hi

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- Aug 9th 2008, 12:34 AMperashconditions
hi

- Aug 9th 2008, 01:53 AMnikhil
If a.b=0 then (it should be 0 unlike your examples)

a=0 and b=0

this will be true always

a and b be are any mathematical expressions - Aug 9th 2008, 03:12 AMwingless
Firstly, you should understand how this works.

$\displaystyle x \cdot y = 7$

What can x and y be? There are infinitely many answers in the rational set.

So, what can x and y be if x and y are natural numbers?

The answer is not hard to see, 7 has two factorizations, 7.1 and 1.7. So (x,y) = (7,1) or (x,y) = (1,7) will be correct.

But!..

Even when we assume that x and y are natural numbers, you have to think of every factorization of the RHS. When the RHS was 7 it wasn't a problem, there were only two possible factorizations.

But what happens if the question was $\displaystyle x \cdot y = 120$ ?

There are lots of factorizations,

(x,y) = (1,120), (2,60), (3,40), (4,30), (5,24), (6,20), (8,15), (10,12), (12,10), (15,8), (20,6), (24,5), (30,4), (60,2), (120,1)

So if you have an equation where the factors are natural numbers, you have to try every single possible factorization in order to get the correct answer.

If the factors x and y are not natural numbers, we can't use this approach. - Aug 9th 2008, 03:55 AMJaneBennet
Sigh. Why can’t people just answer the question directly? (Headbang)

Okay, when does $\displaystyle (x+a)(b-x)=c\ \Rightarrow\ x+a=c\ \mbox{or }b-x=c$?

Obviously $\displaystyle c=0$ is one possibility.

Now suppose $\displaystyle c\ne0$.

$\displaystyle x+a=c\ \Rightarrow\ x=c-a$

$\displaystyle \color{white}.\hspace{18mm}.$$\displaystyle \Rightarrow\ (c-a+a)(b-(c-a))=c$

$\displaystyle \color{white}.\hspace{18mm}.$$\displaystyle \Rightarrow\ c(b-c+a))=c$

$\displaystyle \color{white}.\hspace{18mm}.$$\displaystyle \Rightarrow\ b-c+a=1$ if $\displaystyle c\ne0$

Do the same for $\displaystyle b-x=c$. We get the same result, $\displaystyle b-c+a=1$ if $\displaystyle c\ne0$.

Hence the conditions are $\displaystyle c=0$ or $\displaystyle a+b-c=1$.