|x+3|+x/x+2>1 give me the solution pleas fast
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Given {|x+3| + x}/(x+2) > 1
WE know that
|x| = x if x > 0
and
|x| = -x if x < 0
Case 1. if |x+3| is positive then
{|x+3| + x}/(x+2) > 1
=> (x+3 + x)/(x+2) > 1
=> (2x + 3) > x+2
=> 2x - x > 2-3
=> x > -1
Case 2. if |x+3| is negative then
{|x+3| + x}/(x+2) > 1
=> {-(x+3) + x}/(x+2) > 1
=> (-x - 3 + x) > x+2
=> -3 > x+ 2
=> x + 2 < -3
=> x < -3-2
=> x < -5
So value of x lies between -5 and -1.
hope it helps...
WE know that
|x| = x if x > 0
and
|x| = -x if x < 0
Case 1. if |x+3| is positive then
{|x+3| + x}/(x+2) > 1
=> (x+3 + x)/(x+2) > 1
=> (2x + 3) > x+2
=> 2x - x > 2-3
=> x > -1
Case 2. if |x+3| is negative then
{|x+3| + x}/(x+2) > 1
=> {-(x+3) + x}/(x+2) > 1
=> (-x - 3 + x) > x+2
=> -3 > x+ 2
=> x + 2 < -3
=> x < -3-2
=> x < -5
So value of x lies between -5 and -1.
hope it helps...
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