mathlogic(shsat)
posted by samantha .
Question:
If 15  2x < 7, then for all possible values of x that make the inequality true,
(a) x<8
(b) x>8
(c) x>4
(d) x<4
(e) x>1/4
Answer:
i chose the answer (d) but the answer of the book said (c).
i understand where they are coming from we have to change the inequality sign due to the division we do. but i don't understand the logic. can someone kindly, please, explain the logic?
Great Thank in advance =)

15  2x < 7
Add 2x to each side
152x+2x < 7+2x
or
15 < 7+2x
Switch sides (switch inequality sign also)
7+2x > 15
subtract 7
77+2x > 157
2x > 8
x > 4
So it's (c)
What's important is when you multiply inequalities by 1 (probably what you did), you have to reverse the direction of the inequality. 
oh okay so instead of starting with the integer(15) you're saying i should start with the number and variable(2x). then flip the equation?
im still confused i learned that you should start with the integer and leave the variable for the end.
i also didn't get the part when you said i multiplied my inequality with 1 because i didn't
BTW: thanks a lot for your explanation. 
Basically, you can start with either one, as long as you finish with all variables on one side, and numbers (or integers) on the other side of the inequality.
HOWEVER, if you multiply each side by a negative number (or 1), you must flip the direction of the inequality.
For example,
5x<20, then x<4 (divide each side by 5)
BUT
3x > 18
Divide by NEGATIVE 3 to get
x < (18)/(3)
x < 6