Solve (2-a)(10-3a)<0

$$( 2 - a ) ( 10 - 3 a ) < 0$$

Solve for a

$$a\in \left(2,\frac{10}{3}\right)$$

Solution Steps

For the product to be negative, 2-a and 10-3a have to be of the opposite signs. Consider the case when 2-a is positive and 10-3a is negative.

$$2-a>0 10-3a<0$$

This is false for any a.

$$a\in \emptyset$$

Consider the case when 10-3a is positive and 2-a is negative.

$$10-3a>0 2-a<0$$

The solution satisfying both inequalities is a\in \left(2,\frac{10}{3}\right).

$$a\in \left(2,\frac{10}{3}\right)$$

The final solution is the union of the obtained solutions.

$$a\in \left(2,\frac{10}{3}\right)$$