91Ó°ÊÓ

To graph the endpoint with strict inequality < or >, use the open parenthesis or a hollow circle at that endpoint.

To graph the endpoint with inequality symbol $≤\text{or}â‰\yen$, use the bracket or a solid circle at that endpoint.

Properties of inequality:

width="189">1.b≤c⇒b±a≤c±a2.b≤c⇒ab≤ac,ifa>03.b≤c⇒ab≥ac,ifa<0

In order to solve the inequality:

$\frac{d}{4}+1â‰\yen −3$

Subtract both sides of the inequality by 1, as

$$\begin{gathered} \frac{d}{4}+1â‰\yen −3 \\ \frac{d}{4}+1−1â‰\yen −3−1 \\ \frac{d}{4}â‰\yen −4 \\ \frac{d}{4}×4â‰\yen −4×4 \\ dâ‰\yen −16 \end{gathered}$$

Thus, the inequality is true for d greater than or equal to $-16$, thus the solution of given inequality in interval form is, as $\left[−16,\mathrm{∞}\right)$.

To check the answer solution for the given inequality.

First check the solution at end point by converting it into an equation.

If the solution is less than m, then pick any value of x less than m and check whether it satisfies the inequality.

If the solution is greater than m, then pick any value of x greater than m and check whether it satisfies the inequality.

Now, to check the answer solution for the given inequality first check the solution at end point by converting it into an equation. Putting dequals to$-16$in the left hand side as

$$\begin{gathered} \frac{−16}{4}+1=−3 \\ =\text{RightHandside} \end{gathered}$$

Now put any value any value of dgreater than$-16$, say4and check whether it satisfies the inequality, as

$$\begin{gathered} \frac{d}{4}+1â‰\yen −3 \\ \frac{−4}{4}â‰\yen −3 \\ −1â‰\yen −3\left[\text{True}\right] \end{gathered}$$

Since, both conditions are satisfied thus solution of inequality is correct.