Crucial Doubt about Inequalities...!

[girish3131]

Let's consider a scenario ->

| x - y | > | a + b |

now is it right to open like

x + y > + (a + b ) or x + y > -( a +b ) -> EQUALTION 1

OR

+(x + y) > (a + b) or - (x +y ) > (a + b) -> EQUATION 2

which equation is right.... 1 or 2 ?


this is very crucial ques 4 me... plz reply only if u r pretty confident abt it... otherwise it will become more confusing 4 me...


Thanks!

[sanju09]

Let's consider a scenario ->

| x - y | > | a + b |

now is it right to open like

x + y > + (a + b ) or x + y > -( a +b ) -> EQUALTION 1

OR

+(x + y) > (a + b) or - (x +y ) > (a + b) -> EQUATION 2

which equation is right.... 1 or 2 ?


this is very crucial ques 4 me... plz reply only if u r pretty confident abt it... otherwise it will become more confusing 4 me...


Thanks!

Modulus Inequalities can be worked out by squaring both sides of the inequalities or by means of graphs. It is valuable to note that squaring both sides is legitimate only when both sides of the inequality are either zero or positive for all values of the variable(s) used, or else it is optional to work loose the inequalities using the graphical method.

Moderators are requested to move the problem to the right bin with a reminder.

[ajith]

moved to GMAT Math Section

[rockeyb]

Let's consider a scenario ->

| x - y | > | a + b |

now is it right to open like

x + y > + (a + b ) or x + y > -( a +b ) -> EQUALTION 1

OR

+(x + y) > (a + b) or - (x +y ) > (a + b) -> EQUATION 2

which equation is right.... 1 or 2 ?


this is very crucial ques 4 me... plz reply only if u r pretty confident abt it... otherwise it will become more confusing 4 me...


Thanks!

When solving or removing the mod we need to consider two cases as we dont know sign of the variable inside the mod .

|x+y| > |a+b|

Ideally you should solve it like this

1. case (x+y) > (a+b)
2.case (x+y) > -(a+b)

But before you finalize the answer you must put back the value of X back in to the orignal equation to see if it satisfies .

The value that satisfies is the correct one .

Ex : |x-2| = |2x-3|

1 .case
(x-2) = (2x-3)
x=1

2.Case
(x-2) = -(2x-3)
x=5/3

Now put x=1 back in the original equation.

|1-2| = |2(1)-3|
= 1

satisfies.

Now put x= 5/3 in the equation

|5/3 -2| = |2(5/3) - 2|

= 1/3

satisfies .

Hence both are the solution for the equation.