Chapter 5.8, Problem 3WE

To determine

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

Answer to Problem 3WE

The value of open sentence is $t=12$

Explanation of Solution

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

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

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

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

  $\begin{array}{c}12\left(\frac{2t-1}{6}\right)=12\left(\frac{t+2}{4}+\frac{1}{3}\right)\\ 2(2t-1)=3(t+2)+4\end{array}$

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

  $4t-2=3t+6+4$

Add 2 both the sides

  $\begin{array}{c}4t-2+2=3t+6+4+2\\ 4t=3t+12\end{array}$

Subtract $3t$ both the sides

  $\begin{array}{c}4t-3t=3t-3t+12\\ t=12\end{array}$

Hence, the open sentence value is $t=12$