Algebra solution diffuculties

bjuveges

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Given that and are distinct nonzero real numbers such that , what is [FONT=Verdana, Geneva, sans-serif]?[/FONT]

[FONT=Verdana, Geneva, sans-serif]I understand the question and without anyone needing to solve it, the answer is (2) but I am actually having a different issue other then finding the answer, but deciphering one of the possible solutions for the problem.

[/FONT]



Since



Where I get confused at is:

How did they factor the 2 fractions from the right side into this form? and next:



when the equation attempts to simplify by multiplying 2 to the fraction and gets (1-2/xy) I also do not understand.

Just to be clear I generally don't understand anything past these parts except for the solution which I found with another method, I just want to be able to understand this specific solution.

Any help is greatly appreciated and thanks in advance

Brandon
 
\(\displaystyle x\, -\, y\, +\, \dfrac{2}{x}\, -\, \dfrac{2}{y}\, =\, 0\)

\(\displaystyle (x\, -\, y)\, +\, 2\left(\dfrac{y\, -\, x}{xy}\right)\, =\, 0\)

How did they factor the 2 fractions from the right side into this form?
They factored in the usual manner:

. . . . .\(\displaystyle \dfrac{2}{x}\, -\, \dfrac{2}{y}\, =\, 2\left(\dfrac{1}{x}\right)\, -\, 2\left(\dfrac{1}{y}\right)\, =\, 2\left(\dfrac{1}{x}\, -\, \dfrac{1}{y}\right)\)

Then they converted to a common denominator, and combined the two fractions.

\(\displaystyle (x\, -\, y)\, +\, 2\left(\dfrac{y\, -\, x}{xy}\right)\, =\, 0\)

\(\displaystyle (x\, -\, y)\left(1\, -\, \dfrac{2}{xy}\right)\, =\, 0\)

when the equation attempts to simplify by multiplying 2 to the fraction and gets (1-2/xy) I also do not understand.

They skipped a few steps. What did you get when you (1) reversed the subtraction within the second parentheses and kicked the "minus" sign out front and then (2) factored the common \(\displaystyle (x\, -\, y)\) out front?

Please show all your steps. Thank you! ;)
 
Given that and are distinct nonzero real numbers such that , what is ?

I understand the question and without anyone needing to solve it, the answer is (2) but I am actually having a different issue other then finding the answer, but deciphering one of the possible solutions for the problem.





Since



Where I get confused at is:

How did they factor the 2 fractions from the right side into this form? and next:



when the equation attempts to simplify by multiplying 2 to the fraction and gets (1-2/xy) I also do not understand.

Just to be clear I generally don't understand anything past these parts except for the solution which I found with another method, I just want to be able to understand this specific solution.

Any help is greatly appreciated and thanks in advance

Brandon


Hi Brandon:

Try these steps, to go from x - y + 2/x - 2/y = 0

to (x - y) + 2*(y - x)/(xy) = 0


Factor out 2 from 2/x - 2/y

Combine 1/x - 1/y into a single ratio


To go from (x - y) + 2*(y - x)/(xy) = 0

to (x - y)(1 - 2/(xy)) = 0


Multiply each side by -1

Factor out (y - x) and simplify


Then, as x and y are different, we know that y-x is not zero.

Because the product (on the left-hand side) is zero, the factor 1-2/(xy) must equal zero.

Solve 1 - 2/(xy) = 0


Multiply both sides by xy

Add 2 to both sides


Cheers :)
 
Hi Brandon:

Try these steps, to go from x - y + 2/x - 2/y = 0

to (x - y) + 2*(y - x)/(xy) = 0


Factor out 2 from 2/x - 2/y

Combine 1/x - 1/y into a single ratio


To go from (x - y) + 2*(y - x)/(xy) = 0

to (x - y)(1 - 2/(xy)) = 0


Multiply each side by -1

Factor out (y - x) and simplify


Then, as x and y are different, we know that y-x is not zero.

Because the product (on the left-hand side) is zero, the factor 1-2/(xy) must equal zero.

Solve 1 - 2/(xy) = 0


Multiply both sides by xy

Add 2 to both sides


Cheers :)

Ok thanks for the explanation guys, but how in the heck does (x-y) cancel out? that completely flew over my head!
 
They skipped a few steps. What did you get when you (1) reversed the subtraction within the second parentheses and kicked the "minus" sign out front and then (2) factored the common \(\displaystyle (x\, -\, y)\) out front?

Please show all your steps. Thank you!
...how in the heck does (x-y) cancel out?
It doesn't.

Instead, try answering the questions asked of you, showing all of your steps, so we can try to see where you're getting lost. Thank you.
 
It doesn't.

Instead, try answering the questions asked of you, showing all of your steps, so we can try to see where you're getting lost. Thank you.

(x-y)(1-2/xy)
to
1=2/xy

Where did the (x-y) go? the solution states that x cannot equal y and then gets rid of it.
 
your question gives me a headache :confused:

Let's go step by step; the original equation:
X + 2/x = y + 2/y

(x^2 + 2) / x = (y^2 + 2) / y

(x^2 + 2) / x - (y^2 + 2) / y = 0

[y(x^2 + 2) - x(y^2 + 2)] / (xy) = 0 ; if numerator / denominator = 0, then numerator = 0; so:

Y(x^2 + 2) - x(y^2 + 2) = 0

x^2y + 2y - xy^2 - 2x = 0

x^2y - xy^2 = 2x - 2y

xy(x - y) = 2(x - y) ; cancel out the (x - y): \(\displaystyle \ \ \ \ \)

xy = 2

do you now "see" why x-y "disappears"?

By the way, tell your teacher there is another solution: \(\displaystyle \ \ \ \ \)There are no other solutions, because the statement of
of the problem *includes* the restriction in the quote box below.


x - y = 0
x = y
so xy = x^2 or xy = y^2

given that and are distinct nonzero real numbers ...
.
 
Ok thanks for the explanation guys, but how in the heck does (x-y) cancel out? that completely flew over my head!

The number x-y (whatever it is) does not get cancelled; it is still there. The only importance (to us) of that number is the fact that it's not zero. Once we realize this fact, we immediately focus on the other factor (that is, 1-2/[xy]), because we know that this other factor must be zero.

In other words, after we realize that 1-2/(xy)=0, we can ignore the factor (x-y).

After all, we're not interested in the value of either x or y, only the product xy.

Ciao :)
 
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