Modulus Doubt

[madhur_ahuja]

Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x + 1| = 2x – 1

OA : D

I have the partial solution to this question, which I have a doubt on.

When x+3 >0
x + 3 = 4x – 3
6 = 3x
2 = x

When X+ 3 <0

-1(x + 3) = 4x – 3
-x – 3 = 4x – 3
0 = 5x
0 = x

But if we plug 0 into the original equation, it is not a valid solution:
|0 + 3| = 4(0) – 3
|3| = 0 – 3
3 = -3


Can we reject the second solutoin x=0 on the basis that (X+3) should be negative. However, since x=0, it is positive.

Do we always need to plug back the values to determine the validity of the solution in modulus questions ?

[scoobydooby]

yes you are spot on. the possible solutions must be checked back to see if they really satisfy the given statements.

[yezz]

Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x + 1| = 2x – 1


from 1

x+3 = 4x-3 ie: x = 2 or x+3 = 3-4x ie: x = 0,,,insuff

from 2

x+1 = 2x-1 ie: x = 2 or x+1 = 1-2x ie: x = 0..insuff

both
insuff
E

[yezz]

Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x + 1| = 2x – 1


from 1

x+3 = 4x-3 ie: x = 2 or x+3 = 3-4x ie: x = 0,,,insuff

from 2

x+1 = 2x-1 ie: x = 2 or x+1 = 1-2x ie: x = 0..insuff

both
insuff
E

[pandeyvineet24]

Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x + 1| = 2x – 1

OA : D

I have the partial solution to this question, which I have a doubt on.

When x+3 >0
x + 3 = 4x – 3
6 = 3x
2 = x

When X+ 3 <0

-1(x + 3) = 4x – 3
-x – 3 = 4x – 3
0 = 5x
0 = x

But if we plug 0 into the original equation, it is not a valid solution:
|0 + 3| = 4(0) – 3
|3| = 0 – 3
3 = -3


Can we reject the second solutoin x=0 on the basis that (X+3) should be negative. However, since x=0, it is positive.

Do we always need to plug back the values to determine the validity of the solution in modulus questions ?

Actually Madhur, you can also do the following

from stmt 1

(1) |x + 3| = 4x – 3

since |x + 3| will always be positive and hence > 0, therefore 4x - 3 > 0 which leads to x >3/4
hence suff

now Stmt 2

(2) |x + 1| = 2x – 1
Same as above, gives x > 1/2
hence suff

Answer D

there

[tohellandback]

Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x + 1| = 2x – 1

OA : D

I have the partial solution to this question, which I have a doubt on.

When x+3 >0
x + 3 = 4x – 3
6 = 3x
2 = x

When X+ 3 <0

-1(x + 3) = 4x – 3
-x – 3 = 4x – 3
0 = 5x
0 = x

But if we plug 0 into the original equation, it is not a valid solution:
|0 + 3| = 4(0) – 3
|3| = 0 – 3
3 = -3


Can we reject the second solutoin x=0 on the basis that (X+3) should be negative. However, since x=0, it is positive.

Do we always need to plug back the values to determine the validity of the solution in modulus questions ?

My simple solution:
|X| always >=0

(1) |x + 3| = 4x – 3
4x-3>=0, x>=3/4
SUFFICIENT
(2) |x + 1| = 2x – 1
2x-1>=0
x>=1/2 SUFFICIENT

so D