If x is a positive integer and l2x+6l > 10, what is the least possible value of x?
NOTE: l2x+6l means absolute value of 2x+6
l2x+6l > 10
---> 2x+6 > 10 OR -2x-6 > 10
2x > 4 or -2x > 16
x > 2 or x < -8
think of the positive number line, which integer x do you have so that x > 2 ?
this is kind of a solve by inspection (and thought) problem
the "magic" value is 10 ... 2x + 6 must be GREATER than 10 (not equal to)
how small can x (a positive integer) be , if 2x must be greater than 4?
would it be 3??
of course
thank you
To find the least possible value of x, we need to solve the inequality:
|2x+6| > 10
To do this, we'll consider the two cases: when 2x+6 is positive and when it is negative, since the absolute value is always positive.
Case 1: 2x+6 > 10
This inequality can be solved as follows:
2x+6 > 10
Subtract 6 from both sides:
2x > 4
Divide both sides by 2 (since 2 is positive):
x > 2
Case 2: - (2x+6) > 10
We put the expression - (2x+6) in parentheses since the absolute value could be equal to negative values as well. Now, solve this inequality:
- (2x+6) > 10
Multiplying by -1 (which reverses the inequality):
2x+6 < -10
Subtract 6 from both sides:
2x < -16
Divide both sides by 2 (since 2 is positive):
x < -8
Therefore, the possible values of x are x > 2 and x < -8. However, since x is a positive integer, the least possible value of x is the smallest positive integer greater than 2: x = 3.