Order of writing equation

[Guest]

What is the best way to come up with equation
Consider the problem

How many liters of oil must be added to x liters of an oil-water solution that is y percent oil to produce a solution that is z percent oil?

Solution 1: this is what you guys have mentioned

(y/100)x + n = (z/100)(x + n)
xy + 100n = xz + zn
xy - xz = nz - 100n
xy - xz = n(z - 100)

n = xy - xz
_________
z - 100


Solution 2: this is the one i came up with


(z/100)(x+n) - (y/100)x = n
zx + zn - yx = 100n
zx - yx = 100n - zn
zx - yx
_______ = n
100 - n

Can you tell me what approach did i took wrong while designing equation. How can we be sure as what order to pick.

[RonPurewal]

zx - yx
_______ = n
100 - n

Can you tell me what approach did i took wrong while designing equation. How can we be sure as what order to pick.

well, your denominator is actually supposed to be "100 - z", but nothing else is wrong.

your numerator is the opposite of our numerator, and your denominator is the opposite of our denominator. the two negatives cancel, leaving identical fractions.

it's actually the same reason that, say, -3/-7 is the same fraction as 3/7, or, slightly more complicatedly, (x - y) / (a - b) is the same fraction as (y - x) / (b - a).

if you don't buy this or don't see how it works, try plugging in a few concrete numbers and seeing what happens for yourself.

[Guest]

I clearly see what you are saying and thats what i reach as well, but when you reach my solution and don't find that option, it gets confusing.
And at that point you don't think as you should take negative common. So my question is not if they are different but question is what is the right approach to avoid such situations.

[RonPurewal]

I clearly see what you are saying and thats what i reach as well, but when you reach my solution and don't find that option, it gets confusing.
And at that point you don't think as you should take negative common. So my question is not if they are different but question is what is the right approach to avoid such situations.

well, the best approach is to learn "such situations", from every problem that you ever study.

after every single problem you ever solve, stop and ask yourself the following question:
"what have i learned from this problem that i can turn around and apply to another problem?"

in the case of this problem, you've learned about equivalent representations of fractions involving differences and/or negatives: if you flip the signs in both the numerator and denominator of a fraction, you get a different-looking fraction that is in fact equal to the original fraction.