Question

Solve the following inequation and represent the solution set on the number line, $-2+10x\le 13x+10<24+10x,x\in \mathrm{\mathbb{Z}}$.

Answer 1

Step 1: Given inequation: $-2+10x\le 13x+10<24+10x,x\in \mathrm{\mathbb{Z}}$

Consider the inequality: $-2+10x\le 13x+10$

Subtract both sides of the inequation by $10x$.

$$\begin{gathered} -2+10x-10x\le 13x+10-10x \\ -2\le 3x+10 \end{gathered}$$

Step 2: Subtract both sides of the inequation by $10$.

$$\begin{gathered} -2-10\le 3x \\ 3x\ge -12 \end{gathered}$$

Step 3: Multiply both sides of the inequality by $\frac{1}{3}$.

$$\begin{gathered} 3x\times \frac{1}{3}\ge -12\times \frac{1}{3} \\ x\ge -4 \end{gathered}$$

Thus, the solution set of the given inequation $[-4,\infty )$.

Step 4: Consider the inequality: $13x+10<24+10x$

Subtract both sides of the inequation by $10x$.

$$\begin{gathered} 13x+10-10x<24+10x-10x \\ 3x+10<24 \end{gathered}$$

Step 5: Subtract both sides of the inequation by $10$.

$$\begin{gathered} 3x+10-10<24-10 \\ 3x<14 \end{gathered}$$

Step 6: Multiply both sides of the inequality by $\frac{1}{3}$.

$$\begin{gathered} 3x\times \frac{1}{3}<14\times \frac{1}{3} \\ x<\frac{14}{3} \end{gathered}$$

Thus, the solution set of the given inequation $(-\infty ,\frac{14}{3})$.

Therefore, the solution set of $[-4,\infty )$ and $(-\infty ,\frac{14}{3})$ is $[-4,\frac{14}{3})$.

Step 7: Represent the solution set on the number line

The solution set of the given inequality can be plotted on the number line as,