Chapter 2.5, Problem 31E

To determine

To fill in:The given compound inequality.

Answer to Problem 31E

Here, the compound inequality $4\left(x-6\right)\ge 2\left(x-10\right)and5\left(x+2\right)\ge 2\left(x+8\right)$ has only value $y\le -1$ .

Explanation of Solution

Given: $4\left(x-6\right)\ge 2\left(x-10\right)and5\left(x+2\right)\ge 2\left(x+8\right)$

Calculation:

Given the inequalities:

  $4\left(x-6\right)?2\left(x-10\right)and5\left(x+2\right)\ge 2\left(x+8\right)$

Solve the inequality on the right:

  $\begin{array}{l}5x+10\ge 2x+16\\ 5x+10-2x\ge 2x+16-2x\\ 3x+10\ge 16\\ 3x+10-10\ge 16-10\\ 3x\ge 16\\ x\ge 2\end{array}$

The two inequalities have only one solution if there solutions are in the form:

  $\left(-\infty ,a\right]$

  $\left[a,\infty \right),$

Any other combination would lead to more than one solution. Since it is already known that the solution for the second inequality is $\left[2,\infty \right)$ , it means the solution for the first one must be $\left(-\infty ,2\right]$ . Thus the possible inequality signs are $"\le ","\ge ".$

The solve the first inequality using the sign $"\le "$ .

  $\begin{array}{l}4\left(x-6\right)\le 2\left(x-10\right)\\ 2\left(x-6\right)\le x-10\\ 2x-12\le x-10\\ 2x-12-x\le x-10-x\\ x-12\le -10\\ x-12+12\le -10+12\\ x\le 2\\ x\in \left(\infty ,2\right]\end{array}$

Finally found the unknown sign of the first inequality (the other must not be checked because the inequality would have the complimentary solution $\left[2,\infty \right)$ ).

Conclusion:

Therefore, the compound inequality $4\left(x-6\right)\ge 2\left(x-10\right)and5\left(x+2\right)\ge 2\left(x+8\right)$ has only value $y\le -1$ .