91Ó°ÊÓ

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$−\frac{2}{3}tâ‰\yen 24$ is:

$\begin{array}{c}−\frac{2}{3}tâ‰\yen 24\\ \left(\frac{−3}{2}\right)\left(−\frac{2}{3}t\right)≤\left(\frac{−3}{2}\right)\left(24\right)\left(\text{byusingmutiplicationpropertyofinequality}\right)\\ \left(\frac{−3}{2}×\frac{−2}{3}\right)t≤−\frac{3}{2}×24\\ t≤−36\end{array}$

Therefore, the solution of the given inequality$−\frac{2}{3}tâ‰\yen 24$ is $t≤−36$.

To perform the check of the solution, substitute a number less than or equal to $−36$as $t≤−36$in the given inequality $−\frac{2}{3}tâ‰\yen 24$, if the condition obtained is true, the solution is correct and if the condition obtained is false, the solution is incorrect.

Let $t=−39$, as $−39≤−36$.

Now substitute $−39$for $t$in the inequality $−\frac{2}{3}tâ‰\yen 24$.

$\begin{array}{c}−\frac{2}{3}tâ‰\yen 24\\ \frac{−2}{3}\left(−39\right)â‰\yen 24\\ −2×−13â‰\yen 24\\ 26â‰\yen 24\end{array}$

As, the condition obtained$26â‰\yen 24$ is true, therefore the solution is correct.