Solve each inequality. Check your solution.

$\frac{\mathbf{1}}{\mathbf{5}}\mathit{h}\mathbf{>}\mathbf{3}$

The multiplication property of inequalities states that if both sides of the inequality are multiplied by a positive number the sign of the inequality remains the same and if both sides of the inequality are multiplied by a negative number then the sign of the inequality changes that is:

1. If$a>b$ and$c$ is a positive number then $ac>bc$.
2. If$a<b$ and$c$ is a positive number then $ac<bc$.
3. If$a>b$ and$c$ is a negative number then $ac<bc$.
4. If$a<b$ and$c$ is a negative number then $ac>bc$.

The solution of the given inequality$\frac{1}{5}h>3$is:

$\begin{array}{c}\frac{1}{5}h>3\\ \left(5\right)\left(\frac{1}{5}h\right)>\left(5\right)\left(3\right)\left(\text{byusingmutiplicationpropertyofinequality}\right)\\ h>15\end{array}$

Therefore, the solution of the given inequality$\frac{1}{5}h>3$is $h>15$.

To perform the check of the solution, substitute a number greater than 15 as $h>15$in the given inequality $\frac{1}{5}h>3$, if the condition obtained is true, the solution is correct and if the condition obtained is false, the solution is incorrect.

Let $h=20$, as $20>15$.

Now substitute 20 for $h$in the inequality $\frac{1}{5}h>3$.

$\begin{array}{c}\frac{1}{5}h>3\\ \frac{1}{5}\left(20\right)>3\\ 4>3\end{array}$

As, the condition obtained$4>3$ is true, therefore the solution is correct.