Question

If $\frac{-2(y-2)}{5}>10$ holds for all $y,$ then which of the following is true for any $y$?

Answer 1

Given: $\frac{-2(y-2)}{5}>10$

We multiply both sides by 5. Since 5 is a positive number, the sign of the inequality remains unchanged.

$\therefore \frac{-2(y-2)}{5}\times 5>10\times 5$

i.e., $-2(y-2)>50$

i.e., $-2y+4>50$

Subtracting 4 from both sides of the above inequality, we get:

$-2y+4-4>50-4$

i.e., $-2y>46$

Dividing both sides by -2 to isolate $y,$ the sign of the inequality reverses as -2 is a negative number.

$\therefore \frac{-2y}{-2}<\frac{46}{-2}$

i.e., $y<-23$