equations and inequallities

alternative · #1 $Old$ October 4th, 2009, 05:38 AM

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

Respectfully
Alternative

mathaddict · #2 $Old$ October 4th, 2009, 08:19 AM

Quote:

Originally Posted by alternative $View Post$

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

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

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

(c) |x|>x+1

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

$2x+1<0$

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

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

$|x+1|=2+|x+2|$

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

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

Then square again ..

$4x^2+28x+49=16(x^2+4x+4)$
carry on here .

alternative · #3 $Old$ October 4th, 2009, 08:35 AM

Thank You Very Veryyyyy Much, And Sorry, I Didn't Meant To Break The Rules