Is q > t

[cdashw]

Is q > t ?

(1) qp^2 < tp^2

(2) qp^3 > tp^3

the solution states you can divide "p" expression from both sides of the inequality since p#0. However this method is only applied to statement 1 and not statement 2? Why

[esledge]

It is because of the positive and negative possibilities.

For (1) we have p^2 in the expression. If p #0, then it is either negative or positive. However, p^2 will always be positive because it has an even exponent. We are allowed to divide by p because it is not 0, but we also know that we don't need to "flip the sign" of the inequality because p^2 is positive.

For (2), we cannot just divide by p^3 because there are two cases:

If p<0, then qp^3 > tp^3 becomes q < t (we flip the sign).
If p>0, then qp^3 > tp^3 becomes q > t (don't flip the sign).
Without knowing p's sign, we don't know which option applies.

Try some numbers to see this.

If p = -1, then:
q(-1)^3 > t(-1)^3
-q > -t
q < t
One possibility: q = 2 and t = 3 works because -2>-3 and 2<3.

If p = +1, then:
q(1)^3 > t(1)^3
q > t
One possibility: q = 3 and t = 2 works.