Chapter 5.8, Problem 23WE

To determine

To find the open sentence of $\frac{t^2}{6}+\frac{t-2}{4}\ge \frac{t+1}{3}$

Answer to Problem 23WE

The value of open sentence is $t:t\ge -2ort\ge \frac{5}{2}$

Explanation of Solution

Given open sentence is $\frac{t^2}{6}+\frac{t-2}{4}\ge \frac{t+1}{3}$

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

  $\frac{t^2}{6}+\frac{t-2}{4}\ge \frac{t+1}{3}$

Multiply both the sides by 12, the LCM of 6,4 and 3.

  $12\left(\frac{t^2}{6}+\frac{t-2}{4}\right)\ge 12\left(\frac{t+1}{3}\right)$

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

  $\begin{array}{l}2t^2+3(t-2)\ge 4(t+1)\\ 2t^2+3t-6\ge 4t+4\end{array}$

Subtract $4t+4$ both sides

  $\begin{array}{c}2t^2+3t-6-4t-4\ge 0\\ 2t^2-t-10\ge 0\end{array}$

Use factorization and simplify

  $\begin{array}{c}2t^2-t-10\ge 0\\ 2t^2-t(5-4)-10\ge 0\\ 2t^2-5t+4t-10\ge 0\\ t(2t-5)+2(2t-5)\ge 0\\ (2t-5)(t+2)\ge 0\end{array}$

  $\begin{array}{c}(2t-5)\le 0\\ 2t\le 5\\ t\le \frac{5}{2}\\ \text{Similarly}\\ (t+2)\le 0\\ t\le -2\end{array}$

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

Both factor values satisfied in the given condition, therefore, need to summarize the table:

	$t>\frac{5}{2}$	$t=\frac{5}{2}$	$t>-2$	$t=-2$
$2t-5$	+	0	-	-
$t+2$	+	+	-	0
$(2t-5)(t+2)$	+	0	-	0

Hence, the open sentence variable values $t:t\ge -2ort\ge \frac{5}{2}$ satisfy the required condition $\ge 0$ where both factor have the same sign.