Chapter 5.8, Problem 21WE

To determine

To find the open sentence of $\frac{y^2+4}{6}+\frac{y+1}{3}<\frac{3}{2}$

Answer to Problem 21WE

The value of open sentence is $-3<y<1$

Explanation of Solution

Given open sentence is $\frac{y^2+4}{6}+\frac{y+1}{3}<\frac{3}{2}$

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

  $\frac{y^2+4}{6}+\frac{y+1}{3}<\frac{3}{2}$

Multiply both the sides by 6, the LCM of 2,3 and 6

  $6\left(\frac{y^2+4}{6}+\frac{y+1}{3}\right)<\frac{3}{2}(6)$

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

  $\begin{array}{l}{y}^2+4+2(y+1)<9\\ y^2+2y+6<9\end{array}$

  $\begin{array}{l}{y}^2+4+2(y+1)<9\\ y^2+2y+6<9\end{array}$

Subtract 9 both sides

  $\begin{array}{l}{y}^2+2y+6-9<9-9\\ y^2+2y-3<0\end{array}$

Use factorization and simplify

  $\begin{array}{l}{y}^2+2y-3<0\\ y^2+(3-1)y-3<0\\ y^2+3y-y-3<0\\ y(y+3)-1(y+3)<0\\ (y+3)(y-1)<0\end{array}$

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

  $\begin{array}{c}(y+3)=0\\ y=-3\\ \text{Similarly}\\ (y-1)=0\\ y=1\end{array}$

Hence, the open sentence satisfied only on variable values are $-3<y<1$