If we turn the division into the multiplication of the reciprocal, we get:
(201/15) (1/x) < 1
Now multiply the fractions:
201 / (15x) < 1
From here, let's multiply both sides by 15:
201 / x < 15
At this point, we don't know if x is positive or negative. We need to know since we have to flip the sign if it's negative. So let's split this up into two inequalities, one where x > 0 and the other where x < 0. We flip the sign when x < 0, so we have:
201 < 15x when x > 0 and 201 > 15x when x < 0
Now let's flip the left and right sides so "x" is on the left side. We flip the signs to compensate:
15x > 201 when x > 0 and 15x < 201 when x < 0
Now divide both sides by 15:
x > 201/15 when x > 0 and x < 201/15 when x < 0
This reduces:
x > 67/5 when x > 0 and x < 67/5 when x < 0
So when x > 67/5, all values are also > 0, so this entire range can be included in the solution set.
When x < 67/5, since x must also be < 0, only the values that overlap the ranges must be included. So that overlap makes x < 0 for this part of the solution set.
Answers & Comments
201 / 15x < 1
201 < 15 x
x > 201 / 15
x > 67 / 5
x > 13 . 4
(201/15)/x <1
201/(15x) < 1
201 < 15 x
15 x > 201
x > 201/15
Any number greater than 201/15 = 13.4 will give a result < 1
Graph it and see.
201/15 ÷ x < 1
If we turn the division into the multiplication of the reciprocal, we get:
(201/15) (1/x) < 1
Now multiply the fractions:
201 / (15x) < 1
From here, let's multiply both sides by 15:
201 / x < 15
At this point, we don't know if x is positive or negative. We need to know since we have to flip the sign if it's negative. So let's split this up into two inequalities, one where x > 0 and the other where x < 0. We flip the sign when x < 0, so we have:
201 < 15x when x > 0 and 201 > 15x when x < 0
Now let's flip the left and right sides so "x" is on the left side. We flip the signs to compensate:
15x > 201 when x > 0 and 15x < 201 when x < 0
Now divide both sides by 15:
x > 201/15 when x > 0 and x < 201/15 when x < 0
This reduces:
x > 67/5 when x > 0 and x < 67/5 when x < 0
So when x > 67/5, all values are also > 0, so this entire range can be included in the solution set.
When x < 67/5, since x must also be < 0, only the values that overlap the ranges must be included. So that overlap makes x < 0 for this part of the solution set.
So there are two ranges that will satisfy:
x < 0 and x > 67/5
Divide by number greater than 201/15.
201/15 ÷ x < 1
Multiply both sides of the inequality by x. Since x is positive, the direction of the inequality does not change.
201/15 < x
201/15 = 13.4, so x is any number greater than 13.4
201/15 ÷ 14 ≅ 0.96 < 1
201/15 ÷ 15 ≅ 0.83 < 1
etc.