Chapter A, Problem 41E

To determine

To solve the below inequality in terms of intervals where a, b and care positive constants-

  $a\left(bx-c\right)\ge bc$

Answer to Problem 41E

The solution of the inequality is $x\ge c\left(\frac{1}{a}+\frac{1}{b}\right)$

Explanation of Solution

Given: Inequality: $a\left(bx-c\right)\ge bc$

Formula Used:

An inequality compares two values, showing if one is less than, greater than, or simply not equal to another value.

Real number line is the line whose points are the real numbers. 

Calculation:

Given : Inequality equation is $a\left(bx-c\right)\ge bc$

Solving, we have:

  $\begin{array}{l}a\left(bx-c\right)\ge bc\\ abx-ac\ge bc\end{array}$

Adding $ac$ to both the sides, we have:

  $\begin{array}{l}abx-ac+ac\ge ac+bc\\ abx\ge ac+bc\end{array}$

Divide both the sides by $ab$ , we have:

  $\begin{array}{l}\frac{abx}{ab}\ge \frac{ac+bc}{ab}\\ x\ge \frac{ac+bc}{ab}\\ x\ge \frac{c\left(a+b\right)}{ab}\\ x\ge c\left(\frac{1}{a}+\frac{1}{b}\right)\end{array}$

Conclusion:

Hence, the solution of the inequality is $x\ge c\left(\frac{1}{a}+\frac{1}{b}\right)$