Chapter 7, Problem 37PT

To determine

To Solve:

  $\sqrt{y-7}+5\ge 10$

Answer to Problem 37PT

Solution region is

  $y\ge 32$

Explanation of Solution

Given Information:

Given inequality:

  $\sqrt{y-7}+5\ge 10$

Concept Used :

Solve Radical inequalities:

• If the index of the root is even , identify the values of the variable for which the radicand is nonnegative.
• Solve the inequality algebraically.
• Test values to check.

Calculation:

• Given $\sqrt{y-7}+5\ge 10$
• $\begin{array}{l}\text{Since the radicand of a square root must be greater than or equal to zero, }\\ \text{first solve y-7}\ge 0\text{ to identify the values of y for which the left hand side}\\ \text{of the inequality is defined}\text{.}\\ y-7\ge \mathrm{0....................}[\text{set the radicand }\ge \text{ 0]}\\ y\ge 0+\mathrm{7....................}[add\text{ 7 each side]}\\ y\ge 7\end{array}$
• $\begin{array}{l}Now,solve\\ \sqrt{y-7}+5\ge 10\\ \sqrt{y-7}\ge 10-\mathrm{5.................}[subtract\text{ 5 from each side]}\\ \sqrt{y-7}\ge 5\\ {\left(\sqrt{y-7}\right)}^2\ge {\left(5\right)}^2\mathrm{................}[\text{eliminate the radical]}\\ y-7\ge 25\\ y\ge 25+\mathrm{7..........................}[\text{add 7 each side]}\\ y\ge 32\end{array}$

  $\begin{array}{l}Combining\\ y\ge 7\\ and\\ y\ge 32\\ \text{The solution region is }y\ge 32,\text{ the intersection of all the regions}\text{.}\\ \end{array}$