Question

is the solution of the following inequation.
$2y-3<y+2<3y+5$ if $y\in \text{R}.$

• A. $\frac{-3}{2}<y<5$
• B. $\frac{-3}{2}<y<4$
• C. $\frac{-3}{2}<y<3$

Answer 1

The correct option is A $\frac{-3}{2}<y<5$

Given: $2y-3<y+2<3y+5$
$\begin{array}{c}2y-3<y+2\\ \\ 2y-y<2+3\\ y<5\end{array}|\begin{array}{c}y+2<3y+5\\ -2y<3\\ -y<\frac{3}{2}\\ \text{Dividing by -1 and reversing the inequality sign we get}\\ y>-\frac{3}{2}\\ -\frac{3}{2}<y\end{array}$
Hence the solution set is $=y:yϵR,-\frac{3}{2}<y<5$
The number line is given by: