Chapter 5, Problem 7CST

To determine

Tofind:the correct answer of expression $\frac{2}{3}x+\frac{1}{6}\ge 1\frac{17}{18}$

Answer to Problem 7CST

The option (A): 2 is the correct answer of expression $\frac{2}{3}x+\frac{1}{6}\ge 1\frac{17}{18}$

Explanation of Solution

Given information: Consider, the expression $\frac{2}{3}x+\frac{1}{6}\ge 1\frac{17}{18}$ given in the question

Calculation:

To evaluate the expression $\frac{2}{3}x+\frac{1}{6}\ge 1\frac{17}{18}$ provided, it is needed to first simplify the mixed fraction

Therefore,

  $\begin{array}{l}\Rightarrow \frac{2}{3}x+\frac{1}{6}\ge 1\frac{17}{18}\\ \Rightarrow \frac{2}{3}x+\frac{1}{6}\ge \frac{35}{18}\end{array}$

  $\text{Multiply}\text{each}\text{side}\text{by}\text{least}\text{common}\text{denominator}\text{of fractions}$

  $\Rightarrow 18\times (\frac{2}{3}x+\frac{1}{6})\ge \frac{35}{18}$

  $\begin{array}{l}\text{By}\text{using}\text{distributive}\text{property}\\ \Rightarrow 18\times (\frac{2}{3}x)+18\times (\frac{1}{6})\ge 18\times (\frac{35}{18})\\ \Rightarrow 12x+3\ge 35\end{array}$

  $\begin{array}{l}\text{Sutract}\text{3}\text{from}\text{each}\text{side}\\ \Rightarrow 12x+3-3\ge 35-3\\ \Rightarrow 12x\ge 32\end{array}$

  $\begin{array}{l}\text{Divide}\text{each}\text{term}\text{by}\text{12}\\ \Rightarrow \frac{12x}{12}\ge \frac{32}{12}\\ \Rightarrow x\ge \frac{8}{3}\end{array}$

$x\ge \frac{8}{3}$ is the solution of the given expression so the values of $x$ less than $\frac{8}{3}$ are not a solution of the given expression.

Hence,

The option (A): 2 is the correct answer of expression $\frac{2}{3}x+\frac{1}{6}\ge 1\frac{17}{18}$