Question 91998: I need massive help, I have a problem that I have been unable to find an example of in my books, all five of them and no one of my other coworkers who are algebra smart have a clue on how to approach it. This is a online course for military and my student adivsor never answers my emails. So here it goes: Find all values of x satisfying the given conditions. y1(lower)=1/x+7,y2(lower)=2/x+3,y3(Lower)=-4/x^2+10x+21, and y1(lower) + y2(lower)=y3(lower) Thank very much
Found 3 solutions by stanbon, homeworkhelpanytime, bucky:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Find all values of x satisfying the given conditions.
y1(lower)=1/x+7
y2(lower)=2/x+3,
y3(Lower)=-4/x^2+10x+21,
and y1(lower) + y2(lower)=y3(lower)
------------
EQUATION:
1/(x+7) + 2/(x+3) = 4/(x^2+10x+21)
1/(x+7) + 2/(x+3) = 4/[(x+7)(x+3)]
-----------------
Multiply thru by (x+7)(x+3) to get:
(x+3) + 2(x+7) = 4
3x + 10 = 4
3x = -6
x = -2
============
Cheers,
Stan H.
Answer by homeworkhelpanytime(21) (Show Source):
You can put this solution on YOUR website! Given
y(1) = 1/x + 7
y(2) = 2/x + 3
y(3) = -4/x^2 + 10x + 21
now
y(1) + y(2) = y(3)
1/x + 7 + 2/x + 3 = -4/x^2 + 10x + 21
3/x + 10 = -4/x^2 + 10x + 21
4/x^2 - 10x - 21 + 3/x + 10 = 0
taking L.c.m we get
[4 -10x^3 + 3x -21x^2 + 10x^2]/x^2 = 0
-10x^3 -11x^2 +3x +4 =0
(x+1)(-10x^2 - x +4) = 0
so 3 values of x satisfying this will be
x = -1,
x = [(1 +sqrt(161)]/20
x = [(1 -sqrt(161)]/20
contact at info@homeworkhelpanytime.com for nay further details
Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website! I've looked at the two answers posted, and I think they both have mistakes. I think the
first answer interpreted the problem incorrectly for the reason that I will discuss at
the end of my work. And, although Stan H read between the lines and interpreted the problem
the way you meant it to be, he also made a couple of mistakes ... one in multiplying
2(x + 7) and another one in forgetting that the 4 on the right side of the equation has
a minus sign.
You can compare my answer (closer to Stan H's than the other party) and maybe come up with
your own version of what is correct. Anyhow, here goes:
.
I think the problem you intended to write is:
.
 and
,
With the added condition that 
.
Let’s go right to the “added condition” equation and substitute for  , for  ,
and for  to get:
.
.
Notice that on the right side, the denominator can be factored to  .
This makes the equation become:
.
.
Now suppose we multiply the first term on the left side by  and the second
term on the left side by  .
.
Note that since the two multipliers each have the same numerator as they do denominator.
Therefore, they are each equivalent to 1. So we are multiplying both terms by 1 and are
not changing the equation. The multiplication of the two terms makes the equation become:
.
.
In this equation, every term has the denominator  . So we can get rid
of the denominators entirely by multiplying both sides of the equation (all terms) by
. This in effect cancels the denominators out and the equation simplifies
to just the numerators … or to:
.
.
Do the multiplications on the left side and the equation becomes:
.
.
Combine the terms on the left side to get:
.
.
Subtract 17 from both sides to reduce the equation to:
.
.
And solve for x by dividing both sides by 3 to get:
.
.
All those who think x = -7 is the right answer, take one step forward. Not so fast ...
.
Go back to the original problem and look at 
.
What happens to the denominator of if you substitute -7 for x. The denominator
becomes zero … and division by zero is not allowed in algebra. Therefore, our answer
of x = -7 will not work. So the answer to this problem is that there is no value for
x that will satisfy this problem.
.
Hope this helps you get on to bigger and better problems than this one … and hope that
you work through all three of the answers you got and pick the one that seems to work for you.
An added note … according to the rules of algebra, when you write 1/x +7 you mean 
because the rules say to do multiplications and divisions first, then go back and do
the additions and subtractions. To make sure both the x and the +7 are in the denominator,
you need to write 1/(x+7). Then when you do the division you divide by the quantity (x+7).
The way you wrote it is a very common practice in this forum, but it often makes it difficult
to interpret the problem correctly. I hope I got your problem interpreted correctly in form …
.
|
|
|
