Chapter 5.8, Problem 19WE

To determine

To find the open sentence of $\frac{y(2y-1)}{4}+\frac{3}{10}=\frac{y(y+2)}{5}$

Answer to Problem 19WE

The value of open sentence is $y=\left\{\frac{3}{2},\frac{2}{3}\right\}$

Explanation of Solution

Given open sentence is $\frac{y(2y-1)}{4}+\frac{3}{10}=\frac{y(y+2)}{5}$

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

  $\frac{y(2y-1)}{4}+\frac{3}{10}=\frac{y(y+2)}{5}$

Multiply both the sides by 20, the LCM of 4,5 and 10

  $20\left(\frac{y(2y-1)}{4}+\frac{3}{10}\right)=20\left(\frac{y(y+2)}{5}\right)$

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

  $\begin{array}{l}5y(2y-1)+6=4y(y+2)\\ 10y^2-5y+6=4y^2+8y\end{array}$

Arrange both sides

  $\begin{array}{c}10y^2-4y^2-5y-8y+6=0\\ 6y^2-13y+6=0\end{array}$

Use factorization and simplify

  $\begin{array}{l}6y^2-13y+6=0\\ \text{Apply}\text{quadratic}\text{root}\text{equation}\\ \frac{-b\pm \sqrt{b^2-4ac}}{2a}wherea=6,b=-13andc=6\\ y_{1,2}=\frac{-(-13)\pm \sqrt{{(-13)}^2-4(6)(6)}}{2(6)}\\ y_{1,2}=\frac{13\pm \sqrt{169-144}}{12}=\frac{13\pm \sqrt{25}}{12}=\frac{13\pm 5}{12}\\ y_1=\frac{13+5}{12}=\frac{18}{12}=\frac{3}{2}\\ \text{Similarly}\\ y_2=\frac{13-5}{12}=\frac{8}{12}=\frac{2}{3}\end{array}$

Hence, the open sentence variable values are $y=\left\{\frac{3}{2},\frac{2}{3}\right\}$