I dont know if the your approach is right

Best way is to
find out roots of individual expression
x=-3/2 and x=2/7..at these points the expression changes sign from +ve to negative


consider -3/2<x<2/7
=> 2x+3>0 and 7x-2<0
now solve
2x+3>-(7x-2)
=> 9x>-1=>x>-1/9

soln is -1/9<x<2/7------(A)

now x>2/7=> both expression are positive

2x+3>7x-2
=>5x<5
=>x<1
therefore soln is

2/7<x<1-----(B)

now consider x<-3/2

both expression are negative
-(2x+3)>-(7x-2)
=>2x+3<7x-2
=>5x>5=>x>1

contradiction since x<-3/2 assumed but soln set says x>1

therefore no soln for x<-3/2

therefore using A & B we have
-1/9<x<1

E cannot be a soln. Just plug x=100 or -100 and see the inequality does not hold

Just remebered that squaring both sides will provide a solution but it may also give soln that will not satisy original equation. So need to check by plugging in original eqn

|2x+3|>|7x-2|
squaring we get

9x^2-8x-1<0
=>(9x+1)*(x-1)<0
=>-1/9<x<1

check for x=0 and x=-10 and x=10 so ensure that soln is correct

you have made mistake in finding the last inequality

What Is the OA??

It has to be C...........

am getting C as well

rah_pandey wrote:I dont know if the your approach is right

Best way is to
find out roots of individual expression
x=-3/2 and x=2/7..at these points the expression changes sign from +ve to negative


consider -3/2<x<2/7
=> 2x+3>0 and 7x-2<0
now solve
2x+3>-(7x-2)
=> 9x>-1=>x>-1/9

soln is -1/9<x<2/7------(A)

now x>2/7=> both expression are positive

2x+3>7x-2
=>5x<5
=>x<1
therefore soln is

2/7<x<1-----(B)

now consider x<-3/2

both expression are negative
-(2x+3)>-(7x-2)
=>2x+3<7x-2
=>5x>5=>x>1

contradiction since x<-3/2 assumed but soln set says x>1

therefore no soln for x<-3/2

therefore using A & B we have
-1/9<x<1

E cannot be a soln. Just plug x=100 or -100 and see the inequality does not hold

I couldn't agree more with your explanation. congrats

What is the OA??

Naruto wrote:What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < (-1/9) or x > 5

B. -1 < x < (1/9)

C. (-1/9) < x < 1

D. (-1/9) < x < 5

E. x < (-1/9) or x > 1



[spoiler]Answer:E [/spoiler]

My approach was solving by squaring both sides:
which leads to

45x^2-40x-5<0
i.e 9x^2-8x-5<0
i.e (9x+1)(x-1)<0
i.e x<(-1/9) or X<1

I eventually chose the correct answer but i cant understand how x>1 , is there another easier approach or alternative, please ecplain. Thanks.

Guys the OA is E. And well in a way thanks to you, i had to solve this with all possibilities and choose the greatest range. Here is the explanation.
|2x + 3| > |7x - 2|
So 4 conditions to be considered.
I------2X+3>7X-2
i.e x<1

II-----2X+3>-(7X-2)
2X+3>-7X+2
i.e x> -1/9

III--- -(2X+3)> 7X-2
-2X-3>7X-2
i.e x< -1/9

IV---- -(2X+3)> -(7X-2)
i.e x>1

Now with the 4 results consider the largest range of x....
since x<1 and x<-1/9, therefore x<-1/9 is chosen
similarly x>-1/9 and x>1 therefore x>1 is chosen.
Hence the answer E.
Let me know your comments.

I dont think the soln as posted by Naruto is right. The four conditions are not possible.Only 3 make sense. any 2 points will divide the number line into three regions and not four. See comments inline

I------2X+3>7X-2
i.e x<1
Possible for x>2/7
II-----2X+3>-(7X-2)
2X+3>-7X+2

Possible for x>-3/2 and x<2/7
III--- -(2X+3)> 7X-2
-2X-3>7X-2

Not Possible for any x. The reason is obvious. Left expression is negative for x<-3/2 but 7x-2 is also negative for x<-3/2. Try out with x=-2
IV---- -(2X+3)> -(7X-2)

Possible for x<-3/2

I hope this is not OG problem and soln. As discussed earlier x>1 is not a soln. We can all plug and see if the equation is satisfied