Fun Inequality

[scott.yin]

I understand the answer to this question is C if you assume X is positive. But if you work out both cases (pos and neg), shouldnt the answer be E?

If x is an integer, what is the value of x?

(1) -5x > -3x + 10

(2) -11x - 10 < 67

[mithunsam]

No, both statements together give x = -6. There is no +ve case for x.

Statement 1
-5x > -3x + 10 => -2x > 10 => -x > 5 => x < -5 [This is true only when x is -ve]

Statement 2
-11x - 10 < 67 => -11x < 77 => -x < 7 => x > -7 [x could be +ve or -ve]

Together we get -7 < x < -5 => x = -6

[scott.yin]

-5x > -3x + 10 => -2x > 10 => -x > 5 => x < -5 [This is true only when x is -ve]

Don't you need to test both +ve and -ve cases (flip the initial inequality)?

So wouldnt you also go ahead and find -5x < -3x + 10 => -x < 5 => x >5 ??

[mithunsam]

-5x > -3x + 10 => -2x > 10 => -x > 5 => x < -5 [This is true only when x is -ve]

Don't you need to test both +ve and -ve cases (flip the initial inequality)?

So wouldnt you also go ahead and find -5x < -3x + 10 => -x < 5 => x >5 ??

One point to note - For -ve values, we don't just flip the inequality sign alone. We also have to multiply both left and right sides of the inequality with -1. [In this case, when you do that, you will end up in the same equation].

For addition and subtraction, you don't have to do the work, which we do for squares, square roots etc... This is because addition and subtraction do not hide signs. Multiplication hide signs (-ve*-ve = +ve). Therefore, we have to consider both scenarios. (For example, x^2 will always be +ve, but x could be +ve or -ve. Therefore, we have to consider both the scenarios.)

If you still have question, substitute 0 or +ve values in the original equation.

For x=0
-5x > -3x + 10 => -5*0 > -3*0 + 10 => 0 > 10 ---> False
For x=1
-5x > -3x + 10 => -5*1 > -3*1 + 10 => -5 > 7 ---> False
For x=2
-5x > -3x + 10 => -5*2 > -3*2 + 10 => -10 > 4 ---> False

Only -ve values could satisfy this equation.

[jnelson0612]

Good stuff, thanks all.