,
// and the -1 times the -2... I don't know how to explain
// this in proper mathematical terms anymore, though...lol
5(x^2-3x+2)
5x^2-15x+10


Now back to our full equation, simplifying everything, and moving it all to the left side, leaving only 0 on the right:

(6x-12) + (6x-6) = 5x^2-15x+10
12x-18 = 5x^2-15x+10
-5x^2+27x-28 = 0


At this point, we can simplify things ever-so-slightly by factoring out a -1:

5x^2-27x+28 = 0


I'll leave the rest up to you (don't want to do all of this poblem for you...lol), but by plugging in the correct answer of 4, it works out, so at least you know it's right so far...lol
In response to SuperSaiyanGokuX
Thanks. That was a tremendous help.

You se, I forgot to FOIL the two binomials first, and leave the 5 until the end.

Thanks again,
P_S
In response to Prodigal Squirrel
Hey, no problem... I'd hate to see my AP Calc going to waste (and it really has since graduation)...lol
Prodigal Squirrel wrote:
This is some review work for my first math class in college. I've had no problems with the rest of the problems, but I can't figure this one out.

>   1         1        5
> ----- + ----- = -
> x - 1 x - 2 6
>

I multiplied each side by (x-1)(x-2)(6) to get rid of all the fractions, but once I do that, I get the following equation.

(x-2)(6) + (x-1)(6) = 5(x-1)(x-2)

From here, I'm stuck. No matter how I solve the eqution, x doesn't equate to 4, which is the correct answer (I found that out in my head). What am I missing here? I believe I did something wrong on the right side of the equation.

The equation is right; it just needs to be taken further.

(x-2)(6) + (x-1)(6) = 5(x-1)(x-2)
6(2x-3) = 5(x-1)(x-2)
12x-18 = 5x2-15x+10
5x2-27x+28 = 0

From here you can complete the square or just factor.

5x2-7x-20x+28 = 0
(5x-7)(x-4)=0

Hence there are two answers: x=7/5 and x=4.

Lummox JR
   1         1        5
----- + ----- = -
x - 1 x - 2 6

Starting here is simple problem:


First raise both sides to the negative first power (invert both sides). You will get:
   x - 1       x - 2      6
-------- + ------- = -
1 1 5

Which simplifies to:
   2x - 3  =  6/5


Solve from there.

--Vito
In response to Vito Stolidus
Vito Stolidus wrote:
First raise both sides to the negative first power (invert both sides). You will get:
>    x - 1       x - 2      6
> -------- + ------- = -
> 1 1 5
>


There's a difference between raising a polynomial to a given power, and raising each term in a polynomial to a power. If you have the equation 2 + 2 = 4 and take the inverse of both sides, you don't get one-half plus one-half equals one-fourth, you get

   1     1
----- = -
2 + 2 4


Similarly, in the case of Prodigal Squirrel's question, if you were to take the inverse of both sides you'd actually end up with

        1
----------------- 6
1 1 = -
----- + ----- 5
x - 1 x - 2


which can be simplified, but you just end up with the same thing Prodigal Squirrel got, 5(x - 1)(x - 2) = 6(2x - 3).
In response to Vito Stolidus
Augh, no. Please don't offer help when you don't know what you're saying; you only make things worse. In this case the answer was already supplied correctly twice over.

Your answer would've earned a 0 mark on this problem. The technique Squirrel used, multiplying all terms by the least common denominator, was the correct one. What you did was invert each term, which is completely wrong. As Leftley demonstrated, you can't invert 2+2=4 to get 1/2+1/2=1/4. You have to invert the two terms as a whole, as if there were parentheses around them, because what you're really doing is 1/(sideA)=1/(sideB). Similarly if you square both sides of the equation, you can't just square each term; you have to square the sum.

For the same reason, you can't cancel individual terms in a fraction like so:
 x + y
-------
x^2+y^2

Some might try to cancel the x's and the y's, to get 1/(x+y) (or 2/(x+y), which is also wrong), but this is really equivalent to (x+y)/(x^2+y^2), and the parentheses get in the way. The only way you can cancel is if you can find a common factor between the two sides. If the bottom had been x^2+2xy+y^2, then you could get 1/(x+y) with no problem.

So it suffices to say you still have some learning to do in algebra. Best not to offer advice in it until you learn a lot more.

Lummox JR