Chapter 6.7, Problem 57PPS

To determine

To Solve:

  $\sqrt{2y+5}+3\le 6$

Answer to Problem 57PPS

Solution region is

  $-\frac{5}{2}\le y\le 2$

Explanation of Solution

Given Information:

Given inequality:

  $\sqrt{2y+5}+3\le 6$

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 $\sqrt{2y+5}+3\le 6$
• $\begin{array}{l}\text{Since the radicand of a square root must be greater than or equal to zero, }\\ \text{first solve 2y+5}\ge 0\text{ to identify the values of }x\text{ for which the left hand side}\\ \text{of the inequality is defined}\text{.}\\ 2y+5\ge \mathrm{0....................}[\text{set the radicand }\ge \text{ 0]}\\ 2y\ge 0-\mathrm{5....................}[subtract\text{ 5 from each side]}\\ \frac{2y}{2}\ge -\frac{5}{2}\mathrm{.....................}[\text{Divide each side by 2]}\\ y\ge -\frac{5}{2}\end{array}$
• $\begin{array}{l}Now,solve\\ \sqrt{2y+5}+3\le 6\\ \sqrt{2y+5}\le 6-\mathrm{3.................................}[subtract\text{ 3 from each side]}\\ \sqrt{2y+5}\le 3\\ {\left(\sqrt{2y+5}\right)}^2\le {\left(3\right)}^2\mathrm{................}[\text{eliminate the radical]}\\ 2y+5\le 9\\ 2y\le 9-\mathrm{5.........................}[subtract\text{ 5 from each side]}\\ 2y\le 4\\ \frac{2y}{2}\le \frac{4}{2}\mathrm{........................................}[\text{divide 3 each side]}\\ y\le 2\end{array}$
• $\begin{array}{l}Combining\\ y\ge -\frac{5}{2}\\ and\\ y\le 2\\ \text{The solution region is }-\frac{5}{2}\le y\le 2,\text{ the intersection of two regions}\text{.}\end{array}$