Question

Solve the following inequation and represent the solution set on the number line, $13x-5<15x+4<7x+12,\forall x\in \mathrm{ℝ}$.

Answer 1

Step 1: Given inequality: $13x-5<15x+4<7x+12,\forall x\in \mathrm{ℝ}$.

Consider the inequality: $13x-5<15x+4$.

Subtract $15x$ to both sides of the inequality.

$$\begin{gathered} 13x-5-15x<15x+4-15x \\ -2x-5<4 \end{gathered}$$

Step 2: Add $5$ to both sides of the inequality.

$$\begin{gathered} -2x-5+5<4+5 \\ -2x<9 \end{gathered}$$

Step 3: Multiply $\left(-\frac{1}{2}\right)$ to both sides of the inequality and change the sign of the inequality.

$$\begin{gathered} -2x\times \left(-\frac{1}{2}\right)>9\times \left(-\frac{1}{2}\right) \\ x>-\frac{9}{2} \end{gathered}$$

Thus, the solution set of the inequality can be given by, $\left(-\frac{9}{2},\infty \right)$.

Step 4: Consider the inequality: $15x+4<7x+12$.

Subtract $7x$ to both sides of the inequality.

$$\begin{gathered} 15x+4-7x<7x+12-7x \\ 8x+4<12 \end{gathered}$$

Step 5: Subtract $4$ to both sides of the inequality.

$$\begin{gathered} 8x+4-4<12-4 \\ 8x<8 \end{gathered}$$

Step 6: Multiply both sides of the inequation by $\frac{1}{8}$.

$$\begin{gathered} 8x\times \frac{1}{8}<8\times \frac{1}{8} \\ x<1 \end{gathered}$$

Thus, the solution set of the inequality can be given by, $\left(-\infty ,1\right)$.

Therefore, the solution set of $\left(-\frac{9}{2},\infty \right)$ and $\left(-\infty ,1\right)$ is $\left(-\frac{9}{2},1\right)$

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,