Chapter 3.4, Problem 38PPE

To determine

To Solve: The given inequality, $\frac{4}{3}s-3<s+\frac{2}{3}-\frac{1}{3}s$ .

Answer to Problem 38PPE

  $s<\frac{11}{2}$

Explanation of Solution

Given information:

  $\frac{4}{3}s-3<s+\frac{2}{3}-\frac{1}{3}s$

Concept Used:

We use the algebra of operations by adding or subtracting variable terms on both the sides.

Also, multi-step inequality is solved by simplifying it into equivalent one step inequality.

Calculation:

As the given inequality is $\frac{4}{3}s-3<s+\frac{2}{3}-\frac{1}{3}s$ , we solve it as follows:

Simplifying right hand side by subtracting like terms, we get

  $\begin{array}{l}\frac{4}{3}s-3<s+\frac{2}{3}-\frac{1}{3}s\\ \frac{4}{3}s-3<\left(s-\frac{1}{3}s\right)+\frac{2}{3}\\ \frac{4}{3}s-3<\frac{3s-s}{3}+\frac{2}{3}\\ \frac{4}{3}s-3<\frac{2}{3}s+\frac{2}{3}\end{array}$

Subtracting $\frac{2}{3}s$ from both the sides, we get

  $\begin{array}{l}\frac{4}{3}s-3-\frac{2}{3}s<\frac{2}{3}s+\frac{2}{3}-\frac{2}{3}s\\ \left(\frac{4}{3}s-\frac{2}{3}s\right)-3<\frac{2}{3}+\left(\frac{2}{3}s-\frac{2}{3}s\right)\\ \frac{2}{3}s-3<\frac{2}{3}\end{array}$

Adding $3$ both the sides we get

  $\begin{array}{l}\frac{2}{3}s-3+3<\frac{2}{3}+3\\ \frac{2}{3}s<\frac{2}{3}+\frac{9}{3}\\ \frac{2}{3}s<\frac{2+9}{3}\\ \frac{2}{3}s<\frac{11}{3}\end{array}$

Multiplying both the sides by $\frac{3}{2}$ , we get

  $\begin{array}{l}\frac{2}{3}s\times \frac{3}{2}<\frac{11}{3}\times \frac{3}{2}\\ s<\frac{11}{2}\end{array}$