Chapter 5.8, Problem 24WE

To determine

To find the open sentence of $\frac{u^2}{15}\le \frac{u+2}{6}-\frac{2}{5}$

Answer to Problem 24WE

The value of open sentence is $\frac{1}{2}\le u\le 2$

Explanation of Solution

Given open sentence is $\frac{u^2}{15}\le \frac{u+2}{6}-\frac{2}{5}$

Value of open sentence means a sentence is it true and depend upon the variable values .

  $\frac{u^2}{15}\le \frac{u+2}{6}-\frac{2}{5}$

Multiply both the sides by 30, the LCM of 6,5 and 15

  $30\left(\frac{u^2}{15}\right)\le 30\left(\frac{u+2}{6}-\frac{2}{5}\right)$

Apply distributive property and simplify $\left[a(b\pm c)=a\ast b\pm a*c\right]$

  $2u^2\le 5(u+2)-6(2)$

Simplify the fractions

  $\begin{array}{l}2u^2\le 5u+10-12\\ 2u^2\le 5u-2\\ \text{Subtract}\text{5u-2}\text{from}\text{both}\text{sides}\\ 2u^2-5u+2\le 0\end{array}$

Use factorization and simplify

  $\begin{array}{c}2u^2-5u+2\le 0\\ 2u^2-(4+1)u+2\le 0\\ 2u^2-4u-u+2\le 0\\ 2u(u-2)-1(u-2)\le 0\\ (2u-1)(u-2)\le 0\end{array}$

Now, check the variable value satisfied the sentence with zero product property with inequality .

  $\begin{array}{c}(2u-1)\le 0\\ 2u\le 1\\ u\le \frac{1}{2}\\ \text{Similarly}\\ (u-2)\le 0\\ u\le 2\end{array}$

Both factor values satisfied in the given condition, therefore need to summarize the table to satisfy the given condition $\le 0$:

	$u<\frac{1}{2}$	$u=\frac{1}{2}$	$\frac{1}{2}<u<2$	$u=2$
$2u-1$	-	0	+	+
$u-2$	-	-	-	0
$(2u-1)(u-2)$	+	0	-	0

Hence, the open sentence variable values $\frac{1}{2}\le u\le 2$ satisfy the required condition $\le 0$ .