Which of the inequalities has the solution whose graph is shown?

F. $2x+10<16$

G. $-4x-5<7$

H. $-3x+8<-1$

I. $4-2x>10$

In this, use the concept of solving the linear inequalities by doing mathematical operations.

Here,

First expression is $2x+10<16$.

$\begin{array}{l}2x+10<16\\ 2x+10-10<16-10\\ 2x<6\\ \frac{2x}{2}<\frac{6}{2}\\ x<3\end{array}$

Second Expression is $-4x-5<7$

$\begin{array}{l}-4x-5<7\\ -4x-5+5<7+5\\ -4x<12\\ \frac{-4x}{-4}<\frac{12}{-4}\\ x>-3\end{array}$

Third expression is $-3x+8<-1$.

$\begin{array}{l}-3x+8<-1\\ -3x+8-8<-1-8\\ -3x<-9\\ \frac{-3x}{-3}<\frac{-9}{-3}\\ x>3\end{array}$

Fourth expression is $4-2x>10$.

$\begin{array}{l}4-2x>10\\ 4-2x-4>10-4\\ -2x>6\\ \frac{-2x}{-2}>\frac{6}{-2}\\ x<-3\end{array}$

Now, from the above four solved expression, it is clearly seen that the correct solution for the graph is second expression whose solution is $x>-3$.

F. $2x+10<16$:It is not the correct option because it’s solution is$x<3$ which is different from the correct solution that is $x>-3$.

G. $-4x-5<7$: It is the correct option because it’s solution is $x>-3$ which is same from the correct solution that is $x>-3$.

H. $-3x+8<-1$: It is not the correct option because it’s solution is $x>3$ which is different from the correct solution that is $x>-3$.

I. $4-2x>10$: It is not the correct option because it’s solution is $x<-3$ which is different from the correct solution that is $x>-3$.