# equations and inequallities

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• Oct 4th 2009, 04:38 AM
alternative
equations and inequallities
Hi

I apologize If I didn't choose the right forum, for my new thread, but I really need some help with my homework

/ / Used for absolute value

Solve Inequalities:

a. /2x-3/<5
b. /x^2-5x+6/>2
c. /x/>x+1
d. (x+1)(3x-5)<0
e. 2/x/+3>5

Solve equations:

a. /x+1/-/x+2/=2
b. /2x+3/=x^2

I'd appreciate the way of solving too, not just the results :D:D:D:D:D

Respectfully
Alternative
• Oct 4th 2009, 07:19 AM
mathaddict
Quote:

Originally Posted by alternative
Hi

I apologize If I didn't choose the right forum, for my new thread, but I really need some help with my homework

/ / Used for absolute value

Solve Inequalities:

a. /2x-3/<5
b. /x^2-5x+6/>2
c. /x/>x+1
d. (x+1)(3x-5)<0
e. 2/x/+3>5

Solve equations:

a. /x+1/-/x+2/=2
b. /2x+3/=x^2

I'd appreciate the way of solving too, not just the results :D:D:D:D:D

Respectfully
Alternative

HI

next time , do not post so many questions in one thread . Its against the rules of the forum . Post one or two , then try to solve the rest and come back if you are really lost .

I will show you a few , then u try out the rest . OK ?

(b) |x^2-5x+6|>2

The properties of absolute value tells us that |x|>a , then x>a , x<-a

so here $\displaystyle x^2-5x+6>2\Rightarrow (x-1)(x-4)>0$ .. try continue here

then the other case would be $\displaystyle x^2-5x+6<-2$ .. this is complex so we can just ignore .

(c) |x|>x+1

square both sides .. $\displaystyle x^2>x^2+2x+1$

$\displaystyle 2x+1<0$

$\displaystyle x<-\frac{1}{2}$

(a) $\displaystyle |x+1|-|x+2|=2$

$\displaystyle |x+1|=2+|x+2|$

square both sides $\displaystyle x^2+2x+1=4+4|x+2|+x^2+4x+4$

$\displaystyle -2x-7=4|x+2|$

Then square again ..

$\displaystyle 4x^2+28x+49=16(x^2+4x+4)$
carry on here .
• Oct 4th 2009, 07:35 AM
alternative
Thank You Very Veryyyyy Much, And Sorry, I Didn't Meant To Break The Rules(Angel)