Chapter 2, Problem 15CT

To determine

To describe: the possible solution sets of the given inequality.

Answer to Problem 15CT

The possible solution sets is

  $0<d-b$

  $0<d-b\le 0$

  $x<\frac{d-b}{a-c}$

  $x>\frac{d-b}{a-c}$

Explanation of Solution

Given:

  $ax+b<cx+d$

Calculation:

Given the inequality

  $ax+b<cx+d$

Solve

  $\begin{array}{l}ax+b-cx<cx+d-cx\\ ax+b-cx<d\\ ax+b-cx-b<d-b\\ x\left(a-c\right)<d-b\end{array}$

Case $1$ : $a=c$

Subcase $1$

  $a:d>b$

  $0<d-b$

  $\Rightarrow$ The solution is all real numbers

Subcase $1b:d\le b$

  $0<d-b\le 0$

  $\Rightarrow$ there is no solution

Case $2:a>c$

Divide by $a-c:$

  $\begin{array}{l}x\left(a-c\right)<d-b\\ x<\frac{d-b}{a-c}\end{array}$

Case $3:a<c$

Divide by $a-c$ and the sense of the inequality changes:

  $\begin{array}{l}x\left(a-c\right)<d-b\\ x>\frac{d-b}{a-c}\end{array}$

Conclusion:

Hence, the solution is

  $0<d-b$

  $0<d-b\le 0$

  $x<\frac{d-b}{a-c}$

  $x>\frac{d-b}{a-c}$