# Solve diffrent

• Dec 4th 2012, 09:45 AM
Petrus
Solve diffrent
Its old exam im working with:) the problem is solve diffrent so the picture nr 1 at begining is the problem and rest is how u solve it but on third step (the step before last) acording to me it should be wrong:S they wanna get same in the bottom so u multiplicerar with x+1 and x+3 so acording to me it should be (x+1)^2(x+3)^2 can some1 plz explain:)?
• Dec 4th 2012, 10:22 AM
topsquark
Re: Solve diffrent
Quote:

Originally Posted by Petrus
Its old exam im working with:) the problem is solve diffrent so the picture nr 1 at begining is the problem and rest is how u solve it but on third step (the step before last) acording to me it should be wrong:S they wanna get same in the bottom so u multiplicerar with x+1 and x+3 so acording to me it should be (x+1)^2(x+3)^2 can some1 plz explain:)?

$\displaystyle \frac{7}{4} - \frac{6}{13}$

You want to get a common denominator, which is 4*13. So
$\displaystyle \frac{7}{4} - \frac{6}{13} = \frac{7*13 - 6*4}{4 \cdot 13}$

The point to note here is that we only need one of each factor (4 and 13) in the denominator.

So if we have
$\displaystyle \frac{x - 2}{x + 1} - \frac{x - 1}{x + 3}$

the common denominator will be (x + 1)(x + 3). We only need one of each factor.

-Dan
• Dec 4th 2012, 12:40 PM
Deveno
Re: Solve diffrent
in the field of rational functions of x, the quotient:

f(x)/f(x) = 1 (so long as f(x) is not the constant 0-function) (strictly speaking this isn't really true, because we might have to "leave out points where f(x) = 0" whereas the constant function c(x) = 1 is defined for ALL x...but it's true "almost everywhere"...there's only a finite number of x's if we're talking about polynomials where the denominator might be undefined. these points might have to be looked at separately).

so:

$\displaystyle \frac{f(x)}{g(x)} + \frac{h(x)}{k(x)} = \left(\frac{f(x)}{g(x)}\cdot\frac{k(x)}{k(x)} \right) + \left(\frac{h(x)}{k(x)}\cdot\frac{g(x)}{g(x)} \right) = \frac{f(x)k(x) + g(x)h(x)}{g(x)k(x)}$

see? no squares.

your example is just the same as this with:

f(x) = x - 2
g(x) = x + 1

h(x) = -(x - 1)
k(x) = x + 3

the points -1 and -3 make the one of the two denominators "blow up". at these two values for x, the inequality "makes no sense", on one side of -1, (x - 2)/(x + 1) is large negative (the right side), on the other side (the left side) (x - 2) is large positive, while the fraction (x - 1)/(x + 3) is negative, and a similar consideration holds for x = 3.

so we can only establish the truth of the inequality on these 3 intervals:

(-∞,-3), (-3,-1), and (-1,∞).